Answers
Answer:
93 m
Explanation:
We need to find the resultant displacement.
Let's take north as positive y-direction and east as positive direction. We have
- Displacement 1 is 51 m to the north, so the two components are
[tex]x_1 =0\\y_1 = 51 m[/tex]
- Displacement 2 is 45 m [tex]60^{\circ}[/tex] north of east, so the two components are
[tex]x_1 = (45)cos 60^{\circ}=22.5 m\\y_1 = (45) sin 60^{\circ}=39.0 m[/tex]
So to find the resultant displacement we have to sum the components along each direction:
[tex]x=x_1 + x_2 = 0+22.5 m = 22.5 m\\y = y_1 + y_2 = 51 m +39.0 m = 90.0 m[/tex]
And the magnitude of the resultant displacement is
[tex]m=\sqrt{x^2+y^2}=\sqrt{(22.5 m)^2+(90.0 m)^2}=92.8 m \sim 93 m[/tex]
Related Questions
8. A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth?
Answers
The force of gravity on earth is 9.807, therefore you multiply earth’s gravitational force by the 5.7 kilograms which will give you 5.7 x 9.807=55.86 Newtons
What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔy, where vi and vf are the particle’s initial and final speeds, respectively, and a is the particle’s acceleration.
Answers
Answer:
3.2 m
Explanation:
The equation to use to solve this problem is:
[tex]v_f^2 = v_i^2 + 2 a \Delta y[/tex]
where
[tex]v_f[/tex] is the final velocity
[tex]v_i[/tex] is the initial velocity
a is the acceleration
[tex]\Delta y[/tex] is the distance covered
For the particle in free-fall in this problem, we have
[tex]v_i = 0[/tex] (it starts from rest)
[tex]v_f = 7.9 m/s[/tex]
[tex]g=9.8 m/s^2[/tex] (acceleration due to gravity)
By re-arranging the equation, we can find the distance travelled:
[tex]\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m[/tex]
A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially uncharged. A point charge q = 5.00 C is placed at the center of the shell. What is the electric field strength in the region r < a? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region a < r < b? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region b < r? Express your answer in terms of 1/r2. Tries 0/8 What is the induced charge density at r = a? (in C/m^2) Tries 0/8 What is the induced charge density (in C/m2) at r = b?
Answers
(a) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
We can solve the different part of the problem by using Gauss theorem.
Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:
[tex]E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}[/tex]
where q is the charge contained in the spherical surface, so
[tex]q=5.00 C[/tex]
Solving for E(r), we find the expression of the field for r<a:
[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
(b) 0
The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.
So, if we use Gauss theorem for the region a < r < b, we get
[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex]
however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so
q' = + q - q = 0
And so we find
E(r) = 0
(c) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
We can use again Gauss theorem:
[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex] (1)
where this time r > b (outside the shell), so the gaussian surface this time contained:
- the charge +q at the centre
- the inner surface, with a charge of -q
- the outer surface, with a charge of +q
So the net charge is
q' = +q -q +q = +q
And so solving (1) we find
[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]
which is identical to the expression of the field inside the shell.
(d) [tex]-12.3 C/m^2[/tex]
We said that at r = a, a charge of -q is induced. The induced charge density will be
[tex]\sigma_a = \frac{-q}{4\pi a^2}[/tex]
where [tex]4 \pi a^2[/tex] is the area of the inner surface of the shell. Substituting
q = 5.00 C
a = 0.18 m
We find the induced charge density:
[tex]\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2[/tex]
(e) [tex]-1.9 C/m^2[/tex]
We said that at r = b, a charge of +q is induced. The induced charge density will be
[tex]\sigma_b = \frac{+q}{4\pi b^2}[/tex]
where [tex]4 \pi b^2[/tex] is the area of the outer surface of the shell. Substituting
q = 5.00 C
b = 0.46 m
We find the induced charge density:
[tex]\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2[/tex]
The electric field strength between the plates of a simple air Capacitor is equal to the voltage across the plates divided by the distance between them. When a voltage of 112 V is put across the plates of such a capacitor an electric field strength of 1.1 kV/cm is measured. Write an equation that will let you calculate the distance d between the plates. Your equation should contain only symbols. Be sure you define each symbol
Answers
Answer:
[tex]d=\frac{V}{E}[/tex]
Explanation:
The electric field strength between the plates of a parallel plate capacitor is given by:
[tex]E=\frac{V}{d}[/tex]
where
V is the potential difference
d is the distance between the plates
If we re-arrange the formula, we find a new equation that lets us calculate d:
[tex]d=\frac{V}{E}[/tex]
We can at this point substitute the values:
V = 112 V
E = 1.1 kV/cm = 1100 V/cm = 110,000 V/m
To find the distance:
[tex]d=\frac{112 V}{110,000 V/m}=1.02\cdot 10^{-3} m[/tex]
The distance between capacitor plates, using the formula d = V/E, is approximately 1 cm
To calculate the distance d between the plates of a parallel plate capacitor, we use the relationship between the electric field strength E, the voltage V, and the distance d: E = V/d. Given:
V = 112 V (voltage across the plates)
E = 1.1 kV/cm = 1.1 * 10⁵ V/m (electric field strength)
The formula can be rearranged to solve for d:
d = V/E
Where:
V is the potential difference across the plates (in volts, V)E is the electric field strength (in volts per meter, V/m)d is the separation distance between the plates (in meters, m)Substituting the given values into the equation will allow us to calculate the distance between the plates.
Given:
Voltage (V) = 112 V
Electric field strength (E) = 1.1 kV/cm = 1.1 × 10⁴ V/m
Distance (d) = ?
