Answer:
Can a room be gravitationally shielded? No, it can't.
Explanation:
the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.
Archimedes' principle can be used not only to determine the specific gravity of a solid using a known liquid; the reverse can be done as well. As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid in kg/m^3
Derive a formula for determining the density of a liquid using this procedure.
Express your answer in terms of the variables mobject, mapparent, and rhoobject.
The density of a fluid in which an object is submerged can be determined using the original mass of the object, the apparent submerged mass of the object, and the density of the object, using the formula provided above. Applying this formula to a 4.00-kg aluminum ball submersed in a liquid and appearing to be 2.10 kg yields a liquid density of 1247 kg/m^3.
Explanation:According to Archimedes' Principle, the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. From this principle, if an object is fully submerged in a fluid then the volume of the fluid displaced is equal to the volume of the object.
To calculate the density of a liquid, using the mass of the object, the apparent mass, and the density of the object, the formula commonly used is:
ρliquid = (mobject - mapparent) / (mobject / ρobject - mapparent / ρobject).
Let's apply this formula to the situation you described. An aluminum ball of mass 4.00 kg is submerged in a liquid and has an apparent mass of 2.10 kg. The density of aluminum is approximately 2700 kg/m3. Therefore, the density of the liquid is (4.00kg - 2.10kg) / (4.00kg / 2700kg/m3 - 2.10kg / 2700kg/m3) = 1247 kg/m3.
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Archimedes' principle states that when a body is submerged in a liquid, it experiences an upward force equal to the weight of the liquid displaced by the object. The magnitude of the buoyant force is equal to the weight of the displaced liquid and is directed upward. The buoyant force can be used to determine the density of an object as well as the density of a liquid. The formula for Archimedes' principle is as follows:
Buoyant force = Weight of displaced fluid = density of fluid x volume of fluid displaced x g
where g is the acceleration due to gravity.
In this case, we can use the formula to determine the density of the liquid as follows:
Buoyant force = Weight of the object - Apparent weight of the object
Density of fluid x Volume of fluid displaced x g = m_object x g - m_apparent x g
Density of fluid = (m_object - m_apparent) / volume of fluid displaced
Now, substituting the given values, we get:
Density of fluid = (4.00 kg - 2.10 kg) / [(4.00 kg - 2.10 kg)/1000 kg/m³]
Density of fluid = 1900 kg/m³
Therefore, the density of the liquid is 1900 kg/m³.
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 363 kg, mB = 517 kg, and mC = 154 kg. Find the magnitude and direction of the net gravitational force acting on each of the three particles (let the direction to the right be positive).
Particle A is 0.5m from B and B is .25m from C... All in a astraight line
Answer:
A: 5.67*10^-5 N
B: 3.49*10^-5 N
C: -9.16*10^-5 N
Explanation:
[tex]\\A\\F_{A} = F_{BA} + F_{CA} =G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(\frac{363*517}{0.5^2} + \frac{154*363}{0.75^2})\\\\= 5.67*10^-5 N \\\\B\\\\F_{B} = -F_{BA} + F_{CB} =-G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{b}}{r^2_{BC}} \\\\= (6.67*10^(-11))*(-\frac{517*363}{0.5^2} + \frac{517*154}{0.25^2})\\\\= 3.49*10^-5\\\\[/tex]
[tex]F_{C} = -F_{CA} - F_{CB} =-G*\frac{m_{b}*m_{c}}{r^2_{BC}} - G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(-\frac{517*154}{0.25^2} - \frac{154*363}{0.75^2})\\\\= -9.16*10^-5[/tex]
At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.(a) At how many rev/min are the tires rotating?(b) What is the centripetal acceleration at the edge of the tire?(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?(d) Take the ratio of this force to the bacterium's weight.
Answer:
(a) Angular speed = 1344.8rev/min
(b) Centripetal acceleration = 8473.4m/s^2
(c) Force = 8.4734×10^-12N
(d) Ratio of force to bacterium's weight is 864.6
Explanation:
(a) Angular speed (w) = v/r
v = 60m/s, d = 0.850m, r = 0.850m/2 = 0.425m
w = 60/0.425 = 141.2rad/s = 141.2×9.524rev/min = 1344.8rev/min
(b) Centripetal acceleration = w^2r = 141.2^2 × 0.425 = 8473.4m/s^2
(c) Force = mass × centripetal acceleration = 1×10^-15 × 8473.4 = 8.4734×10^-12N
(d) Bacterium's weight = mass × acceleration due to gravity = 1×10^-15 × 9.8 = 9.8×10^-15N
Ratio of force to bacterium's weight = 8.4734×10^-12N/9.8×10^-15N = 864.6
The commercial jet's tires' rotation speed is calculated using their speed and diameter. The centripetal acceleration at the tire's edge is calculated using the speed and radius. The force with which a tiny bacterium clings to the rim is found using its mass and the centripetal acceleration.
Explanation:We start by calculating the rotation speed of the tires. To do this, we use the known values of tire speed (60 m/s) and diameter (0.85 m). The first step is to convert speed into distance travelled per revolution, and diameter into circumference, then divide speed by circumference to get revolutions per second. We convert this into revolutions per minute for the answer to part (a). For part (b), we use the formula for centripetal acceleration, which involves the square of speed and the radius (half of the tire's diameter) to find acceleration at the rim. For part (c), we find the force exerted by the bacterium by multiplying its mass by the centripetal acceleration. Part (d) compares this force to the bacterium's weight, which is found by multiplying its mass by the acceleration due to gravity.
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Consider two waves moving past you at the same speed. The wavelength of wave A is half that of wave B. You then know that ____.
