To develop this problem we will proceed to use the principle of energy conservation. For this purpose we will have that the change in the electric potential energy and kinetic energy at the beginning must be equal at the end. Our values are given as shown below:
[tex]m = 9.75g = 0.00975kg[/tex]
[tex]r = 1.3cm = 0.013m[/tex]
[tex]q_1 = 0.1 C/m^3 * \frac{4}{3} \pi r^3[/tex]
[tex]q_1 = 9.2*10^{-7}C[/tex]
[tex]q_2 = -9.2*10^{-7}C[/tex]
Applying energy conservation equations
[tex]U_1+K_1 = U_2+K_2[/tex]
[tex]\frac{k q_1q_2}{d} +0 = \frac{kq_1q_2}{2r}+ \frac{1}{2} (2m)v^2[/tex]
Replacing,
[tex]9*10^{9} (9.2*10^{-7})^2(\frac{1}{0.026}-\frac{1}{0.8}) = v^2[/tex]
Solving for v,
[tex]v = 9.2*10^{-7} (\frac{9*10^9}{0.00975}(\frac{1}{0.026}-\frac{1}{0.8}))^{1/2}[/tex]
[tex]v = 5.4 m/s[/tex]
Therefore the speed of each sphere at the moment they collide is 5.4m/s
A wheel with a 0.10-m radius is rotating at 35 rev/s. It then slows uniformly to 15 rev/s over a 3.0-s interval.
a. What is the angular acceleration of a point on the wheel?
b. how to do and explain will give lifesaver!
Answer:
The angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating.
Explanation:
It is given that,
Radius of the wheel, r = 0.1 m
Initial angular velocity of the wheel, [tex]\omega_i=35\ rev/s=219.91\ rad/s[/tex]
Final angular velocity of the wheel, [tex]\omega_f=15\ rev/s=94.24\ rad/s[/tex]
Time, t = 3 s
We need to find the angular acceleration of a point on the wheel. It is given by the rate of change of angular velocity divided by time taken. It is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]\alpha =\dfrac{(94.24-219.91)\ rad/s}{3\ s}[/tex]
[tex]\alpha =-41.89\ rad/s^2[/tex]
So, the angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating. Hence, this is the required solution.
Final answer:
The angular acceleration of a point on the wheel can be found using the formula: angular acceleration = change in angular velocity / time interval. Given the initial and final angular velocities and the time interval, we can calculate the angular acceleration.
Explanation:
The angular acceleration of a point on the wheel can be found using the formula:
angular acceleration = change in angular velocity / time interval
Given that the initial angular velocity is 35 rev/s, the final angular velocity is 15 rev/s, and the time interval is 3.0 s, we can substitute these values into the formula:
angular acceleration = (15 rev/s - 35 rev/s) / 3.0 s
Simplifying the equation, we get:
angular acceleration = -20 rev/s / 3.0 s
Converting rev/s to rad/s, we have:
angular acceleration = -20 rev/s ×(2π rad/rev) / 3.0 s
angular acceleration ≈ -41.89 rad/s²
Therefore, the angular acceleration of a point on the wheel is approximately -41.89 rad/s².
A U.S. Department of Energy report estimates that over 100 billion kWh/year can be saved in the United States by various energy-conservation techniques applied to the pump-driven systems. Calculate (a) how many 1000-MW generating plants running constantly supply this wasted energy and (b) the annual savings in dollars if the cost of electricity is 0.10 $/kWh.
Answer:
(a). 12 plants
(b). 3171 $
Explanation:
(a)first convert units of 100 billion kWh/year into Watts(W)
also convert the units of 1000 MW into Watts(W)
1 billion = 10^9
1 year = 365*24 = 8760 hrs
so
100 billion kWh/year = 1[tex]\frac{100*(10^9)*(10^3)}{8760}[/tex]
= [tex]1.142*10^{10}[/tex]W
1000 MW = [tex]1000*10^{6} = 10^{9}W[/tex]
no. of plants = [tex]\frac{1.14155*10^{10} }{10^9}[/tex] = 11.4
So 12 plants required
(b)
savings = unit price*total units
= [tex]0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )[/tex]
= 3170.9 =3171 $
Answer:
a) Number of generating plants N = 11.42
That means N > 11
N = 12
b) annual savings S = $1×10^10
S = $10 billion
Explanation:
Given;
Amount of energy to be saved A=100 billion kWh/year
Capacity of each generating plant C= 1000 MW
Rate in dollars of cost of electricity R= $0.10/kWh
The number N of generating plants with capacity C that can supply the Amount A of of energy cam be given as;
N = A/C ......1
And also the Annual savings S in dollars if the rate of electricity cost R is used and amount of energy A is saved is:
S = AR .....2
But we need to derive the value of A in Watts
A = 100 billion kWh/year
There are 8760hours in a year,
A = 1×10^14 ÷ 8760 W
A = 11415525114.1W or 11415525.1141kW
C = 1000MW = 1× 10^9 W
a) Using equation 1,
N = 11415525114.1/(1×10^9)
N = 11.42
That means N > 11
N = 12
b) using equation 2
S = 1×10^11 kWh × $0.10/kWh
S = $1×10^10
S = $10 billion
If you were designing a room in a house, where would be the better place to put a heater, near the floor or near the ceiling? Why? Consider your answer in the context of convection.1. Near the ceiling, because the air warmed by the heater would quickly spread down and across the room, replacing the cold air.2. Near the floor, because the heater would warm the air close to it. After being heated, the now-warm air would rise, and be replaced by cool air, which the heater would warm. The cycle would continue until the room is heated.3. Near the ceiling, because the heater would warm the air close to it. After being heated, the now-warm air would descend, and be replaced by cool air, which the heater would warm. The cycle would continue until the room is heated.4. Near the floor, because the air warmed by the heater would quickly spread up and across the room, replacing the cold air.