Formula: E = V/d
Substitute: 1.1 × 10⁴ = 112/d
Solve for d: d = 112 / (1.1 × 10⁴)
[tex]\approx[/tex] 0.01 m [tex]\approx[/tex] 1 cm
The distance between the capacitor plates is thus approximately 1 cm.
The Stanford Linear Accelerator Center (SLAC) accelerates electrons such that the 3 km proper length tube appears to be only 8 cm long due to Lorentz contraction. a. Find the values of gamma and the electrons’ speed. b. How much time does it take an electron to travel down the tube according to someone observing the proper tube length? c. How much time does the journey take from the electron’s point of view?
Answers
a. 37500; 299,792,457.9 m/s
The formula for the length contraction for a particle moving close to the speed of light is:
[tex]L' = \frac{L}{\gamma}[/tex]
where
L' is the length observed by the particle moving
L is the length observed by an observer at rest
In this problem, we have
[tex]L = 3 km = 3000 m[/tex] is the length of the SLAC measured by an observer at rest
[tex]L' = 8cm = 0.08 m[/tex] is the length measured by the electrons moving
Substituting into the formula, we find the gamma factor
[tex]\gamma = \frac{L}{L'}=\frac{3000 m}{0.08 m}=37,500[/tex]
The formula for the gamma factor is
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:
[tex]1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}\\v=c \sqrt{ 1-\frac{1}{\gamma^2}}=(299 792 458 m/s)\sqrt{1-\frac{1}{(37500)^2}}=299,792,457.9 m/s[/tex]
b. [tex]1.0\cdot 10^{-5}s[/tex]
For an observer at rest in the laboratory, the electron is moving at a speed of
v = 299,792,457.9 m/s
and it covers a total distance of
L = 3000 m
which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is
[tex]t=\frac{L}{v}=\frac{3000 m}{299,792,457.9 m/s}=1.0\cdot 10^{-5}s[/tex]
c. [tex]2.67\cdot 10^{-10}s[/tex]
From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of
[tex]L' = 0.08 m[/tex]
And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:
[tex]t' = \frac{t}{\gamma}[/tex]
where
[tex]t=1.0\cdot 10^{-5}s[/tex] is the time measured by the observer at rest
[tex]\gamma=37500[/tex]
Substituting,
[tex]t' = \frac{1.0\cdot 10^{-5}s}{37500}=2.67\cdot 10^{-10}s[/tex]
The Schwarzschild radius is the distance from an object at which the escape velocity is equal to the speed of light. A black hole is an object that is smaller than its Schwarzschild radius, so not even light itself can escape a black hole. The Schwarzschild radius ???? depends on the mass ???? of the black hole according to the equation ????=2????????????2 where ???? is the gravitational constant and ???? is the speed of light. Consider a black hole with a mass of 5.7×107M⊙. Use the given equation to find the Schwarzschild radius for this black hole. Schwarzschild radius: m What is this radius in units of the solar radius
Answers
1. [tex]1.69\cdot 10^{11} m[/tex]
The Schwarzschild radius of an object of mass M is given by:
[tex]r_s = \frac{2GM}{c^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the object
c is the speed of light
The black hole in the problem has a mass of
[tex]M=5.7\cdot 10^7 M_s[/tex]
where
[tex]M_s = 2.0\cdot 10^{30} kg[/tex] is the solar mass. Substituting,
[tex]M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg[/tex]
and substituting into eq.(1), we find the Schwarzschild radius of this black hole:
[tex]r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m[/tex]
2) 242.8 solar radii
We are asked to find the radius of the black hole in units of the solar radius.
The solar radius is
[tex]r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]
Therefore, the Schwarzschild radius of the black hole in solar radius units is
[tex]r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8[/tex]
If a wheel falls from an airplane that is flying horizontally at an altitude of 500 m, how long will it take for the wheel to strike the ground?
10 s
50 s
80 s
100 s
Answers
In this case the only vertical force is gravity, which has an acceleration of 9.8m/s^2. We are also given height (distance) and starting velocity (0 m/s). We need to solve for time. The equation that incorporates these variables is:
d = vt + ½ at^2
500 = 0t + ½ (9.8)t^2
500 = 4.9t^2
t^2 = 102
t = 10.1 seconds
A ball has a mass of 0.0935 kg just before it strikes the Earth after being dropped from a building 39.8 m tall. What is its de Broglie wavelength? The acceleration of gravity is 9.8 m/s 2 and Planck’s constant is 6.62607 × 10−34 J · s. Answer in units of m.
Answers
Answer:
[tex]2.54\cdot 10^{-34}m[/tex]
Explanation:
First of all, we need to find the final velocity of the ball just before it reaches the ground. Since the ball is in free-fall motion, its final velocity is given by
[tex]v^2 = u^2 +2gh[/tex]
where
v is the final velocity
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration due to gravity
h = 39.8 m is the height of the building
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(39.8 m)}=27.9 m/s[/tex]
Now we can calculate the ball's momentum:
[tex]p=mv=(0.0935 kg)(27.9 m/s)=2.61 kg m/s[/tex]
And now we can calculate the De Broglie's wavelength of the ball:
[tex]\lambda = \frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{2.61 kg m/s}=2.54\cdot 10^{-34}m[/tex]
The temperature of an object is changed when heat is added to or extracted from it. Determine the final temperature (in °C) of a 65.0 g mass of aluminum initially at 60.0°C if 1,050 J of heat energy is extracted from it. The specific heat of aluminum is 900 J/(kg · °C).