The frequency of wave B is half the frequency of wave A
Explanation:
The wavelength, the speed and the frequency of a wave are related by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed of the wave
f is its frequency
[tex]\lambda[/tex] is its wavelength
The equation can also be rewritten as
[tex]f=\frac{v}{\lambda}[/tex]
In this problem, we have wave A with wavelength [tex]\lambda_A[/tex] and speed v, so its frequency is
[tex]f_A=\frac{v}{\lambda_A}[/tex]
Then we have wave B, whose wavelength is twice that of wave A:
[tex]\lambda_B = 2 \lambda_A[/tex]
And its speed is the same; Therefore, its frequency is
[tex]f_B = \frac{v}{\lambda_B}=\frac{v}{2\lambda_A}=\frac{1}{2}(\frac{v}{\lambda_A})=\frac{f_A}{2}[/tex]
So, the frequency of wave B is half that of wave A.
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When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the third object is 0.9 C. What was its initial charge (in C)?
Answer:
- Contact 1 with 3 , initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3 , first body should have 3.6 C
Explanation:
The excess charge on a body is distributed evenly throughout the body.
We can have two different configurations:
- Contact 1 with 3
When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3
Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.
Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C
How long would it take to obtain a temperature increase of 2000 degrees celcius, which could cause some metals holding the radioactive materials to melt?
Answer:
Incomplete question.
Complete question given below
a) 2.7933 C/s
b) 714.45 s
Explanation:
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails {1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt.
(a) Calculate the rate of temperature increase in degrees Celsius per second if the mass of the reactor core is 1.60105*10^6 kg and it has an average specific heat of 0.3349 kJ/kgº.
(b) How long would it take to obtain a temperature increase of which could cause some metals holding the radioactive materials to melt?
Part a)
[tex]Q_{rate} = m*c*\frac{dT}{dt}\\\frac{dT}{dt} = \frac{Q_{rate}}{m*c} \\\\\frac{dT}{dt} = \frac{150*10^6}{1.6 * 10^5*334.9} \\\\\frac{dT}{dt} = 2.7993 C/s[/tex]
Part b)
[tex]dt = \frac{dT}{2.7993} \\\\dt = \frac{2000}{2.7993} \\\\dt = 714.45 s[/tex]
Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.
Answer: FR=2.330kN
Explanation:
Write down x and y components.
Fx= FSin30°
Fy= FCos30°
Choose the forces acting up and right as positive.
∑(FR) =∑(Fx )
(FR) x= 5-Fsin30°= 5-0.5F
(FR) y= Fcos30°-4= 0.8660-F
Use Pythagoras theorem
F2R= √F2-11.93F+41
Differentiate both sides
2FRdFR/dF= 2F- 11.93
Set dFR/dF to 0
2F= 11.93
F= 5.964kN
Substitute value back into FR
FR= √F2(F square) - 11.93F + 41
FR=√(5.964)(5.964)-11.93(5.964)+41
FR= 2.330kN
The minimum force is 2.330kN
To determine the magnitude of force F to minimize the resultant FR of three forces, we can use the rule for finding the magnitude of a vector. By taking the square root of the sum of the squares of the components, we can find the magnitude of F.
The question is asking to determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. To find the magnitude of the resultant force, we need to use the rule for finding the magnitude of a vector. We take the square root of the sum of the squares of the components. In this case, we have F = F₁ + F₂, and we can plug in the values given to find the magnitude of F.
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A sphere completely submerged in water is tethered to the bottom with a string. The tension in the string is one-fourth the weight of the sphere.
What is the density of the sphere?
Answer:
ρ = 800 kg/m³
Explanation:
Let the volume of the sphere = V
and the density of the sphere = ρ
density of water , ρ_w = 1000 Kg/m^3
tension in the string is one-fourth the weight of the sphere.
Tension in the rope ,
[tex]T = \rho V \dfrac{g}{4}[/tex]
for the sphere to in equilibrium
T + Weight = buoyant force
[tex]\rho V\dfrac{g}{4} + \rho V g = \rho_w V g[/tex]
[tex] \dfrac{\rho}{4}+\rho= \rho_w[/tex]
[tex] \dfrac{5\rho}{4}= 1000[/tex]
ρ = 800 kg/m³
density of the sphere is equal to 800 kg/m³
Final answer:
To calculate the density of the sphere tethered underwater, we apply Archimedes' principle and equilibrium of forces. Since the tension is one-fourth the sphere's weight, the sphere's density is found to be one-third more than the water's density.
Explanation:
To find the density of the sphere, let's consider the forces acting on the sphere when it is submerged in water. According to Archimedes' principle, the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.
The weight of the sphere (W) can be expressed as the product of its volume (V), its density (p_sphere), and the acceleration due to gravity (g). It can be written as W = V * p_sphere * g. Similarly, the weight of the water displaced by the sphere, which is also the buoyant force (B), is B = V * p_water * g, where p_water is the density of water.
Since the tension (T) in the string is one-fourth the weight of the sphere, we can write T = 1/4 * W. The sphere is in equilibrium, so the sum of the forces equals zero, leading to B = W - T. Substituting in the values, we get (V * p_water * g) = V * p_sphere * g - (1/4 * V * p_sphere * g). From this equation, we can solve for the density of the sphere p_sphere.
After simplifying, p_sphere = 4/3 * p_water, so the sphere's density is one-third more than the density of water.
All valid equations in physics have consistent units. Are all equations that have consistent units valid?
a. No. Any equation can be made to have consistent units through unit conversion.
b. Yes. Consistent units indicate that the equation was derived correctly.
c. No. An equation may have consistent units but still be numerically invaid.
d. Yes. In physics, consistent units guarantee that both sides of an equation represent the same physical quanity
Answer:
c. No. An equation may have consistent units but still be numerically invaid.