Answer: The correct explanation is 2.
Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.
On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.
Final answer:
The heater should be placed near the floor due to convection, where warm air rises after being heated, circulates, and evenly distributes throughout the room as it cools and descends. Therefore, option 2 is the correct answer.
Explanation:
Proper Placement of a Heater in a Room
When considering the placement of a heater in a room in the context of convection, the better option is to place the heater near the floor. This is because, as per the principles of convection, warm air will rise after being heated by the heater. The now-warm air, having become less dense, will rise to the ceiling, creating a convective loop that circulates the warm air throughout the room. As the air cools down near the ceiling and outside walls, it contracts, becoming denser, and subsequently sinks back to the floor, where it will be heated again by the heater. This cycle continues, leading to an efficient distribution of heat throughout the room. Therefore, option 2 is correct: placing the heater near the floor allows it to warm the air close to it, after which the warm air rises and is replaced by cool air that the heater warms again, maintaining a continuous cycle of heating.
What would you expect the barometric pressure outside of a jet flying at 9.6 km to be?
Answer:
The pressure is expected to decrease because as we move up in the atmosphere the is lesser mass of atmosphere weight above us.
Explanation:
The atmospheric pressure is the the force exerted by the layers of the atmosphere above a point. Being a fluid and due to the gravity of earth there is a pressure due to the weight of the atmospheric columns above us.
So, at a height h above the earth surface in the atmosphere the pressure will decrease and this can be calculated as:
[tex]P_h=101325\times (1-2.25577\times 10^{-5}\times h)^{5.25588}[/tex]
putting h=9.6 meter
[tex]P_h=101209.7268\ Pa[/tex]
At an elevation of 9.6 km, the pressure is approximately 254 millibars.
The pressure is inversely proportional to elevation, higher the elevation lower will be the pressure while on the other hand, lower the elevation higher will be the pressure. At an elevation of 10 km pressure drops to 265 millibars then the elevation of 9.6 km, the pressure is approximately 254 millibars.
The atmospheric pressure at sea level is about 1,014 millibars which is equal to the pressure of 14.7 pounds/inch2 so we can conclude that at the elevation of 9.6 km, the pressure is approximately 254 millibars.
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In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop)?
vavg = ______ m/s
(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
Δt = ______ s
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
Fwall, avg = ______ N
(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.
Fwall, avg / mg =
(e) What approximations were necessary in making this analysis? (Select all that apply.)
Neglect the horizontal component of the force of the road on the truck tires.
Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
The deceleration of the truck is approximately equal to g.
Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]
Part b
[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]
(a) The required average speed of the truck during collision is 11 m/s.
(b) The required time interval for the collision is 0.058 s.
(c) The required magnitude of the average force exerted by the wall on the truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e) The required approximations are:
Neglect the horizontal component of the force of the road on the truck tires.Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.
Given data:
The mass of the truck is, m = 2100 kg.
The speed of the truck is, v = 22 m/s.
The distance crumpled by the truck is, d = 0.62 m.
(a)
Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,
[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]
Thus, the required average speed of the truck during collision is 11 m/s.
(b)
Now, apply the first kinematic equation of motion as,
[tex]v' = v + at[/tex]
Here, a is the linear acceleration and t is the time interval for the collision.
Solving as,
[tex]0 = 22 + at\\\\ a = -22/t[/tex]
Now, apply the second kinematic equation as,
[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]
Thus, we can conclude that the required time interval for the collision is 0.058 s.
(c)
The expression for the magnitude of average force exerted by wall on truck is,
[tex]F_{av.} = \dfrac{mv}{t}[/tex]
Solving as,
[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]
Thus, the required magnitude of the average force exerted by the wall on truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d)
The ratio of force of the truck and the gravitation force on the truck is,
[tex]= \dfrac{F_{av}}{mg}[/tex]
Here, g is the gravitational acceleration.
Solving as,
[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]
Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e)
The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.
This is because:
The horizontal force may act as an unbalanced force.And the constant force by wall results in the variation of the speed of the truck at the constant rate.Learn more about the linear kinematics here:
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Unless otherwise specified in the question, responses to numerical question must be within 5% of the correct answer. Try out different responses to the following question, which has an answer of 20, where a correct response is between 19 and 21.
What is the average velocity (in kilometers per hour) of a cart that travels 100 km in 5 hours?
2.An exception to the 5% tolerance rule arises for simple calculations that can be done in your head that come out as integers or or simple decimals. So if the calculation involves dividing 5 by 2 with the result of 2.5, an answer of 2.6 may not be graded as correct.
Give the value of 6.5 times 2.
Try it several times giving the answer (13) in different forms (say, 13 or 1.3e1) and with different accuracies (say, 13.001 or 13.2). Since this is a simple calculation, two of these four will be graded as incorrect.