Answers
Answer:
[tex]42.1^{\circ}C[/tex]
Explanation:
The heat extracted from the aluminium is given by:
[tex]Q=mC_s (T_f-T_i)[/tex]
where we have
Q = -1050 J is the heat extracted
m = 65.0 g = 0.065 kg is the mass
Cs = 900 J/kgC is the specific heat of aluminum
[tex]T_i = 60.0^{\circ}[/tex] is the initial temperature
By solving for [tex]T_f[/tex], we find the final temperature:
[tex]T_f = \frac{Q}{m C_s}+T_i=\frac{-1050 J}{(0.065 kg)(900 J/kg C)}+60.0^{\circ}C=42.1^{\circ}C[/tex]
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, and specific heat value. Given that 1050J of heat is extracted, which is negative heat transfer (-1050 J), plug in the values to get ΔT and then add this to the initial temperature to get the final temperature.
Explanation:The subject of the question is the concept of heat transfer and specific heat in Physics. The specific heat of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius, which is a property intrinsic to the substance. Different substances require different amounts of heat to change their temperature.
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, where Q is heat transferred, m is mass, c is the specific heat, and ΔT is the change in temperature (final temperature - initial temperature). Here, we're asked to find the final temperature of a 65 g (or 0.065 kg) mass of aluminum initially at 60.0°C when 1050 J of heat energy is extracted.
The sign of Q needs to represent that heat energy is extracted from the aluminum, so Q is -1050 J. Now, you can plug in the values and solve the equation for ΔT first, then add the initial temperature to find the final temperature.
Learn more about Heat Transfer here:https://brainly.com/question/13433948
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Which of the following is a reason why microbes were found to be problematic?
They were found everywhere.
They could cause infection.
They killed specific bacteria.
They did not respond to antibiotics.
Answers
Answer:
I believe the answer is B.
Explanation:
Microbes could cause infectious diseases like for Ex: Flu and measles or heart diseases.
Hope my answer has helped you! :)
Microbes are found to be problematic for two reasons that they cause infection and sometime these organisms does not respond to antibiotics.
Explanation:
The microbes can cause infections in humans is a problem. Fungi, helminths and protozoans are the major cause of infectious diseases. Resistance to an antibiotic occurs when an ability to defeat drugs created to kill them are developed in a microbe. When a microbe becomes resistant, the antibiotics cannot fight against them and hence they multiply.
Antibiotic resistance is an urgent threat to the health of the public. They become a threat, when a microbe become resistant to antibiotic and it will be very difficult to destroy them. The resistance to antibiotics can cause serious issues like disability or even cause death.
Ultraviolet (UV) radiation is a part of the electromagnetic spectrum that reaches the Earth from the Sun. It has wavelengths shorter than those of visible light, making it invisible to the naked eye. These wavelengths are classified as UVA, UVB, or UVC, with UVA the longest of the three at 317 nm to 400 nm. Both the U.S. Department of Health and Human Services and the World Health Organization have identified UV as a proven human carcinogen. Many experts believe that, especially for fair-skinned people, UV radiation frequently plays a key role in melanoma, the deadliest form of skin cancer, which kills more than 8000 Americans each year. UVB has a wavelength between 280 nm and 317 nm. Determine the frequency ranges of UVA and UVB. UVA Hz (smaller value) Hz (larger value) UVB Hz (smaller value) Hz (larger value)
Answers
Final answer:
To determine the frequency ranges for UVA and UVB, we use the equation f = c/λ. The UVA frequency range is 7.5 × 10^14 Hz to 9.375 × 10^14 Hz, and the UVB frequency range is 9.375 × 10^14 Hz to 1.0345 × 10^15 Hz.
Explanation:
The frequency of electromagnetic radiation, including UV radiation, can be calculated using the formula c = λf, where c is the speed of light (approximately 3 × 10^8 m/s), λ is the wavelength, and f is the frequency. To find the frequency ranges for UVA and UVB, we can rearrange the equation to f = c/λ.
Calculating UVA Frequency
For UVA with a wavelength range of 320-400 nm (or 3.2 × 10^-7 m - 4 × 10^-7 m), we use the formula to calculate the frequency as:
• Lower frequency limit: f = (3 × 10^8 m/s) / (4 × 10^-7 m) = 7.5 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
Calculating UVB Frequency
For UVB with a wavelength range of 290-320 nm (or 2.9 × 10^-7 m - 3.2 × 10^-7 m), the frequency range is:
• Lower frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (2.9 × 10^-7 m) = 1.0345 × 10^15 Hz
The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·C°. The latent heat of fusion of water is 80 cal/g. a. How much heat is required to raise the ice to 0°C and completely melt the ice? b. How much additional heat is required to heat the water (obtained by melting the ice) to 25°C? c. What is the total heat that must be added to convert the 80 g of ice at ????20°C to water at ????25°C? d. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C? Explain.
Answers
1. 13,500 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 150 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal[/tex]
Now we have to find the amount of heat needed to melt the ice, which is
[tex]Q_2 = m \lambda_f[/tex]
where
m = 150 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (150 g)(80 cal/g)=12,000 cal[/tex]
So the total heat required is
[tex]Q_3 = 1500 cal + 12,000 cal = 13,500 cal[/tex]
2. 3750 cal
The additional amount of heat required to heat the water to 25°C is
[tex]Q_4 = m C_w \Delta T[/tex]
where
m = 150 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal[/tex]
3. 9200 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 80 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal[/tex]
Now we have to find the amount of heat needed to melt the ice:
[tex]Q_2 = m \lambda_f[/tex]
where
m = 80 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (80 g)(80 cal/g)=6,400 cal[/tex]
Finally, the amount of heat required to heat the water to 25°C is
[tex]Q_3 = m C_w \Delta T[/tex]
where
m = 80 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal[/tex]
So the total heat required is
[tex]Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal[/tex]
4. No
Explanation:
The total heat required for this process consists of 3 different amounts of heat:
1- The heat required to bring the ice at melting temperature
2- The heat required to melt the ice, while its temperature stays constant
3- The heat required to raise the temperature of the water
However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.
