Person X pushes twice as hard against a stationary brick wall as person Y. Which one of the following statements is correct?
1 answer:
Answer:
D) Both do zero work
Explanation:
The work done by a force is given by:
where
F is the force applied
d is the displacement
is the angle between the direction of the force and the displacement
From the formula, we notice that work is done online when the displacement is non-zero, so when the object is moving.
In this problem, the wall is stationary: this means that the displacement is zero, d = 0, so no work is done.
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Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =
=
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .
Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:
It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:
The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.
So, let's isolate :
The Cp of the air at 27ºC is 1007 (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:
Surface area:
Then, the heat Q is:
Finally, find the exit temperature:
=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.