Part of the problem involves not only answering the question with the correct degree of accuracy, but also making approximations to the correct answer. For example, for the first case we know that the speed is equivalent to the distance traveled in a given time. Therefore it would be defined as
]1) [tex]V = \frac{x}{t}[/tex]
x = Displacement
t = Time
The displacement value is 100km and the time value is 5 hours. Therefore the speed value would be
[tex]V = \frac{100km}{5h} = 20km/h[/tex]
We know that the answer margin is within 5% of the value, then 5% of 20 would be 1. That is, we have a margin of error of '1km / h' to answer the question. Any value that falls within that range can be added or subtracted from the response and the response will be valid. Values included within this value would be
[tex]V_1 = 20km/h +1km/h = 21km/h[/tex]
[tex]V_2 = 20km/h - 1km/h = 19km/h[/tex]
[tex]V_3 = 20km/h + 0.5km/h = 20.5km/h[/tex]
[tex]V_4 = 20km/h -0.5km/h = 19.5km/h[/tex]
[tex]V_5 = 20km/h +0.01km/h = 20.01km/h[/tex]
2) For the second case the margin of tolerance for the response is 5%, so if we multiply the given value we would have a response of.
[tex]x = 6.5*2 = 13[/tex]
5% of 13 is 0.65. Therefore, any value that falls within that range will be a correct answer. The value could then be
[tex]x_1 = 13+0.65 = 13.65[/tex]
[tex]x_2 = 13-0.65 =12.35[/tex]
Incorrect values will be
[tex]x_3 = 13+1 = 14[/tex]
[tex]x_4 = 13-1=12[/tex]
You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. The force acting on you that causes you to fall forward is A) the normal force due to your contact with the floor of the bus. B) the force due to static friction between you and the floor of the bus. C) the force of gravity. D) the force due to kinetic friction between you and the floor of the bus. E) No forces were acting on you to cause you to fall.
Answer:
D
Explanation:
Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction.
When a bus suddenly stops, a passenger tends to keep moving forward due to inertia (Newton's first law). The force causing you to fall forward in this instance is the insufficient static friction between you and the floor of the bus, which isn't enough to counteract your forward momentum.
Explanation:The force acting on you that causes you to fall forward when the bus comes to an immediate stop is B) the force due to static friction between you and the floor of the bus. This is because of Newton's first law, also known as the law of inertia, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless acted upon by an external force.
While you're standing in the moving bus, both you and the bus are moving forward. When the bus suddenly stops, your body tends to keep moving forward due to inertia, causing you to fall forward. Here, the static friction between you and the floor of the bus isn't enough to counteract your momentum, leading to your forward fall.
It's important to note that other forces such as gravity and normal force are in effect too. Gravity pulls you downward, and the bus floor exerts an upward normal force to counterbalance it. These two forces are balanced and don't contribute to your forward movement. The force causing the unbalanced motion (falling forward) is the lack of sufficient friction to oppose your inertia.
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A projectile of mass 5 kg is fired with an initial speed of 176 m/s at an angle of 32◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 2 kg and 3 kg . The 3 kg fragment lands on the ground directly below the point of explosion 4.1 s after the explosion. The acceleration due to gravity is 9.81 m/s 2 . Find the magnitude of the velocity of the 2 kg fragment immediatedly after the explosion. Answer in units of m/s.
Answer:
v1 = 377.98 m/s
Explanation:
m = 5 Kg
v0 = 176 m/s
v0x = v0*Cos 32° = 176 m/s*Cos 32° = 149.256 m/s
m1 = 2 Kg
m2 = 3 Kg
t = 4.1 s
g = 9.81 m/s²
Before the explosion
pix = m*v0x = 5 Kg*149.256 m/s = 746.282 Kgm/s
piy = 0
After the explosion
pfx = m1*v1x
knowing that pix = pfx
we have
746.282 = 2*v1x
v1x = 373.14 m/s
v2y = g*t
pfy = m1*v1y + m2*v2y
pfy = 2*v1y + 3*(9.81*4.1)
pfy = 2*v1y + 120.663
knowing that piy = pfy = 0
we have
0 = 2*v1y + 120.663
v1y = 60.33 m/s
Finally we apply
v1 = √(v1x² + v1y²)
v1 = √(373.14² + 60.33²)
v1 = 377.98 m/s
Final answer:
To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we use the principle of conservation of momentum. By setting the momentum before the explosion equal to the momentum after the explosion, we can solve for the velocity of the 2 kg fragment. The magnitude of the velocity is approximately 56.8 m/s.
Explanation:
Projectile Motion
To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we need to use the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion is equal to the total momentum after the explosion.
Before the explosion, the momentum of the projectile can be calculated using the formula:
momentum = mass * velocity
After the explosion, the momentum of the 2 kg fragment can be calculated as:
momentum = mass * velocity
Setting the two equations equal to each other, we can solve for the velocity of the 2 kg fragment.
Plugging in the given values:
Initial velocity of the projectile = 176 m/s
Initial angle = 32°
Mass of the 2 kg fragment = 2 kg
Mass of the 3 kg fragment = 3 kg
Time after explosion = 4.1 s
Acceleration due to gravity = 9.81 m/s^2
Solving the equation, we find that the magnitude of the velocity of the 2 kg fragment immediately after the explosion is approximately 56.8 m/s.
Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are shorted together. (Round the final answers to two decimal places.) With terminals c-d open, Req = Ω. With terminals c-d shorted, Req = Ω.
Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
To find the equivalent resistance Req between terminals a and b, we can follow a series of steps to simplify the circuit. The final equivalent resistance can be calculated by adding the resistors in series. With terminals c and d open, Req is 12.22 ohms, and with terminals c and d shorted, Req is 22.00 ohms.
Explanation:To find the equivalent resistance between terminals a and b, we need to consider the circuit shown in Figure 10.15. To simplify the circuit, we can apply a series of steps to reduce it to a single equivalent resistance. Step 1 involves reducing resistors R3 and R4 in series. Step 2 involves reducing resistors R2 and the equivalent resistance R34 in parallel. Finally, Step 3 involves reducing resistor R1 and the equivalent resistance R234 in series. The final equivalent resistance Req can be calculated by adding the resistors in series.