Explanation:
For an equation to be corrected, it should have consistent units and also be numerically correct.
Most equation are of the form;
(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)
From the above, without the dimensionless constant the equation would be numerically wrong.
For example; Kinetic energy equation.
KE = 0.5(mv^2)
Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.
Answer:
No. An equation may have consistent units but still be numerically invalid.
Explanation:
A quarterback throws a football with an angle of elevation 40° and speed 80 ft/s. Find the horizontal and vertical components of the velocity vector. (Round your answers to one decimal place.)
Answer:
Horizontal component of velocity will be 61.28 ft/sec
Vertical component of the velocity will be 51.423 ft/sec
Explanation:
We have given velocity of football v = 80 ft/sec'
Angle at which football is thrown [tex]\Theta =40^{\circ}[/tex]
Now we have to fond the horizontal and vertical component of the velocity
Horizontal component of velocity [tex]v_x=vcos\Theta =80\times cos40^{\circ}=80\times 0.766=61.28ft/sec[/tex]
Vertical component of the velocity [tex]v_y=vsin\Theta =80\times sin40^{\circ}=51.423ft/sec[/tex]
At 20 m from a localized sound source you measure the intensity level as 75 dB. How far away must you be for theperceived loud-ness to drop in half [i.e., to an intensity level of 65 dB)?
Answer: 21.48m
Explanation:
Using the inverse square law, the intensity of sound is inversely proportional to the square of the distance.
Intensity ∝ 1/distance²
let l represent intensity and d represent distanc
l₁/l₂ = d₂²/d₁² ......1
given;
l₁ = 75dB
l₂ = 65dB
d₁ = 20 m
substituting into eqn 1, we have;
75/65 = d₂²/20²
d₂² = 75 × 20²/65
d₂² = 461.54
d₂ = √461.54
d₂ = 21.48m
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about 2.17 cm. The speed of sound is 348 m/s. What is the fundamental frequency of the canal?
Answer:
4009.21658986 Hz
Explanation:
v = Speed of sound in air = 348 m/s
L = Length of tube = 2.17 cm
Fundamental frequency for a tube where one end is closed and the other end open is given by
[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{348}{4\times 2.17\times 10^{-2}}\\\Rightarrow f=4009.21658986\ Hz[/tex]
The fundamental frequency of the canal is 4009.21658986 Hz
Callisto, one of Jupiter's moons, has an orbital period of 16.69 days and its distance from Jupiter is 1.88*10^6 km. What is Jupiter's mass?
Answer:
The Jupiter´s mass is approximately 1.89*10²⁷ kg.
Explanation:
The only force acting on Calisto while is rotating around Jupiter, is the gravitational force, as defined by the Newton´s Universal Law of Gravitation:
Fg = G*mc*mj / rcj²
where G = 6.67*10⁻¹¹ N*m²/kg², mc= Callisto´s mass, mj= Jupiter´s mass, and rcj = distance from Jupiter for Callisto= 1.88*10⁹ m.
At the same time, there exists a force that keeps Callisto in orbit, which is the centripetal force, that actually is the same gravitational force we have already mentioned.
This centripetal force is related with the period of the orbit, as follows:
Fc = mc*(2*π/T)²*rcj.
In order to be consistent in terms of units, we need to convert the orbital period, from days to seconds, as follows:
T = 16.69 days* 86,400 (sec/day) = 1.44*10⁶ sec.
We have already said that Fg= Fc, so we can write the following equality:
G*mc*mj / rcj² = mc*(2*π/T)²*rcj
Simplifying common terms, and solving for mj, we get:
mj = 4*π²*(1.88*10⁹)³m³ / ((1.44*10⁶)² m²*6.67*10⁻11 N*m²/kg²)
mj = 1.89*10²⁷ kg.
Answer: Mass of Jupiter ~= 1.89 × 10^23 kg
Explanation:
Given:
Period P= 16.69days × 86400s/day= 1442016s
Radius of orbit a = 1.88×10^6km × 1000m/km
r = 1.88 × 10^9 m
Gravitational constant G= 6.67×10^-11 m^3 kg^-1 s^-2
Applying Kepler's third law, which is stated mathematically as;
P^2 = (4π^2a^3)/G(M1+M2) .....1
Where M1 and M2 are the radius of Jupiter and callisto respectively.
Since M1 >> M1
M1+M2 ~= M1
Equation 1 becomes;
P^2 = (4π^2a^3)/G(M1)
M1 = (4π^2a^3)/GP^2 .....3
Substituting the values into equation 3 above
M1 = (4 × π^2 × (1.88 × 10^9)^3)/(6.67×10^-11 × 1442016^2)
M1 = 1.89 × 10^27 kg
An engine does 25 J of work and exhausts 20 J of waste heat during each cycle. If the cold-reservoir temperature is 30 ∘C, what is the minimum possible temperature in ∘C of the hot reservoir?
Final answer:
The minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat in each cycle, with a cold-reservoir temperature of 30°C is 408.87°C.
Explanation:
To find the minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat during each cycle, with a cold-reservoir temperature of 30 °C, we have to use the concept of efficiency and the Carnot engine.