An electron moving at right angles to a 0.14 T magnetic field experiences an acceleration of 6.5×1015 m/ s 2 . What is the electron's speed? Express your answer using two significant figures. By how much does its speed change in 1 ns( 10 −9 s) ?
Answers
Answer:
[tex]2.64\cdot 10^5 m/s[/tex], the speed does not change
Explanation:
The magnetic force on the electron is equal to the product between its mass and its acceleration:
[tex]qvB = ma[/tex]
where
q is the electron charge
v is the electron speed
B = 0.14 T is the magnetic field
m is the electron's mass
[tex]a=6.5\cdot 10^{15}m/s^2[/tex] is the acceleration (centripetal acceleration)
Solving for v, we find
[tex]v=\frac{ma}{qB}=\frac{(9.11\cdot 10^{-31} kg)(6.5\cdot 10^{15} m/s^2)}{(1.6\cdot 10^{-19} C)(0.14 T)}=2.64\cdot 10^5 m/s[/tex]
The speed of the electron does not change, because the acceleration is a centripetal acceleration, so it acts perpendicular to the direction of motion of the electron; therefore, no work is done on the electron by the magnetic force, and therefore the electron does not gain kinetic energy, which means that its speed does not change.
A projectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizontal. Ignore air drag. A.) Determine the MAXIMUM HEIGHT (above where it was thrown) the projectile will reach. B.) Determine how LONG the projectile will be in the air. Assume it lands at the same height it was launched. C.)Assuming it lands at the same height it was launched at, determine how FAR the projectile will land from where it was launched. D.) Assuming it lands at the same height it was launched at, determine how FAST the projectile will be going upon impact. E.) Assuming it lands at the same height it was launched at, determine the ANGLE (relative to the horizontal) of impact. °
Answers
A) 1296.3 m
The initial velocity of the projectile is
[tex]u = 165 m/s[/tex]
and the angle is
[tex]\theta=75^{\circ}[/tex]
So, we can find the initial vertical velocity of the projectile, which is given by
[tex]u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s[/tex]
The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration
[tex]g=-9.8 m/s^2[/tex]
where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation
[tex]v_y ^2 - u_y^2 = 2gh[/tex]
where
[tex]v_y = 0[/tex] is the final vertical velocity
h is the maximum height
Solving for h, we find
[tex]h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m[/tex]
B) 32.5 s
In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.
We can find it by analyzing the vertical motion only. The vertical position at time t is given by
[tex]y(t) = u_y t + \frac{1}{2}gt^2[/tex]
By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:
[tex]0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)[/tex]
which has two solutions:
t = 0 --> beginning of the motion
[tex]u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s[/tex]
C) 1387.8 m
The range of the projectile can be found by analyzing the horizontal motion only.
In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:
[tex]u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s[/tex]
So, the horizontal position at time t is given by
[tex]x(t) = u_x t[/tex]
If we substitute
t = 32.5 s
which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.
So we have:
[tex]x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m[/tex]
D) 165 m/s
The speed of the projectile consists of two independent components:
- The horizontal velocity, which is constant during the motion, and which is equal to
[tex]v_x = u_x = 42.7 m/s[/tex]
- The vertical velocity, which changes during the motion, given by
[tex]v_y = u_y + gt[/tex]
Substituting
[tex]u_y = 159.4 m/s[/tex]
and
[tex]t=32.5 s[/tex]
We find the vertical velocity when the projectile reaches the ground
[tex]v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s[/tex]
which is the same as the initial vertical velocity, but with opposite direction.
Now that we have the two components, we can calculate the actual speed of the projectile:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s[/tex]
and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.
E) [tex]-75^{\circ}[/tex]
The angle of impact is given by
[tex]\theta = tan^{-1} (\frac{|v_y|}{v_x})[/tex]
where
[tex]|v_y| = 159.4 m/s[/tex] is the final vertical velocity
[tex]v_x = 42.7 m/s[/tex] is the final horizontal velocity
We have taken the absolute value of [tex]v_y[/tex] since [tex]v_y[/tex] is negative; this means that the resulting angle will be BELOW the horizontal. So we have:
[tex]\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}[/tex]
which means [tex]-75.0^{\circ}[/tex], below the horizontal.
You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) (a) 2.7 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 4.8 × 1031 kg (a) 2.7 × 1025 kg, (b) 4.8 × 1031 kg
Answers
(a) [tex]2.7\cdot 10^{25} kg[/tex]
The acceleration due to gravity on the surface of the planet is given by
[tex]g=\frac{GM}{R^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
Here we know:
[tex]g=22.4 m/s^2[/tex]
[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is
[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]
So we can re-arrange eq.(1) to find M, the mass of the planet:
[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]
(b) [tex]4.8\cdot 10^{31}kg[/tex]
The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:
[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)
where
m is the mass of the planet
M is the mass of the star
v is the orbital speed of the planet
r is the radius of the orbit
The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:
[tex]v=\frac{2\pi r}{T}[/tex]
where
[tex]T=402 days = 3.47\cdot 10^7 s[/tex]
Substituting into (1) and re-arranging the equation
[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]
And substituting the numbers, we find the mass of the star:
[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]
Can you cool a kitchen by leaving the refrigerator door open while the refrigerator is operating? a. No. Since the refrigerator is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Since this process is not 100% efficient due to the second law, the average temperature of the room will increase over time. b. No. Since the refrigerator is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Due to the second law, there will be no change in temperature due to the perfect efficiency of the heat pump.
