Wherever a river loses energy
As a substance is changing from a liquid to a gas, the distance between its molecules increases, and the temperature of the system remains the same.
Option A
<u>Explanation:</u>
The external energy required to change from one state to another is mostly considered as temperature. So on increase in temperature, the solid changes to liquid and the liquid changes to gases. But the temperature remains constant in the system after changing the phase.
This is because when the temperature is increased on a liquid system, the rise in temperature is utilized for breaking the bonds and thus the molecules will be distanced from each other. If we consider liquid - gas phase transition, the gas molecules are farther distanced compared to liquid molecules.
So the rise in temperature is utilized for breaking the bonds and also to provide the kinetic energy to the gas molecules as they are tend to move more freely compared to liquid. Thus, the distance between the molecules increases, and the temperature of the system remains the same on changing from liquid to gas.
Work done = Force X Distance
3 430 000J = Force X 14m
Force = 3 430 000J / 14m
= 245 000 N
Hope this helps!
The best demonstration that applies to Newton's Third Law of motion would be D) When you walk your foot pushes down on the ground while the ground pushes back on your foot.
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Newton's Third Law states that for every action, there is an equal and opposite reaction. This is actually explains that forces come in pairs and forces are an interaction between two objects. As per the correct option given in the question explains Newton’s Third Law.
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When you walk your foot (say object A) pushes down on the ground while the ground (say object Q) pushes back on your foot with the same force but in the opposite direction.
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, 
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point
, the speakers are out of phase and so the path difference is 
Therefore,




Thus the frequency is :


Hz