Firstly, let's convert the cold-reservoir temperature from Celsius to Kelvin:
Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15
Tc (cold reservoir) = 30°C + 273.15 = 303.15 K
Next, we can calculate the efficiency (ε) of the engine using the formula:
ε = Work done (W) / Heat absorbed (Qh)
Qh (heat absorbed) = W (work done) + Qc (waste heat) = 25 J + 20 J = 45 J
So, ε = 25 J / 45 J = 0.555... (repeating)
The efficiency of a Carnot engine is also given by:
ε = 1 - (Tc/Th)
Now, we solve for Th (the hot reservoir temperature in Kelvin):
Th = Tc / (1 - ε)
Th = 303.15 K / (1 - 0.555...) = 303.15 K / 0.444... = 682.02 K
Finally, we convert the hot reservoir temperature back to Celsius:
Temperature in Celsius (°C) = Temperature in Kelvin (K) - 273.15
Th (hot reservoir) in °C = 682.02 K - 273.15 = 408.87°C
Thus, the minimum possible temperature of the hot reservoir is 408.87°C.
A rocket has a mass 340(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
a) Specify mass of the rocket in SI units.
b) Specify weight of the rocket in SI units.
c) If the rocket is on the moon, where the acceleration due to gravity is gm= 5.30ft/s2, determine to three significant figures its weight in SI units.
d) Determine to three significant figures its mass on the moon in SI units.
Answer:
[tex]49.6\times 10^5\ kg[/tex]
[tex]48.6\times 10^6\ N[/tex]
[tex]80.1\times 10^5\ N[/tex]
[tex]49.6\times 10^5\ kg[/tex]
Explanation:
[tex]1\ slug=14.5939\ kg[/tex]
[tex]340\times 10^3\ slug=340\times 10^3\times 14.5939=4961926\ kg[/tex]
The mass in SI unit is [tex]49.6\times 10^5\ kg[/tex]
Weight would be
[tex]W=mg\\\Rightarrow W=4961926\times 9.81\\\Rightarrow W=48676494.06\ N[/tex]
The weight in SI unit is [tex]48.6\times 10^6\ N[/tex]
[tex]1\ ft/s^2=0.3048\ m/s^2[/tex]
[tex]5.30\ ft/s^2=5.30\times 0.3048=1.61544\ m/s^2[/tex]
[tex]W=mg\\\Rightarrow W=4961926\times 1.61544\\\Rightarrow W=8015693.73744\ N[/tex]
The weight on the moon is [tex]80.1\times 10^5\ N[/tex]
The mass of an object is same anywhere in the universe.
So, the mass of the rocket on Moon is [tex]49.6\times 10^5\ kg[/tex]
An object moves along a coordinate line, its position at each time t≥0 is given by x(t)=4t2t+1. Find the velocity at time t0=3.
The velocity of the object at t=3 seconds is calculated by differentiating the given position function and substituting t=3 into the derived velocity function, resulting in a value of 0.25 m/s.
Explanation:This problem is about finding the velocity of an object given a position function x(t)=4t²/(t+1), which describes the object's position at any time 't'. To find the velocity at a specific time, you need to take the derivative of the position function, which gives you a velocity function v(t).
The derivative of x(t) with respect to 't' is v(t) = 4t(t+1)-4t²/ (t+1)² = 4/(t+1)². So, when t0=3, the velocity v(3) = 4/(3+1)² = 4/16 = 0.25 m/s.
In conclusion, at t = 3 seconds, the velocity of the object is 0.25 m/s.
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The question seeks to determine the velocity of an object at a specific time given its position function. However, the position function provided in the question appears to contain a typo or error, therefore it is not possible to calculate the velocity without the correct function.
Explanation:The question involves calculating the velocity of an object at a given time t based on its position function. To find the velocity at time t = 3 seconds, we'll need to take the derivative of the position function. This is a principle from physics dealing with kinematics. However, the function provided, which is x(t) = 4t2t+1, seems to be incorrect or has a typo as it is not usable in its current form. Assuming that the correct position function might have been x(t) = 4t^2/(2t+1), the derivative or velocity function would be v(t) = d/dt [x(t)]. Nevertheless, since we do not have a correct function to work with, we cannot find a specific numerical value for the velocity at t = 3 seconds.
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.820 and μk = 0.440. The turntable very slowly speeds up.1. What is the angular speed in rpm when the coin slides off?
Answer:
62.64 RPM.
Explanation:
Given that
m= 4.6 g
r= 19 cm
μs = 0.820
μk = 0.440.
The angular speed of the turntable = ω rad/s
Condition just before the slipping starts
The maximum value of the static friction force =Centripetal force
[tex]\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s[/tex]
[tex]\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM[/tex]
Therefore the speed in RPM will be 62.64 RPM.
A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/sm/s and at point B is 5.4 m/sm/s . Part A Use the impulse-momentum theorem to find how long the sled takes to travel from A to B.
Answer:
[tex]\Delta t =1.31\ s[/tex]
Explanation:
given,
coefficient of kinetic friction, μ = 0.25
Speed of sled at point A = 8.6 m/s
Speed of sled at point B = 5.4 m/s
time taken to travel from point A to B.
we know,
J = F Δ t
J is the impulse
where F is the frictional force.
t is the time.
we also know that impulse is equal to change in momentum.