Following these steps, the equivalent resistance Req between terminals a and b with terminals c and d open is 12.22 ohms. With terminals c and d shorted together, the equivalent resistance Req is 22.00 ohms.
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The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus) is v = 3 cos 4πt V.
(a) Find the power delivered to the element at t = 0.3 s.
(b) Calculate the energy delivered to the element between 0 and 0.6 s.
Answer:
(a). The power delivered to the element is 187.68 mW
(b). The energy delivered to the element is 57.52 mJ.
Explanation:
Given that,
Charge [tex]q=5\sin4\pi t\ mC[/tex]
Voltage [tex]v=3\cos4\pi t\ V[/tex]
Time t = 0.3 sec
We need to calculate the current
Using formula of current
[tex]i(t)=\dfrac{dq}{dt}[/tex]
Put the value of charge
[tex]i(t)=\dfrac{d}{dt}(5\sin4\pi t)[/tex]
[tex]i(t)=5\times4\pi\cos4\pi t[/tex]
[tex]i(t)=20\pi\cos4\pi t[/tex]
(a).We need to calculate the power delivered to the element
Using formula of power
[tex]p(t)=v(t)\times i(t)[/tex]
Put the value into the formula
[tex]p(t)=3\cos4\pi t\times20\pi\cos4\pi t[/tex]
[tex]p(t)=60\pi\times10^{-3}\cos^2(4\pi t)[/tex]
[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})[/tex]
Put the value of t
[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})[/tex]
[tex]p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)[/tex]
[tex]p(t)=187.68\ mW[/tex]
(b). We need to calculate the energy delivered to the element between 0 and 0.6 s
Using formula of energy
[tex]E(t)=\int_{0}^{t}{p(t)dt}[/tex]
Put the value into the formula
[tex]E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}[/tex]
[tex]E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}[/tex]
[tex]E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}[/tex]
[tex]E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)[/tex]
[tex]E(t)=57.52\ mJ[/tex]
Hence, (a). The power delivered to the element is 187.68 mW
(b). The energy delivered to the element is 57.52 mJ.
To solve the problem, the power at a certain moment (t=0.3s) is calculated by substitifying the values of voltage and charge into the power formula. The energy delivered to the element can be calculated by integrating the power function over the period from 0 to 0.6 s.
Explanation:The subject of this question is related to electrical power and energy in a circuit, and it requires a knowledge of trigonometry. The instantaneous power in an electrical circuit is given by the product of charge (q) and voltage (v). Hence we can calculate the power at t = 0.3s by substituting the given values into the power formula.
Power, p = qv = (5 sin 4πt) * (3 cos 4πt) = 15 sin 4πt cos 4πt. At t = 0.3s, power p = 15 sin 4π * 0.3 cos 4π * 0.3 = 15 sin 1.2π cos 1.2π.
For part (b), the energy delivered to the element between 0 and 0.6 s can be obtained from the integral of the power function over the given time interval, which requires integration skills in calculus.
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A thin cylindrical shell and a solid cylinderhave the same mass and radius. The two arereleased side by side and roll down, withoutslipping, from the top of an inclined planethat is 1.3 m above the ground.Find the final linear velocity of the thincylindrical shell.The acceleration of gravity is9.8 m/s2
Answer:
v = 5.05m/s
Explanation:
H = 1.3m
initial velocity = 0
final velocity = v = ?
g =9.8 m/s^2
we apply the conservation of energy; all potential energy is comletely converting to kinetic energy
[tex]mgh = \frac{mv^{2}}{2}[/tex]
mass is same; it cancels out
[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]
v = 5.05m/s
Answer:
the final linear velocity is 3.56931 m/s
Explanation:
the solution is in the attached Word file
A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted to 24.84° to the vertical, what is the index ofrefraction of the syrup solution?(b) Suppose the light is red, with wavelength 632.8 nm in a vacuum.Find its wavelength in the solution.(c) What is its frequencyin the solution?(d) What is its speed in the solution?
Answer:
1.29649
488.08706 nm
[tex]6.14644\times 10^{14}\ Hz[/tex]
231715700.28346 m/s
Explanation:
n denotes refractive index
1 denotes air
2 denotes solution
[tex]\lambda_0[/tex] = 632.8 nm
From Snell's law we have the relation
[tex]n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649[/tex]
Refractive index of the solution is 1.29649
Wavelength is given by
[tex]\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm[/tex]
The wavelength of the solution is 488.08706 nm
Frequency is given by
[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz[/tex]
The frequency is [tex]6.14644\times 10^{14}\ Hz[/tex]
[tex]v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s[/tex]
The speed in the solution is 231715700.28346 m/s
The index of refraction can be found using Snell's law. The wavelength of red light within the solution can be calculated from the index and its frequency remains constant. The speed of light in the solution is the product of its frequency and the calculated wavelength.
Explanation:Solution to the Laser Beam Question
To answer the student's question regarding refraction of a laser beam in a corn syrup solution, we can use Snell's law, which relates the incident angle, refracted angle, and indices of refraction of the two media. We can also calculate the wavelength, frequency, and speed of light within the solution using the principles of wave optics.
(a) According to Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n2 is the index of refraction of the corn syrup solution, θ1 is the incident angle, 90° - 33.0° = 57.0°, and θ2 is the refracted angle, 90° - 24.84° = 65.16°. Since the laser light is passing from air (n1=1) to the solution, we have 1 * sin(57.0°) = n2 * sin(65.16°). Solving for n2 gives us the index of refraction of the syrup solution.
