Two electric motors drive two elevators of equal mass in a three-story building 10 meters tall. Each elevator has a mass of 1,00
1 answer:
The work done by each motor is c)
Explanation:
The work done when lifting an object is equal to the change in gravitational potential energy of the object:
where
m is the mass of the object
g is the acceleration of gravity
is the change in height of the object
In this problem, each motor is lifting one elevator, so the mass of each elevator is
m = 1000 kg
and we have
Therefore, the work done by each motor is
Learn more about work:
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Answer:
The average acceleration of the car is 2.143 meters per square second.
Explanation:
Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Time, measured in seconds.
Now, we clear acceleration within expression:
Initial and final velocities are now converted from kilometers per hour into meters per second:
If we know that , then, the average acceleration of the car is:
The average acceleration of the car is 2.143 meters per square second.
The question is not complete and the complete question is;
An insulated piston-cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-Vdiagram with respect to saturation lines.
Answer:
Voltage of source = 175.33 V
I've attached an image of the P-Vdiagram of the process with respect to saturation lines.
Explanation:
To solve this question, we'll adopt the following assumptions.
- The kinetic and potential energy changes are negligible
-The thermal energy stored in the cylinder is negligible
-The cylinder is well insulated and thus heat transfer is negligible.
If the contents of the cylinder is taken as the system, the energy balance is;
Ein - Eout = ΔEsystem
Thus, [W(e-internal) + W(pw-internal)] - Wout = ΔU
Thus, IVΔt + W(pw-internal) = ΔH
= m(h2 - +1)
Now,looking at steam table A-5 attached to this answer, at P =175, x1 is calculated to be 0; dryness fraction, x2 = 0.5; vf = 0.001057m³/kg and hf = 487.01KJ/Kg and hg at saturation vapour = 2700.2
So, from this question v1=vf = 0.001057m³/kg and h1 = hf = 487.01KJ/Kg.
Also, h2 = hf + x2 (hg - hf) = 487.1 + 0.5(2700. 2 - 487.1) = 1593.65 KJ/Kg
The mass of the water is defined as;
M=V/v1
Volume(V) = 4L from the question. Since v1 is in m³/kg, let's convert V to m³. So V=4/1000 = 0.004
So, M = 0.004/0.001057 = 3.784kg
Now the formula for the voltage is ;
V = [M(h2 - h1) - W(pw-internal)] /(IΔt)
Δt=45minutes. We convert it to seconds to get 45x60 =2700 seconds
V = [[3.784 x (1593.65 - 487.1) - 400]/(8 x2700)] x 1000 = 175.33 V