[tex]J = m(v_f - v_i)[/tex]
frictional force
F = μ N
where as N is the normal force
now,
[tex]F\Delta t = m(v_f -v_i)[/tex]
[tex]\mu m g \times \Delta t = m(v_f-v_i)[/tex]
[tex]\mu g \times \Delta t = v_f-v_i[/tex]
[tex]\Delta t =\dfrac{v_f-v_i}{\mu g}[/tex]
[tex]\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}[/tex]
[tex]\Delta t =1.31\ s[/tex]
time taken to move from A to B is equal to 1.31 s
Answer:
Time taken by the sled is 1.31 s
Solution:
As per the question:
Coefficient of kinetic friction, [tex]\mu_{k} = 0.25[/tex]
Velocity at point A, [tex]v_{A} = 8.6\ m/s[/tex]
Velocity at point A, [tex]v_{B} = 5.4\ m/s[/tex]
Now,
To calculate the time taken by the sled to travel from A to B:
According to the impulse-momentum theorem, impulse and the change in the momentum of an object are equal:
Impulse, I = Change in momentum of the sled, [tex]\Delta p[/tex] (1)
[tex]I = Ft[/tex] (2)
where,
F = Force
t = time
p = momentum of the sled
Force on the sled is given by:
[tex]F = \mu_{k}N[/tex]
where
N = normal reaction force = mg
where
m = mass of the sled
g = acceleration due to gravity
Thus
[tex]F = \mu_{k}mg[/tex] (3)
Using eqn (1), (2) and (3):
[tex]\mu_{k}mgt = m\Delta v[/tex]
[tex]\mu_{k}gt = v_{A} - v_{B}[/tex]
[tex]t = \frac{v_{A} - v_{B}}{\mu_{k}g}[/tex]
[tex]t = \frac{8.6 - 5.4}{0.25\times 9.8}[/tex]
t = 1.31 s
A ball is tossed so that it bounces off the ground, rises to a height of 2.50 m, and then hits the ground again 0.90 m away from the first bounce.
1.How long is the ball in the air between the two bounces?2.What is the ball's velocity in the x-direction?3.What is the ball's speed just before the second bounce?4.What is the angle of the velocity vector with respect to the ground right after the first bounce
To solve this problem we will apply the linear motion kinematic equations. With the equation describing the position with respect to acceleration (gravity), initial velocity and time, we will find time.
Our values are,
[tex]h = 2.5m[/tex]
[tex]x = 0.9m[/tex]
1) We know that, the time, the ball in the air depends on the maximum height.
Therefore the height is calculated as
[tex]h = v_{yi}t+\frac{1}{2} gt^2[/tex]
There is not initial velocity then replacing,
[tex]2.5 = 0t + \frac{1}{2}9.8t^2[/tex]
[tex]t = \sqrt{\frac{2(2.5)}{9.8}}[/tex]
[tex]t = 0.714s[/tex]
By symmetry, the time taken to reach ground is equal to the time taken to reach the maximum height, then
[tex]T = t+t[/tex]
[tex]T = 2*0.714s[/tex]
[tex]T = 1.426s[/tex]
2 ) In the time t=1.428s, the ball moves horizontally a distance x = 0.9
Then the horizontal velocity,
[tex]v_x = \frac{x}{t} = \frac{0.9m}{1.428s}[/tex]
[tex]v_x = 0.6302m/s[/tex]
3 ) The ball's speed just before the second bounce is found as
[tex]v_y = v_{yi} +gt_{down}[/tex]
Replacing we have
[tex]v_y = 0+(9.8)(0.714)[/tex]
[tex]v_y = 6.9972m/s[/tex]
4 ) The angle of the velocity vector with respect to the ground right after the first bounce is
[tex]\theta = tan^{-1} (\frac{v_y}{v_x})[/tex]
[tex]\theta = tan^{-1} (\frac{6.9972}{0.6302})[/tex]
[tex]\theta = 84.8\°[/tex]
The helium balloon you got on your birthday has almost entirely deflated, but there is still a very small amount of 4He inside with a total mass of 160 amu. How many protons and how many electrons are present in the balloon (assuming only 4He is there)?
In a helium balloon with a total mass of 160 amu containing helium-4 atoms, there would be 40 helium atoms, resulting in a total of 80 protons and 80 electrons.
Explanation:The question involves calculating the number of protons and electrons in a helium balloon with a total mass of 160 amu, assuming it contains only 4He (helium-4). A helium-4 atom consists of 2 protons, 2 neutrons, and 2 electrons, with a total atomic mass of 4 amu. To find the number of helium atoms, we divide the total mass of helium in the balloon (160 amu) by the mass of a single helium-4 atom, which is 4 amu. This gives us 40 helium atoms.
Since each helium atom has 2 protons and 2 electrons, to find the total number of protons and electrons in 40 helium atoms, we multiply the number of helium atoms by the number of protons or electrons per atom. Therefore, there are 80 protons and 80 electrons in the helium balloon.
A world-class shotputter can put a 7.26 kg shot a distance of 22 m. Assume that the shot is constantly accelerated over a distance of 2 m at an angle of 45 degrees and is released from a height of 2 m above the ground. Estimate the weight that this athlete can lift with one hand.
To solve this problem we will make a free body diagram to better understand the displacement measurements made by the body. From there we will apply the linear motion kinematic equations that describe the position of the body in reference to its vertical displacement, acceleration and speed. With this speed found we will apply the energy conservation theorem that will allow us to find the Force.