(b) The wavelength of light in the solution is given by λ' = λ0/n, where λ0 is the wavelength in a vacuum, and n is the index of refraction. Inserting the red light's wavelength (632.8 nm) in vacuum and the obtained index of refraction will yield the wavelength in the solution.
(c) The frequency of light does not change when it enters another medium, so we use the vacuum frequency calculated by f = c/λ0, and c is the speed of light in a vacuum.
(d) The speed of light in the solution, v, can be found using the equation v = f * λ', which uses the frequency found in part (c) and the wavelength in the solution from part (b).
Consider n equal positive charged particles each of magnitude Q/n placed symmetrically around a circle of radius a. Calculate the magnitude of the electric field at a point a distance x from the center of the circle and on the line passing through the center and perpendicular to the plane of the circle. (Use any variable or symbol stated above along with the following as necessary: ke.) (b) Now consider a ring of radius a that carries a uniformly distributed positive total charge Q. Recall the calculation of the electric field at point a point a distance x from the center of the ring and on the line passing through the center and perpendicular to the plane of the ring. Explain why the result in part (a) is identical to the result for the ring.
Answer:
To make it easier to Understand, consider the circle to be in the usual 3 dimensional x-z plane (the "horizontal plane") and the point of measurement C to be at a distance p (instead of x to avoid confusion with the x-axis) on the y-axis (the "vertical direction").
The answer for a and b is the same. This is because a the horizontal component of a charged portion of a uniform ring is canceled by the opposite portion of the ring since they are placed at an equal distance from the position in which the electric field is being measured or applied, just as are the opposing point charges described in part a.
Explanation:
A.) Using Coulomb's law of point charges, each charge on the circle would exert a field Ec at C given by:
(1) Ec = Ke * (Q / n) / d²
where:
Ke is Coulomb's constant,
Q / n is the magnitude of the charge, and
d = the distance between the charge and the point of measurement C, with d² = a² + c²
Since the charges are in a circle in the x-z plane, all force components in both the x- and the z-directions are canceled by symmetry; the vertical force (that in the y-direction) is the only component that does not cancel.
Therefore the resultant vector Ecy points up (+y-direction) and has a magnitude of:
(2) Ecy = Eq * sin(theta)
= (Ke * (Q / n) / d²) * (c / d)
Then, summing the forces from all the charges, the magnitude of the total electric field is given by:
(3) Ey = n * Ecy
= n * [ (Ke * (Q / n) / d²) * (c / d) ]
= c * Ke * Q / d^3
B.) The equation is the same. This is because both the x- and z-components (the two planar components) of a charged portion of a uniform ring are canceled by the opposite portion of the ring, since they lie at an equal distance but opposite direction from c.
This is the same way the opposing point charges described in part A behave.
In both scenarios described, because of symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.
Explanation:The magnitude of the electric field in the scenario described is given by Coulomb's law, which states that the electric field E due to a charge Q at a distance r from a point is given by E = ke*(Q/r^2), where ke is Coulomb's constant.
In this case, the magnitude of each charge is Q/n and we have n charges symmetrically distributed along the circle. Because we are calculating the electric field at a point on the line passing through the center of the circle and perpendicular to the plane of the circle, each pair of charges on the circle will produce an electric field that is symmetric and thus its contribution will be canceled out by another pair's contribution. The final resultant electric field will hence be zero.
(b) Now let's consider a ring of radius a that carries a uniformly distributed positive total charge Q. Because the charge is symmetrically distributed, the electric field at any point on the line passing through the center and perpendicular to the plane of the ring will be sum of electric fields due to all small charged particles which make up this ring, and due to symmetry of the distribution, it will add up to zero.
Therefore, the results in part (a) and (b) are identical because in both cases, due to symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.
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Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are 4.04 times as many A nuclei as there are B nuclei. The half-life of species B is 1.37 days. Find the half-life of species A (in days).
Answer:
The half-life of A is 17.1 days.
Explanation:
Hi there!
The half-life of B is 1.73 days.
Let´s write the elapsed time (3 days) in terms of half-lives of B:
1.37 days = 1 half-life B
3 days = (3 days · 1 half-life B / 1.37 days) = 2.19 half-lives B.
After 3 days, the amount of A in terms of B is the following:
A = 4.04 B
The amount of B after 3 days can be expressed in terms of the initial amount of B (B0) and the number of half-lives (n):
B after n half-lives = B0 / 2ⁿ
Then after 2.19 half-lives:
B = B0 /2^(2.19)
In the same way, the amount of A can also be expressed in terms of the initial amount and the number of half-lives:
A = A0 / 2ⁿ
Replacing A and B in the equation:
A = 4.04 B
A0 / 2ⁿ = 4.04 · B0 / 2^(2.19)
Since A0 = B0
A0 / 2ⁿ = 4.04 · A0 / 2^(2.19)
Dividing by A0:
1/2ⁿ = 4.04 / 2^(2.19)
Multipliying by 2ⁿ and dividing by 4.04 / 2^(2.19):
2^(2.19) / 4.04 = 2ⁿ
Apply ln to both sides of the equation:
ln( 2^(2.19) / 4.04) = n ln(2)
n = ln( 2^(2.19) / 4.04) / ln(2)
n = 0.1756
Then, if 3 days is 0.1756 half-lives of A, 1 half-life of A will be:
1 half-life ·(3 days / 0.1756 half-lives) = 17.1 days
The half-life of A is 17.1 days.
Suppose you hear a clap of thunder 16.2 seconds after seeing the associated lightning stroke. The speed of sound waves in air is 343-m/s and the speed of light in air is 3.00 x 10^8-m/s. How far are you from the lightning stroke in both meters and miles? Assume that the light reaches you instantaneously.