Answers
Answer:
no
Explanation:
because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out
'a' is the correct statement.
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature the pressure is 102.1 kPa at sea level and 87.8 kPa at h = 1,000 m. (Round your answers to one decimal place.) (a) What is the pressure at an altitude of 4500 m? kPa (b) What is the pressure at the top of a mountain that is 6165 m high?
Answers
The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.
Explanation:To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.
(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.
Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.
(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.
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Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e. In the classical model, the electron orbits around the nucleus, being held in orbit by the electromagnetic interaction between itself and the protons in the nucleus, much like planets orbit around the sun, being held in orbit by their gravitational interaction. When the electron is in a circular orbit, it must meet the condition for circular motion: The magnitude of the net force toward the center, Fc, is equal to mv 2/r. Given these two pieces of information, deduce the velocity v of the electron as it orbits around the nucleus.
Answers
Final answer:
In the classical model of the hydrogen atom, the velocity of the electron as it orbits around the nucleus can be deduced using the equation for the centripetal force. The magnitude of the net force towards the center, Fc, is equal to mv^2/r, and solving for v gives the velocity of the electron.
Explanation:
In the classical model of the hydrogen atom, the electron orbits the proton in a circular path. The magnitude of the net force towards the center, also known as the centripetal force, is equal to the mass of the electron times its velocity squared divided by the radius of the orbit (Fc = mv^2 / r). In this case, we have the mass of the electron, the radius of the orbit, and we need to find the velocity of the electron as it orbits around the nucleus.
Using the given equation of the centripetal force, we can rearrange it to solve for the velocity (v). So, v = sqrt(Fc * r / m). Plugging in the values for the mass of the electron, the radius of the orbit, and the known value of the centripetal force, we can calculate the velocity.
For example, if the radius of the orbit is 5.28 x 10^-11 m and the mass of the electron is 9.11 x 10^-31 kg, the velocity of the electron would be v = sqrt(Fc * r / m) = sqrt(m * v^2 / r * r / m) = v = sqrt(v^2) = v = 2.18 x 10^6 m/s.
If a negatively-charged rod is brought close to an electroscope that is negatively charged, what will the two leaves of the electroscope do?
move closer together
move farther apart
not move at all
become positively charged
Answers
Answer:
Move further apart
Explanation:
The negative charge on the rod repels the electrons on the cap thus they concentrate at the leaves causing an increase in the amount of charge at the leaves. Since like charges repel, the two leaves of the electroscope move further apart.
Final answer:
When a negatively-charged rod is brought near a negatively-charged electroscope, the leaves of the electroscope will move farther apart due to increased electrostatic repulsion.
Explanation:
When a negatively-charged rod is brought close to an already negatively-charged electroscope, the electrostatic repulsion between the like charges will cause the two leaves of the electroscope to move farther apart. This is due to an increase in the density of negative charges within the leaves of the electroscope, which causes the leaves to repel each other even more.
Charging by induction occurs when a charged object is brought near, but not in contact with, a conductor. However, since the electroscope in this case is already negatively charged, bringing another negatively charged object close will simply enhance the repulsion between charges.
A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and current leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.
Answers
Answer:
All of the choices are correct
Explanation:
The power dissipated in a single resistor connected to a battery is given by:
[tex]P=VI = I^2 R=\frac{V^2}{R}[/tex]
where
V is the voltage
I is the current
R is the resistance
Let's analyze each case:
A) Doubling the voltage (V'=2V) and reducing the current by a factor of 2 (I'=I/2). The new power dissipated is:
[tex]P'=V'I'=(2V)(\frac{I}{2})=VI=P[/tex] --> the power is unchanged
B) Doubling the voltage (V'=2V) and increasing the resistance by a factor of 4 (R'=4R). The new power dissipated is:
[tex]P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}[/tex] --> the power is unchanged
C) Doubling the current (I'=2I) and reducing the resistance by a factor of four (R'=R/4). The new power dissipated is:
[tex]P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R[/tex] --> the power is unchanged
Final answer:
Option (B), which involves doubling the voltage and increasing the resistance by a factor of four, is the correct scenario where the electric power dissipated in the resistor remains unchanged, explained by the formulas P = V^2/R and P = I^2R.
Explanation:
The question addresses how voltage, current, and resistance affect electric power dissipation in a resistor. Electric power (P) dissipated by a resistor can be calculated using the formula P = V2/R or P = I2R, where V is the voltage across the resistor, I is the current through the resistor, and R is the resistance. To keep the power dissipated unchanged, we need to ensure that any changes in voltage, current, or resistance occur in a way that does not change the final calculation of P.
Option (A) suggests doubling the voltage and reducing the current by a factor of two. This will not keep power dissipated the same due to the direct relationship in the formula P = I2R.
Option (B), doubling the voltage and increasing the resistance by a factor of four, will keep the power dissipated in the resistor unchanged because the power can also be expressed as P = V2/R, and thus doubling V while quadrupling R does not change the P.
Option (C) involves doubling the current and reducing the resistance by a factor of four. According to P = I2R, this change will not keep the power dissipated unchanged as the increase in current will significantly raise power dissipation due to its squared relationship in the formula.
Therefore, Option (B) correctly describes a method to change voltage and resistance in such a way that the power dissipated in the resistor remains unchanged.
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. Part A What is the angular speed of the CD when scanning the innermost part of the track? B What is the angular speed of the CD when scanning the outermost part of the track? C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line? D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.