Equation of trajectory of a projectile is
[tex]y = xtan\theta - x^2 \frac{g}{2u^2cos^2\theta}[/tex]
Here
u = Initial velocity
x = Horizontal displacement
g = Acceleration due to gravity
y = Vertical displacement
We have that
[tex]x = 22m[/tex]
[tex]y = -2\sqrt{2m}[/tex]
Replacing we have that,
[tex]-2\sqrt{2} = 22tan45\° -\frac{9.8}{2u^2cos^2 45}(22)^2[/tex]
[tex]-2\sqrt{2} =22 -\frac{9.8*484}{2u^2(1/2)}[/tex]
[tex]-2\sqrt{2} =22 -\frac{4743.2}{u^2}[/tex]
[tex]u^2 = \frac{4743.2}{22+2\sqrt{2}}[/tex]
[tex]u = 191.03m/s[/tex]
From the work energy theorem
[tex]W_{net} = K_f +K_i[/tex]
Here,
[tex]K_f = \frac{1}{2} mu^2[/tex]
[tex]K_i = \frac{1}{2} m(0)^2 = 0[/tex]
[tex]W_{net} = W_m+W_g[/tex]
Where,
[tex]W_m =\text{Work by man} = F_s[/tex]
[tex]W_{g} = \text{Work by gravity} = -mgh[/tex]
Therefore
[tex]F_s -mgh = \frac{1}{2} mu^2[/tex]
[tex]F_s = \frac{m}{s} (\frac{u^2}{2}+gh)[/tex]
[tex]F_s = \frac{7.26}{2\sqrt{2}}(\frac{191.03}{2}+9.8*2)[/tex]
[tex]F_s = 295.477N[/tex]
Which of the following phenomena are due to the electric interaction? (Select all that apply.) surface tension in water friction between tires and pavement the elliptical orbit of comets dissolution of salt in water the binding of protons in an atomic nucleus
Answer:
Surface tension in water
Friction between tires and pavement
Dissolution of salt in water
Explanation:
Surface tension in water: It is due to the electrostatic force of attraction (cohesive force) between water molecules.
Friction between tires and pavement: It is due to the attractive force between tires and pavement.
Dissolution of salt in water: The ions of [tex]Na ^ +[/tex] and [tex]Cl ^ -[/tex] separate due to the strong attraction of water molecules.
Final answer:
Surface tension in water, friction between tires and pavement, and the dissolution of salt in water are phenomena due to electric interaction, which is rooted in the forces of cohesion, adhesion, and intermolecular attractions.
Explanation:
The phenomena that are due to electric interaction include surface tension in water, friction between tires and pavement, and the dissolution of salt in water. The electric interaction plays a crucial role in these phenomena through the forces of cohesion, adhesion, and intermolecular attractions, which are manifestations of the electromagnetic force. While the elliptical orbit of comets is primarily governed by gravitational forces and the binding of protons in an atomic nucleus by the strong nuclear force, surface tension, friction, and dissolution phenomena are deeply rooted in electrical interactions between atoms and molecules.
You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving for 4.5 h. a. how far is your hometown from school?b. what was your average speed?
To solve this problem we will apply the linear motion kinematic equations, for this purpose we will define the time of each of the sections in which the speed is different. After determining the segments of these speeds we can calculate the average distance and the average speed. Our values are given as
x = 180 km
v = 95 km / h
Speed can be described as the displacement of a body per unit of time, and from that definition clearing the time we would have
[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]
[tex]v = \frac{180}{95}[/tex]
[tex]v = 1.89473 hrs[/tex]
From the statement we have that the total time is 4.5, then he remaining time is
[tex]t' = T-t =4.5-1.89473 = 2.60526 hrs[/tex]
When it starts to rain there is a phase change in the speed which is given by
[tex]v' = 65km/h[/tex]
Then the distance travel in this velocity
[tex]x' = v ' t '[/tex]
[tex]x' = 65*2.60526[/tex]
[tex]x = 169.34 km[/tex]
(a). Distance of your hometown from school ,
[tex]x " = x + x '[/tex]
[tex]x'' = 180+169.34[/tex]
[tex]x= 349.34 km[/tex]
(b) The average speed
[tex]V = \frac{x "}{ T}[/tex]
[tex]v = \frac{349.34 km}{4.5 h}[/tex]
[tex]v = 77.63 km/ h[/tex]
Final answer:
The student's hometown is 350 km away from school, and their average speed for the trip was 77.78 km/h.
Explanation:
To determine how far the student's hometown is from school and their average speed, we need to use their driving speeds and times. Firstly, let's calculate the time taken to travel the first 180 km at 95 km/h.
Time = Distance \/ Speed = 180 km \/ 95 km/h = 1.8947 hours (approximately 1.89 hours).
Since the total travel time is 4.5 hours, the time spent driving in the rain at 65 km/h is 4.5 hours - 1.89 hours = 2.61 hours.
Distance driven in the rain = Speed × Time = 65 km/h × 2.61 hours = 169.65 km (approximately 170 km).
Therefore, the total distance from school to the student's hometown is the sum of both distances: 180 km (before rain) + 170 km (in rain) = 350 km.
Now, to find the average speed, we use the total distance and total time.
Average speed = Total distance \/ Total time = 350 km \/ 4.5 hours = 77.78 km/h.
Thus, the hometown is 350 km away from school, and the student's average speed for the trip was 77.78 km/h.
In Fig. 21-24, the particles have charges q1 q2 100 nC and q3 q4 200 nC, and distance a 5.0 cm. What are the (a) x and (b) y components of the net electrostatic force on particle 3?
Answer:
(a) The x component of the net electrostatic force on charge q3 is 0.195N
(b) The x component of the net electrostatic force on charge q3 is 0.021N
Explanation:
The force of q1 on q3 is a repulsive force and those of q2 and q4 on q3 are attractive forces. From second diagram at the bottom of the page in the attachment below it can be seen that the force of q1 on q3 is directed vertically downward and has on y component while the force of q4 on q3 is directed to the right and has on x component. The force of q2 on q3 has both x and y components. The full solution can be found in the attachment below.
Thank you for reading.
The electric force between two charged bodies at rest is called as electrostatic force.
