#9 because you cant get more work out than you put in and you cant have a machine that is over 100% efficient
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Answer:
22.5 [m] is the distance from the person to the second post.
Explanation:
We can solve this problem graphically, first place the position of the person at the origin of x-y coordinates.
Then we locate the first post (1), tracing a line (or vector) with a length of 52 [m] from the origin of coordinates and with an angle of 37 ° north of East.
And for finding the distance from the second post (2) with respect to the person, we have to draw a line that has a length of 68 [m] and this line must be crossed with the y-axis and in the negative direction, in this way we can find the distance from the post with respect to the person.
Note: we can use set square and protractor to solve this problem
In the attached image we can find the graphic solution.
By this method of construction we can find the distance of post number 2 from the person and this is 22.55 [m] to the South
Galileo discovered during his inclined-plane experiments that a ball rolling down an incline and onto a horizontal surface would roll indefinitely.
Answer:
+16 J
Explanation:
We can solve the problem by using the 1st law of thermodynamics:

where
is the change of the internal energy of the system
Q is the heat (positive if supplied to the system, negative if dissipated by the system)
W is the work done (positive if done by the system, negative if done by the surroundings on the system)
In this case we have:
Q = -12 J is the heat dissipated by the system
W = -28 J is the work done ON the system
Substituting into the equation, we find the change in internal energy of the system:

The work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻² would be
<h3>What is work done?</h3>
The total amount of energy transferred when a force is applied to move an object through some distance.
The work done is the multiplication of applied force with displacement.
Work Done = Force × Displacement
As given in the problem we have to find the work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻²,
Lets us first calculate the displacement of the elevator in 2 seconds if it is accelerating upwards from rest at a rate of 2 ms⁻².
s = ut + 0.5at²
= 0 + 0.5×2×2²
= 4 meters
The work done on the elevator = mgh
=50×9.81×4
=1962 joules
Thus, the work done on the elevator would be 1962 joules.
To learn more about work done, refer to the link;
brainly.com/question/13662169
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