Answer:
D = 3.45 mile
Explanation:
given,
time taken by the sound wave, t = 16.2 s
speed of sound, v = 343 m/s
speed of the light = 3 x 10⁸ m/s
distance where lightning stroke
we know,
distance = speed x time
D = 343 x 16.2
D = 5556.6 m
1 m = 0.000621371 mile
5556.6 m = 5556.6 x 0.0006214
D = 3.45 mile
the distance lightning strike is 3.45 mile away from the observer.
To calculate the distance from a lightning stroke: use the time delay between seeing the lightning and hearing the thunder, and multiply it by the speed of sound (343 m/s). In this case, the lightning is approximately 5562.6 meters or about 3.46 miles away.
Explanation:Your distance from a lightning stroke can be calculated by knowing the speed of sound and the time it takes for the sound from the lightning stroke to reach you. You said you hear the thunder sound 16.2 seconds after seeing the lightning. The speed of sound in air is 343 m/s. Therefore, the distance can be calculated using the formula: distance = speed * time.
In this case, distance = 343 m/s * 16.2s = 5562.6 meters or about 5.56 kilometers.
When this is converted to miles (since 1 mile is about 1609.34 meters), it is approximately 3.46 miles away. Therefore, based on the time difference between the flash and the thunder, you are about 5562.6 meters or 3.46 miles away from the lightning strike.
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If A is the amplitude of a mass on an oscillating spring, then in one period the mass travels a distance of ___________.
a. 0
b. A
c. 2A
d. 4A
Answer:
d. 4A
Explanation:
Given that
Amplitude = A
We know that the distance cover by a particle before coming the the rest from the mean point in the oscillation motion is known as amplitude.
The distance cover by particle from 1 - 2 = A
The distance cover by particle from 2 - 1 = A
The distance cover by particle from 1 - 3 = A
The distance cover by particle from 3-4 = A
Therefore the total distance cover by particle = A+A+A+A = 4 A
Therefore the total distance = 4 A
That is why the answer will be 4 A.
d. 4A
The acceleration due to gravity on the moon is about 5.4 ft/s2 . If your weight is 150 lbf on the earth:
What is your mass on the moon, in slugs.
Answer:
4.662 slugs
Explanation:
Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.
If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is
m = W / g = 150 / 32.174 = 4.662 slugs
which is also your mass on the moon.
While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg⋅m² and its rotational kinetic energy is 175 J.
(a) What is the angular velocity of the leg?
(b) What is the velocity of the tip of the punter's shoe if it is 1.05 m from the hip joint?
Answer:
9.6609 rad/s
10.143945 m/s
Explanation:
I = Moment of inertia = 3.75 kgm²
K = Kinetic energy = 175 J
r = Radius = 1.05 m
Kinetic energy is given by
[tex]K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s[/tex]
The angular velocity of the leg is 9.6609 rad/s
Velocity is given by
[tex]v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s[/tex]
The velocity of the tip of the punters shoe is 10.143945 m/s
During launches, rockets often discard unneeded parts. A certainrocket starts from rest on the launch pad and accelerates upward ata steady 3.25 m/s^2. When it is 230 m above the launch pad, it discards a usedfuel canister by simply disconnecting it. Once it is disconnected,the only force acting on the canister is gravity (air resistancecan be ignored).
A) How high is the rocket when thecanister hits the launch pad, assuming that the rocket does notchange its acceleration?
B) What total distance did the canistertravel between its release and its crash onto the launchpad?
Answer:
915.69549 m
306.1968 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 3.25 m/s²
g = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.25\times 230+0^2}\\\Rightarrow v=38.665\ m/s[/tex]
Height is given by
[tex]h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{38.665^2}{2\times 9.81}\\\Rightarrow h=76.1968\ m[/tex]
Total distance the canister has to travel is 230+76.1968 = 306.1968 m
Time to reach the max height
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{g}\\\Rightarrow t=\dfrac{0-38.665}{-9.81}\\\Rightarrow t=3.9413\ s[/tex]
Time taken to reach the ground is
[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 306.1968=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{306.1968\times 2}{9.81}}\\\Rightarrow t=7.9\ s[/tex]
Total time taken would be 3.9413+7.9 = 11.8413 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=38.665\times 11.8413+\dfrac{1}{2}\times 3.25\times 11.8413^2\\\Rightarrow s=685.69549\ m[/tex]
The rocket is 685.69549+230 = 915.69549 m from the launchpad.
The total distance is 230+76.1968 = 306.1968 m
The gravitational force exerted on a solid object is 5.00N. When the object is suspended from a spring scale and submerged completely in water the scale reads 3.50N. Find the density of the object.
Answer:
3333.33 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a body to its volume.
The unit of density is kg/m³.
From Archimedes principle,
R.d = W/U = D/D'
Where R.d = relative density, W = weight of the object in air, u = upthrust in water, D = Density of the object, D' = Density of water.
W/U = D/D'
making D the subject of the equation
D = D'(W/U).......................... Equation 1
Given: W = 5.0 N, U = 5.0 -3.5 = 1.5 N, D' = 1000 kg/m³
Note: U = lost in weight = weight in air - weight in water
Substitute into equation 1
D = 1000(5/1.5)
D = 3333.33 kg/m³
Thus the density of the object = 3333.33 kg/m³
The first step is to calculate the buoyant force that the water exerts on the solid object. The buoyant force can be found by subtracting the scale reading from the gravitational force, or 5.00N - 3.50N = 1.50N.