The values are the following:
The x component of the net electrostatic force = 0.17 N
The y component of the net electrostatic force = 0.045 N
Solution :For component X
[tex]F_3_y = \dfrac{lql}{4\pi\varepsilon _0q^2 } (\dfrac{-lq_1l+ lq_1l}{\sqrt[2]{2} } )[/tex][tex]= (8.99 \times 10^{9} N.m2/ C^{2} \dfrac{(2.0\times10^{-7} )^2} {(0.050 m)^2}(-2 +\frac{2}{2\sqrt{2} } )\\ = 0.17 N[/tex]
X= 0.17 N
For component Y
[tex]F_3_y = \dfrac{lql}{4\pi\varepsilon _0q^2 } (\dfrac{-lq_1l+ lq_1l}{\sqrt[2]{2} } )[/tex][tex]= (8.99 \times 10^{9} N.m^2/ C^{2} \dfrac{(1.0\times10^{-7} )^2} {(0.050 m)^2}(-1 +\frac{1}{2\sqrt{2} } )\\ = 0.045 N[/tex]Y = 0.045 N.
Hence, X= 0.17 N and Y = 0.045 N.
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Three point charges are located on the positive x-axis of a coordinate system. Charge qı = 1.5 nC is 2.0 cm from the origin, charge q2 = -3.0 nC is 4.0 cm from the origin and charge 93 = 4.5 nC located at the origin. a. What is the magnitude of the net force exerted by the other two charges on charge q1q1q1 = 1.5 nCnC?b. What is the direction of the net force exerted by the other two charges on charge q1q1q1 = 1.5 nCnC?
Answer:
[tex]F_1=2.53\ 10^{-4} \ N[/tex]
The net force goes to the right
Explanation:
Electrostatic Force
Let's consider the situation where 2 point charges q1 and q2 are separated by a distance d. An electrostatic force appears between them whose magnitude can be computed by the Coulomb's formula
[tex]\displaystyle F=\frac{k\ q_1\ q_2}{d^2}[/tex]
Where k is the constant of proportionality
[tex]k=9.10^9\ Nw.m^2/c^2[/tex]
Two equally-signed charges repel each other, two opposite-signed charges attract each other.
We need to find the total net force exerted on q1 by q2 and q3. We're assuming the charges are placed to the right of the origin, so the distribution is shown in the figure below.
Since q3 repels q1, its force goes to the right, since q2 attracts q1, its force goes to the right also, thus the total force on q1 is :
[tex]F_1=F_{31}+F_{21}[/tex]
It's directed to the right
Let's compute the individual forces. q3 is separated 2 cm from q1, so d=0.02 m
[tex]\displaystyle F_{31}=\frac{9.10^9\ 4.5\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}[/tex]
[tex]F_{31}=0.000151875\ N[/tex]
[tex]\displaystyle F_{21}=\frac{9.10^9\ 3\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}[/tex]
[tex]F_{21}=0.00010125\ N[/tex]
[tex]F_1=F_{31}+F_{21}=0.000151875\ N+0.00010125\ N=0.000253125 \ N[/tex]
Expressing the force in scientific notation
[tex]\boxed{F_1=2.53\ 10^{-4}\ N}[/tex]
The net force goes to the right
Solution: The magnitude of the net force acting on q₁ is : Fn = 25.312 [N]
and the direction is in the direction of the positive x axis
The electric force between two charges q₁ and q₂, is according to Coulombs´law as:
F₂₁ = K × q₁ × q₂ / d₁₂²
In that equation
F₂₁ is the force exerted by charge q₂ on the charge q₁
K = 9× 10⁹ [Nm²/C²] Coulomb contantq₁ is the charge upon which the force is actingq₂ is the force acting over q₁d₁₂ is the distance between the chargesIt is pretty obvious that F₂₁ = F₁₂, and the force is of rejection if the charges are of the same sign or attraction if they are of opposite signs
Now in this case we calculate F₂₁
F₂₁ = [9×10⁹ × ( 1.5)×10⁻⁹ × (3)×10⁻⁹] /(0.02)²
F₂₁ = 40.5 × 10⁻⁹/ 4 × 10⁻⁴
F₂₁ = 10.125 × 10⁻⁵ [N]
F₂₁ is in the direction of the positive x ( attraction force)
And
F₃₁ = [9×10⁹ × ( 1.5)×10⁻⁹ × (4.5)×10⁻⁹] /(0.02)²
F₃₁ = 15.188× 10⁻⁵ [N]
F₃₁ is in the direction of the positive x ( rejection force)
Then the net force is Fn = 10.125 + 15.188
Fn = 25.312 [N] in direction of x positive
If three uncharged styrofoam balls are placed together and agitated so that one gains 3 CC of charge and another gains 4 CC of charge, how much charge must there be on the third one
Answer:
-7 C
Explanation:
Assuming that object other than the styrofoam balls was part of the charge transfer. In order to maintain charge balance, the initial charge of the system must equal the ending charge. If all balls were uncharged initially, the ending charge on the third ball must be:
[tex]C_1+C_2+C_3 = 0\\3+4+C_3=0\\C_3=-7[/tex]
There must be -7 C of charge on the third ball.
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
(a) What would be the height of the atmosphere if the air density were constant?
km
(b) What would be the height of the atmosphere if the air density decreased linearly to zero with height?