Answers
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
The maximum playing time of the CD is
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
[tex]d=vt=(1.25 m/s)(4,440 s)=5,550 m[/tex]
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
Answer: (a) 50 rad/s, (b) 21.6 rad/s, (c) 5550 m, (d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
Explanation:A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s
Part
A What is the angular speed of the CD when scanning the innermost part of the track?
B What is the angular speed of the CD when scanning the outermost part of the track?
C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time?
(a) 50 rad/s
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed, v = 1.25 m/s
r is the distance from the centre of the CD, r = 25.0 mm = 0.025 m
Therefore, the angular speed
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
The angular speed of the CD is
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
the linear speed of the track is v = 1.25 m/s
the total length of the track would be:
d=vt=(1.25 m/s)(4,440 s)=5,550 m
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
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An object, located 80.0 cm from a concave lens, forms an image 39.6 cm from the lens on the same side as the object. What is the focal length of the lens?
Answers
Answer:
-78.4 cm
Explanation:
We can solve the problem by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
Here we have
p = 80.0 cm
q = -39.6 cm (negative, because the image is on the same side as the object , so it is a virtual image)
Substituting, we find f:
[tex]\frac{1}{f}=\frac{1}{80.0 cm}+\frac{1}{-39.6 cm}=-0.0128 cm^{-1}[/tex]
[tex]f=\frac{1}{0.0128 cm^{-1}}=-78.4 cm[/tex]
A spring of constant k hangs from a ceiling with a mass m attached at the other end. The natural length of the hanging spring without the mass is L. a) Draw this scenario, including forces and known or calculable displacements. b) What is the equilibrium length of the spring with the mass attached? c) Suppose the spring is pulled down to create a displacement y from this equilibrium position and then released. Write an expression y (t) describing its position as a function of time t. Let "up" be the positive vertical direction. d) What is the period of oscillation for this simple harmonic oscillator? e) Suppose you cut the spring in half. What would be the spring constant of each half-spring? (The two are identical, so they will have the same spring constant.) To determine this, redraw your diagram as two springs of equal length connected "in series," i.e., top to bottom. Then indicate the forces acting on each of the three interesting bodies: the mass m and each halfspring of constant k’. Apply Hooke’s Law to each half-spring individually and then to the two springs together. f) Take the two half-springs from part e) and connect them "in parallel" (that is, side by side, hanging from the ceiling and each connected to the mass m). Draw the new diagram and repeat parts a) - d)
Answers
a) See part a of the attached picture.
There are two forces involved here:
- The weight of the block attached to the spring, of magnitude
W = mg
where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward
- The restoring force of the spring,
F = kx
with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward
b) [tex]L' = L+\frac{mg}{k}[/tex]
The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:
[tex]kx = mg[/tex]
where
x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,
[tex]x=\frac{mg}{k}[/tex]
And since the natural length is L, the equilibrium length of the spring is
[tex]L' = L+x=L+\frac{mg}{k}[/tex]
c) [tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
Let's assume that
y = 0
corresponds to the equilibrium position of the spring (which is stretched by an amount [tex]L+\frac{mg}{k}[/tex]). If doing so, the vertical position of the mass at time t is given by
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, so that the position at time t=0 is negative:
y(0) = -y
So rewriting the angular frequency:
[tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
d) [tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
The period of oscillation is given by
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find an expression for the period
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
e) k' = 2k (see part e of attached picture)
Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:
[tex]x=x_1 + x_2[/tex]
Also, since the two springs are identical, their stretching will be the same:
[tex]x_1 = x_2 = x'[/tex]
so we have
[tex]x=x'+x'=2x'[/tex]
[tex]x'=\frac{x}{2}[/tex]
Substituting x find in part (b),
[tex]x'=\frac{mg}{2k}[/tex] (1)
Hooke's law for each spring can be written as
[tex]F=k' x'[/tex] (2)
where
F = mg (3)
is still the weight of the block
Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring
[tex]mg=k'\frac{mg}{2k}\\k' =2k[/tex]
So, the spring constant of each half-spring is twice the spring constant of the original spring.
fa) See part f) of attached picture
This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:
[tex]k' x_1 + k' x_2 = mg[/tex]
fb) [tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:
[tex]F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg[/tex]
The two springs are identical, so they have same spring constant:
[tex]k_1 = k_2 = k'[/tex]
So (1) can be rewritten as
[tex]k' (x_1 + x_2) = mg[/tex]
And since the two springs are identical, their stretching x' is the same:
[tex]x_1 = x_2 = x'[/tex]
so we can rewrite this as
[tex]k' (2x') = mg[/tex]
[tex]x'=\frac{mg}{2k'}[/tex]
and so the equilibrium length of each spring will be
[tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
fc) [tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by
[tex]k_{eq}=k_1 + k_2 = 2k'[/tex]
where k' is the spring constant of each spring.
So, let's assume again that
y = 0
corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where this time we have
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}[/tex] is the angular frequency of the system
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, which we put so that the position at time t=0 is negative:
y(0) = -y
If we rewrite the angular frequency,
[tex]\omega = \sqrt{\frac{2k'}{m}}[/tex]
the position of the mass is
[tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
fd) [tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
Similarly to part d), the period of oscillation is
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find
[tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
can someone help me?!!!!!
Answers
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
[tex](-3\cdot (-6), -2 \cdot (-6))=(18,12)[/tex]
The magnitude is given by Pythagorean's theorem:
[tex]m=\sqrt{18^2+12^2}=21.6[/tex]
and the direction is given by the arctan of the ratio between the y-component and the x-component:
[tex]\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}[/tex]
Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 8.3 days. If it were to undergo gravitational collapse to a neutron star of radius 15 km , losing 34 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times.