Next, we find the volume of the water displaced by the solid. The buoyant force is equal to the weight of the fluid displaced, so we can use the formula F = ρf * V * g, where ρf is the fluid density, V is the water volume displaced, g is the acceleration due to gravity and F is the buoyant force.
We're given that the density of water is 9.8 * 1000 N/m³ and we've just calculated the buoyant force. Thus, we have 1.50N = 9.8 * 1000 * V * 9.81, which simplifies to V = 1.5 / (9.8 * 1000) = 0.0001530612244897959 m³.
Now we calculate the object's mass using the equation, m = F / g, where F is gravitational force and g is acceleration due to gravity. Substituting the given values, we get m = 5.00 / 9.81 = 0.509683995922528 kg.
Finally, we find the density of the object using the formula ρ = m / V, where m is the mass and V is the volume. Substituting the values calculated earlier, we get ρ = 0.509683995922528 / 0.0001530612244897959 = 3329.935440027183 kg/m³.
Therefore, the density of the object is approximately 3330 kg/m³.
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A piece of tape is pulled from a spool and lowered toward a 190-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance
Final answer:
The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.
Explanation:
When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:
F = k * (Q1 * Q2) / r^2
where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
We know that the gravitational force is equal to the weight of the paper:
Fg = m * g
where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).
At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:
F = Fg
k * (Q1 * Q2) / r^2 = m * g
We can solve this equation for the magnitude of the charge Q1 or Q2:
Q1 * Q2 = (m * g * r^2) / k
Substituting the known values, we get:
Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)
Q1 * Q2 = 1.383 x 10^-12 C^2
To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:
Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C
The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.
You slide a coin across the floor, and observe that it slows down and eventually stops. A sensitive thermometer shows that its temperature increased. What can we conclude? (Select all that apply.)
When you slide a coin across the floor, it slows down and eventually stops due to the force of friction. The friction converts the coin's kinetic energy into thermal energy, resulting in an increase in temperature.
Explanation:When you slide a coin across the floor, it eventually slows down and stops due to the force of friction acting on it. Friction is a force that opposes the motion of objects in contact, and it causes the coin to lose kinetic energy, slowing it down. As the coin slows down, its kinetic energy is converted into thermal energy, increasing the temperature of the coin and the surface it slides on. This is why a sensitive thermometer shows an increase in temperature when the coin slides.
A free electron and a free proton are released in identical electric fields.
(i) How do the magnitudes of the electric force exerted on the two particles compare?
It is millions of times greater for the electron.
It is thousands of times greater for the electron.
They are equal.
It is thousands of times smaller for the electron.
It is millions of times smaller for the electron.
(ii) Compare the magnitudes of their accelerations.
It is millions of times greater for the electron.
It is thousands of times greater for the electron.
They are equal.It is thousands of times smaller for the electron.
It is millions of times smaller for the electron.
Answer:
i) They are equal . ii) It is thousands of times greater for the electron.
Explanation:
i) By definition, the electric field is the electric force per unit charge. If the field is the same, the force will depend on the value of the charge under the influence of the field.
As the magnitude of the charge of the electron and the proton are the same, we conclude that the electric force on both must be equal in magnitude.
ii) The acceleration on both particles must meet the Newton´s 2nd Law, so, if the forces are equal in magnitude (neglecting any other external interaction), the acceleration will only depend on the mass of both particles, according this general expression:
a = F/m
As the mass of the electron is approximately two thousands times smaller than the proton´s, it concludes that the acceleration on the electron must be thousands of times greater for the electron.
The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. The magnitudes of their accelerations are equal.
Explanation:(i) The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. This is because the electron has a much smaller mass compared to the proton, and the electric force depends on the charge and mass of the particle.
(ii) The magnitudes of their accelerations are equal. The acceleration of a particle in an electric field depends only on the magnitude of the electric field and the mass of the particle. Since both the electron and proton experience the same electric field, their accelerations will be the same.
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A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate about their center of mass as if they were a dumbbell connected by a massless rod.What is the orbital frequency for an electron and a positron 1.50 apart?
Explanation:
According to the energy conservation,
[tex]F_{centripetal} = F_{electric}[/tex]
[tex]\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}[/tex]
[tex]v^{2} = \frac{kq^{2}r}{d^{2}m}[/tex]
= [tex]\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}[/tex]
= [tex]8.430 \times 10^{10} m^{2}/s^{2}[/tex]
v = [tex]\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}[/tex]
= [tex]2.903 \times 10^{5} m/s[/tex]
Formula for distance from the orbit is as follows.
S = [tex]2 \pi r[/tex]
= [tex]2 \times 3.14 \times 0.75 \times 10^{-9} m[/tex]
= [tex]4.71 \times 10^{-9} m[/tex]
Now, relation between time and distance is as follows.
T = [tex]\frac{S}{v}[/tex]
[tex]\frac{1}{f} = \frac{S}{v}[/tex]
or, f = [tex]\frac{v}{S}[/tex]
= [tex]\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}[/tex]
= [tex]6.164 \times 10^{13} Hz[/tex]
Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is [tex]6.164 \times 10^{13} Hz[/tex].
The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire
Final answer:
The question pertains to the physics discipline, dealing with calculating the tension in supporting cables using principles of static equilibrium and considering the load's weight and the angles of attachment.
Explanation:
The question involves determining the tension in cables that support a structure or an object. This type of problem is common in physics, specifically in the area of mechanics, and it involves understanding forces and how they are distributed within a system. When multiple cables support a load, tensions in each cable can be found using principles of static equilibrium, where the sum of forces in each direction (horizontal and vertical) equals zero. In addition to the weight of the supported object, angles at which cables are attached can play a crucial role in determining individual tensions.