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
[tex]h = \dfrac{P}{\rho\ g}[/tex]
[tex]h = \dfrac{101300}{1.3\times 9.8}[/tex]
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
[tex]\rho_x = \dfrac{\rho_{sl}}{h}\times x[/tex]
now, Pressure at depth x
[tex]dP = \rho_x g dx[/tex]
[tex]dP = \dfrac{\rho_{sl}}{h}\times x g dx[/tex]
integrating both side
[tex]P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx[/tex]
[tex] P =\dfrac{\rho_{sl}\times g h}{2}[/tex]
now,
[tex]h=\dfrac{2P}{\rho_{sl}\times g}[/tex]
[tex]h=\dfrac{2\times 101300}{1.3\times 9.8}[/tex]
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart when they each acquire 1.00 mC charges. Find their acceleration. (b) What is unreasonable about this result? (c) Which premise or assumption is responsible?
Explanation:
Given that,
Mass of the raindrops, [tex]m=0.5\ g=0.0005\ kg[/tex]
Charges, [tex]q=1\ mC=10^{-3}\ C[/tex]
Distance between raindrops, d = 1 cm = 0.01 m
(a) The force due to motion of raindrop is balanced by the electric force between charges. It is given by :
[tex]ma=\dfrac{kq^2}{r^2}[/tex]
a is the acceleration of the raindrop
[tex]a=\dfrac{kq^2}{r^2m}[/tex]
[tex]a=\dfrac{9\times 10^9\times (10^{-3})^2}{(0.01)^2\times 0.0005}[/tex]
[tex]a=1.8\times 10^{11}\ m/s^2[/tex]
(b) It is clear that the acceleration is too large. It can break apart the rain drop.
(c) If the charge is of the factor of [tex]10^{-8}\ C[/tex], the charge would be more reasonable.
Hence, this is the required solution.
To find the acceleration of the charged raindrops, we can use Coulomb's Law and Newton's second law. The acceleration is found to be 1.8 x 10^13 m/s^2, which seems unreasonable for raindrops. The assumption of the raindrops acquiring charges of 1.00 mC is likely responsible for this result.
To calculate the acceleration of the two charged raindrops, we can use Coulomb's Law, which states that the force between two charges is given by: F = (k*q1*q2)/r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. The force experienced by each raindrop is equal in magnitude but opposite in direction.
The acceleration of each raindrop can be found using Newton's second law: F = m*a, where F is the net force, m is the mass of the raindrop, and a is the acceleration. Rearranging the equation, we have: a = F/m. Substituting the values into the equations and solving for acceleration, we find: a = (k*q^2)/(m*r^2). Plugging in the values, we have: a = (9 x 10^9 N*m^2/C^2 * (1 x 10^-3 C)^2) / (0.500 g * 10^-3 kg/g * (1 x 10^-2 m)^2). After calculating the values, we find that the acceleration of the two raindrops due to their charges is 1.8 x 10^13 m/s^2.
(b) What is unreasonable about this result? The result seems unreasonable because the acceleration is extremely high. Normally, objects with masses like raindrops would experience much lower accelerations. (c) The premise or assumption responsible for this result is that the raindrops have acquired charges of 1.00 mC, which is an extremely high charge for raindrops and likely not possible in reality.
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There is an analogy between rotational and linear kinematics. What rotational quantities are analogous to distance and velocity?
Answer:
1
Explanation:
Two identical hard spheres, each of mass m and radius r, are released from rest in otherwise empty space with their centers separated by the distance R. They are allowed to collide under the influence of their gravitational attraction. (a) Find the magnitude of the impulse received by each sphere before they make contact. (b) Find the magnitude of the impulse each receives during their contact if they collide elastically.
Answer:
a) I = √ (2G m³ (1/2r³ - 1/R)), b) I = √ (8 G m³ (1/2r -1/R))
Explanation:
.a) The relation of the Impulse and the moment is
I = Δp = m [tex]v_{f}[/tex] - m v₀
We can use Newton's second law with force the force of universal attraction
F = ma
G m m / r² = m a
dv / dt = G m / r²
Suppose re the direction where the spheres move is x
dv/dx dx/dt = G m / x²
dv/dx v = G m / x²
v dv = G m dx / x²
We integrate
v² / 2 = Gm (-1 / x)
We evaluate this integra from the lower limit v = 0 for x = R to the upper limit, where the spheres v = v and x = 2r are touched
v² / 2-0 = G M (-1 / R + 1 / 2r)
v = √ [2Gm (1 /2r - 1/ R) ]
The impulse on the sphere is
I = m vf - m v₀
I = m vf - 0
I = m √ (2Gm (1 / 2r-1 / R)
I = √ (2G m³ (1/2r³ - 1/R))
b) during the crash each sphere arrives with a velocity v and leaves with a velocity –v, the same magnitude but opposite direction
I = m [tex]v_{f}[/tex]- m v₀
I = m v - m (-v)
I = 2mv
I = 2m √ (2Gm (1 / 2r-1 / R)
I = √ (8 G m³ (1/2r -1/R))
We calculate the impulse received by each sphere before they make contact and during their elastic collision.
Explanation:In this scenario, we have two identical hard spheres separated by a distance R. The spheres are released from rest and allowed to collide under the influence of their gravitational attraction. We need to find the magnitude of the impulse received by each sphere before they make contact and during their elastic collision.
(a) Before the spheres make contact, the magnitude of the impulse received by each sphere can be found using the equation:
Impulse = Change in momentum = Mass x Change in velocity.
Since both spheres are released from rest, their initial velocities are zero. Therefore, the change in velocity is the final velocity. Using the equation:
Final velocity = sqrt(2 x G x m / R).
Substituting the values, we can calculate the magnitude of the impulse received by each sphere before they make contact.
(b) During their elastic collision, the magnitude of the impulse received by each sphere can be found using the equation:
Impulse = Change in momentum = Mass x Change in velocity.
Since the spheres collide elastically, there is no change in velocity. Therefore, the magnitude of the impulse received by each sphere during their contact is zero.
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