Answers
I don’t know what the answer is I Wish I could help
An object of mass 10 kg is released at point A, slides to the bottom of the 30° incline, then collides with a horizontal massless spring, compressing it a maximum distance of 0.75 m. The spring constant is 500 M/m, the height of the incline is 2.0 m, and the horizontal surface is frictionless. (a) What is the speed of the object at the bottom of the incline? (b) What is the work of friction on the object while it is on the incline? (c) The spring recoils and sends the object back toward the incline. What is the speed of the object when it reaches the base of the incline?
Answers
Answer:
Explanation:
The energy stored in the spring = the kinetic energy at the bottom of the incline
1/2 kx² = 1/2 mv²
kx² = mv²
(500 N/m) (0.75 m)² = (10 kg) v²
v ≈ 5.3 m/s
The energy stored in the spring = the initial potential energy - work done by friction
1/2 kx² = mgh - W
1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W
W ≈ 55 J
Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.
v ≈ 5.3 m/s
Answer:
In this problem we need to use the conservation of energy theorem, which is about to constant exchange between kinetic and potential energy. In this case, we have two important points to analyse. The first one is at the initial point, where the mass is just released at 2 m from the ground. The final point would be when the mass hits the massless spring.
So, in the initial point the total energy of the mass would be only potential energy due to its position and its lack of speed. Then, the mass will be released gaining speed, until reaches the spring, where all the kinetic energy is transformed in potential energy but due to elastic forces (spring). So, in the final point, both energy are the same, the kinetic and the elastic potential energy, which we express like this:
[tex]K_{f}=U_{f}[/tex]
But, the definition of each one is:
[tex]K=\frac{1}{2}mv^{2} \\U=\frac{1}{2}kx^{2}[/tex]
Where k is the constant of the spring. So, we replace this in the initial equation and solve for v:
[tex]\frac{1}{2}mv_{f} ^{2}=\frac{1}{2}kx^{2}\\v_{f}=\sqrt{\frac{kx^{2}}{m}}\\v_{f}=\sqrt{\frac{500(0.75)^{2} }{10}}=5.30m/s[/tex]
Therefore, the speed in the spring point is 5.30 m/s.
Now, to calculate the work due to friction, we have to analyse first, because the problem doesn't offer a friction coefficient, but, we can analyse according to energies again, because work and energy are magnitudes close related, and they have the same dimension.
So, if we thought about it, we'll find that the work due to friction is what slows down the mass, subtracting its potential energy from the initial point, giving as a result the final potential energy:
[tex]U_{f}=U_{i}-W_{k}[/tex]
[tex]\frac{1}{2}kx^{2} =mgy-W_{k} \\W_{k}=10(9.8)(2)-\frac{1}{2}(500)(0.75)^{2}\\W_{k}=196-140.625=55.38J[/tex]
Therefore, the word due to friction is 55.38 J.
Lastly, the option c is asking about the speed of the object when it reaches the vase of the incline, which is the horizontal ground. The problem specify that this portion of the trajectory is frictionless, which mean that nothing slows down the mass.
Therefore, the speed in the horizontal part is the same 5.30 m/s.
A major motor company displays a die-cast model of its first automobile, made from 9.00kg of iron. To celebrate its hundredth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model? The density of iron is 7.86 ✕ 10^3 kg/m3, and that of gold is 19.30 ✕ 10^3 kg/m3
Answers
Answer:
22.10 kg
Explanation:
The worker wants to recast the model from iron to gold, so the volume of the model made of iron and made of gold will be the same:
[tex]V_i = V_g[/tex]
the volume of each model can be rewritten as the ratio between the mass of the model, m, and the density of the material, d:
[tex]V=\frac{m}{d}[/tex]
so we can rewrite the first equation as
[tex]\frac{m_i}{d_i}=\frac{m_g}{d_g}[/tex]
where we have
[tex]m_i = 9.00 kg[/tex] is the mass of the model in iron
[tex]d_i = 7.86\cdot 10^3 kg/m^3[/tex] is the density of iron
[tex]m_g[/tex] is the mass of gold needed
[tex]d_g = 19.30\cdot 10^3 kg/m^3[/tex] is the density of gold
Solving for [tex]m_g[/tex], we find
[tex]m_g = d_g \frac{m_i}{d_i}=(19.30\cdot 10^3 kg/m^3) \frac{9.00 kg}{7.86\cdot 10^3 kg/m^3}=22.10 kg[/tex]
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
Answers
(a) [tex]3.1\cdot 10^7 J[/tex]
The total mechanical energy of the space probe must be constant, so we can write:
[tex]E_i = E_f\\K_i + U_i = K_f + U_f[/tex] (1)
where
[tex]K_i[/tex] is the kinetic energy at the surface, when the probe is launched
[tex]U_i[/tex] is the gravitational potential energy at the surface
[tex]K_f[/tex] is the final kinetic energy of the probe
[tex]U_i[/tex] is the final gravitational potential energy
Here we have
[tex]K_i = 5.0 \cdot 10^7 J[/tex]
at the surface, [tex]R=3.3\cdot 10^6 m[/tex] (radius of the planet), [tex]M=5.3\cdot 10^{23}kg[/tex] (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
[tex]U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J[/tex]
At the final point, the distance of the probe from the centre of Zero is
[tex]r=4.0\cdot 10^6 m[/tex]
so the final potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the final kinetic energy:
[tex]K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J[/tex]
(b) [tex]6.3\cdot 10^7 J[/tex]
The probe reaches a maximum distance of
[tex]r=8.0\cdot 10^6 m[/tex]
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
[tex]K_f = 0[/tex]
At that point, the gravitational potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the initial kinetic energy:
[tex]K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J[/tex]
CP Global Positioning System (GPS). The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a 50.0-W (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is 1575.42 MHz. Assume that a satellite transmits half of its power at each frequency and that the waves travel uni- formly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton’s laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel 1.50 cm on a side that absorbs all of the beam, what average pres- sure does the signal exert on it? (d) What wavelength must the receiver be tuned to?
