False. That's exactly how scientific theories begin
B) The wall pulls with a force of 250 N against the force
Explanation:
We can answer this question by using Newton's third law of motion, which states that:
<em>"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"</em>
In this situation, the student is applying a force of 250 N on the wall. We can identify this force as the action. Therefore, this means (according to Newton's third law) that the wall will also apply a force (the reaction) of equal magnitude (so, 250 N), in the opposite direction to the original force.
Therefore, the correct statement is
B) The wall pulls with a force of 250 N against the force
Learn more about Newton's third law:
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Answer:
25 N in the direction of your push
Explanation:
The friction force acts against your pushing force....it reduces the net amount of force you are exerting
Net force = 40 - 15 = 25 N
Answer:
Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k
Explanation:
Given data
Th=619 K
Tc= 455K
We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.
we know that the change in entropy is given as
ΔStotal=ΔShot+ΔScold .
ΔShot=−Q/Th=−4000J/619K=−6.46J/K
For the cold reservoir,
ΔScold=Q/Tc=4000J/455K=8.79J/K
therefore the total is
ΔStotal=ΔShot+ΔScold
=(−6.46+8.79)J/K
=2.33j/k