A stream of air flowing at 20 liters/min with P = 0.20 MPa and T = 400 K is mixed with a stream ofmethane flowing at 5 liters/min with P = 0.20 MPa and T = 300 K. The combined gas stream exitingthe mixer is at P = 0.10 MPa and T = 370 K. What is the volumetric flow rate and composition of theexiting gas stream?
Answer:
a) the mole fraction of air in the exiting stream is Xa=0.25 (25%) and for methane Xm=0.75 (75%)
b) the volumetric flow rate is 49.33 L/s
Explanation:
Assuming ideal gas behaviour, then
for air
Pa*Va=Na*R*Ta
for methane
Pm*Vm=Nm*R*Tm
dividing both equations
(Pa/Pm)*(Va/Vm)= (Na/Nm)*(Ta/Tm)
Na/Nm = (Pa/Pm)*(Va/Vm) * (Tm/Ta) = (0.2/0.2)*(20/5)*(300/400) = 1*4*3/4 = 3
Na=3*Nm
therefore the moles of gas of the outflowing stream are (assuming that the methane does not react with the air):
Ng= Na+Nm = 4*Na
the mole fraction of A is
Xa= Na/Ng= Na/(4*Na) = 1/4 (25%)
and
Xm= 1-Xa = 3/4 (75%)
also for the exiting gas
Pg*Vg=Ng*R*Tg = Na*R*Tg + Nm*R*Tg = Pa*Va * (Tg/Ta) + Pm*Vm * (Tg/Tm)
Vg = Va * (Pa/Pg)*(Tg/Ta) + Vm *(Pm/Pg)* (Tg/Tm)
Vg = 20 L/min * (0.2/0.1)*(370/400) + 5 L/min * (0.2/0.1)*(370/300) = 49.33 L/s
In a certain region of space, the electric field is zero. From this fact, what can you conclude about the electric potential in this region? (a) It is zero. (b) It does not vary with position. (c) It is positive. (d) It is negative. (e) None of those answers is necessarily true.
Answer:
It is constant.
Explanation:
As we know that the electric field is a change in the electric potential so,
E = -ΔV
E = 0 ∵ V = constant
As we know that the electric field is zero. Which means that the electric potential is constant and it's not changing which results in the zero electric field.
In a region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero.
Explanation:In a certain region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero. This is because electric potential is determined by the distribution of charges and their distances from the region in consideration. While the electric field measures the force experienced by a charge, the electric potential is related to the work required to move a charge in a region with an electric field.
Therefore, the correct answer is (e) None of those answers is necessarily true.
If you wanted to move an electron from the positive to the negative terminal of the battery, how much work W would you need to do on the electron
Answer:
W = qV
Explanation:
Let V be the potential difference of the battery, and q be the charge on the electron.
The work done in moving a charge through a potential difference V =?
Work = force * distance
Work = F.r
F = qE
But E = V / r
F = q. V / r
Work ( W ) = (qV/ r ) * r
Work = qV
Therefore the force required to move a charge through a potential difference V = qV
The amount of work needed to move an electron from the positive to the negative terminal of a battery using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery is 1.92 x 10-18 joules.
Explanation:In order to move an electron from the positive to the negative terminal of a battery, you would need to do work on the electron.
The amount of work, W, can be calculated using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery.
The charge of an electron is approximately 1.6 x 10-19 coulombs.
For example, if the voltage of the battery is 12 volts, the work done on the electron would be:
W = ([tex]1.6[/tex]×[tex]10^{-19}[/tex][tex]C[/tex]) × ([tex]12[/tex] [tex]v[/tex]) = [tex]1.92[/tex]×[tex]10^{-18}[/tex] joules.
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The column has a mass of 600 lb/ft and a force of 50 kips applied at the top. The diagonal brace has a 10 kip axial force that can be either tension or compression (note that in seismic and wind loading on buildings, braces experience both tension and compression). Assume all elements are pin-connected at their ends, including the two supports.
Answer: the maximum compression force = 59.47 kip
the minimum compression force = 43.33 kip
Explanation:
In this we required to do force balance in order to get maximum and minimum compression force in the column.
the picture below explains the steps to solve the question with diagrams to ease understanding.
The bellow of a territorial bull hippopotamus is measured at 106 dB above the threshold of hearing. What is the sound intensity? Hint: The threshold of human hearing is I0 = 1.00 x 10-12 W/m2.
Answer:
0.03981 W/m²
Explanation:
I = Sound intensity
[tex]\beta[/tex] = Intensity level = 106 dB
[tex]I_0[/tex] = Threshold of human hearing = [tex]10^{-12}\ W/m^2[/tex]
Intensity of sound is given by
[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow 106=10log\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{106}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{106}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow I=10^{\dfrac{106}{10}}\times 10^{-12}\\\Rightarrow I=10^{-1.4}\\\Rightarrow I=0.03981\ W/m^2[/tex]
The sound intensity is 0.03981 W/m²
The sound intensity is approximately 1.00 x 10^-4 W/m^2.
Explanation:The sound intensity can be calculated using the formula:
I = I0 * 10^(dB/10)
where I is the sound intensity, I0 is the threshold of human hearing (given as 1.00 x 10^-12 W/m^2), and dB is the decibel level above the threshold. In this case, the decibel level is 106 dB. Plugging in the given values into the formula:
I = (1.00 x 10^-12 W/m^2) * 10^(106/10)
Simplifying the equation:
I ≈ 1.00 x 10^-4 W/m^2
The sound intensity is approximately 1.00 x 10^-4 W/m^2.
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