Answers
(a) [tex]9.66\cdot 10^{-15} W/m^2[/tex]
First of all, we need to find the altitude of the satellite. The gravitational attraction between the Earth and the satellite is equal to the centripetal force that keeps the satellite in circular motion, so
[tex]G\frac{mM}{r^2}=m\omega^2 r[/tex] (1)
where
G is the gravitational constant
m is the satellite's mass
M is the earth's mass
r is the distance of the satellite from the Earth's centre
[tex]\omega[/tex] is the angular frequency of the satellite
The satellite here makes two orbits around the Earth per day, so its frequency is
[tex]\omega = \frac {2 \frac{rev}{day}}{24 \frac{h}{day} \cdot 60 \frac{min}{h}} \cdot \frac{2\pi rad/rev}{s/min}=1.45\cdot 10^{-4} rad/s[/tex]
And by solving eq.(1) for r, we find
[tex]r=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)}{(1.45\cdot 10^{-4} rad/s)^2}}=2.67\cdot 10^{7} m[/tex]
The radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
So the altitude of the satellite is
[tex]h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6m=2.03\cdot 10^7 m[/tex]
The average intensity received by a GPS receiver on the Earth will be given by
[tex]I=\frac{P}{A}[/tex]
where
P = 50.0 W is the power
A is the area of a hemisphere, which is:
[tex]A=4\pi h^2 = 4 \pi (2.03\cdot 10^7 m)^2=5.18\cdot 10^{15} m^2[/tex]
So the intensity is
[tex]I=\frac{50.0 W}{5.18\cdot 10^{15}m^2}=9.66\cdot 10^{-15} W/m^2[/tex]
(b) [tex]2.70\cdot 10^{-6} V/m, 9.0\cdot 10^{-15}T[/tex], 0.068 s
The relationship between average intensity of an electromagnetic wave and amplitude of the electric field is
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(9.66\cdot 10^{-15} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=2.70\cdot 10^{-6} V/m[/tex]
Instead, the amplitude of the magnetic field is given by:
[tex]B=\frac{E}{c}=\frac{2.70\cdot 10^{-6} V/m}{3\cdot 10^8 m/s}=9.0\cdot 10^{-15}T[/tex]
The signal travels at the speed of light, so the time it takes to reach the Earth is the distance covered divided by the speed of light:
[tex]t=\frac{h}{c}=\frac{2.03\cdot 10^7 m}{3\cdot 10^8 m/s}=0.068 s[/tex]
(c) [tex]3.22\cdot 10^{-23}Pa[/tex]
In case of a perfect absorber (as in this case), the radiation pressure exerted by an electromagnetic wave on a surface is given by
[tex]p=\frac{I}{c}[/tex]
where
I is the average intensity
c is the speed of light
In this case, we have
[tex]I=9.66\cdot 10^{-15} W/m^2[/tex]
So the average pressure is
[tex]p=\frac{9.66\cdot 10^{-15} W/m^2}{3\cdot 10^8 m/s}=3.22\cdot 10^{-23}Pa[/tex]
(d) 0.190 m
The wavelength of the receiver must be tuned to the same wavelength as the transmitter (the satellite), which is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
c is the speed of light
f is the frequency of the signal
For the satellite in the problem, the frequency is
[tex]f=1575.42 MHz=1575.42\cdot 10^6 Hz[/tex]
So the wavelength of the signal is:
[tex]\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42 \cdot 10^6 Hz}=0.190 m[/tex]
A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 55 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor. static friction kinetic friction
Answers
1) Static friction coefficient: 0.355
The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.
The maximum static frictional force is given by
[tex]F_s = \mu_s mg[/tex]
where
[tex]\mu_s[/tex] is the static coefficient of friction
m = 21 kg is the mass of the crate
g = 9.8 m/s^2 is the acceleration due to gravity
The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have
[tex]F_s=73 N[/tex]
Using this information into the previous equation, we can find the coefficient of static friction:
[tex]\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355[/tex]
2) Kinetic friction coefficient: 0.267
Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is
[tex]F_k = \mu_k mg[/tex] (1)
where
[tex]\mu_k[/tex] is the coefficient of kinetic friction
There is a horizontal force of
F = 55 N
pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.
So we can write Newton's second law as
[tex]F-F_k = ma = 0[/tex]
And by substituting (1), we can find the value of the coefficient of kinetic friction:
[tex]F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267[/tex]
The coefficients of static and kinetic friction between the crate and floor are;
μ_s = 0.3547
μ_k = 0.2672
1) To find the coefficient of static friction, is given by the formula;
F_s = μ_s*mg
We are given;
Mass; m = 21 kg
Horizontal force to set it to motion; F_h = 73 N
Now, in this static friction equation, we are using this force of 73 N because static friction is friction at rest and 73 N is the force to pull the object from rest. Thus;
73 = μ_s(21 × 9.8)
μ_s = 73/(21 × 9.8)
μ_s = 0.3547
2) To find the coefficient of kinetic friction, is given by the formula;
F_k = μ_k*mg
We will make use of the force of 55N because we are dealing with friction associated with motion and 55N is the force that keeps the object in motion. Thus;
μ_k = F_k/(mg)
μ_k = 55/(21 × 9.8)
μ_k = 0.2672
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