When you hit a hammer on the head of a nail, all the momentum associated with hammer is transferred into nail. And due to this, the surface of nail gets deformed or if you touch it you can feel the rise in temperature. When you hit a hammer on nail, hammer exerts force on nail and by Newton's 3rd law, nail will also exert a force on the hammer perpendicular to the surface of the hammer of same magnitude but its direction will be reversed. Thus, only 2 forces acts in this interaction and they are action-reaction pairs.
Answer:
![-4.9\cdot 10^5 m/s^2](https://tex.z-dn.net/?f=-4.9%5Ccdot%2010%5E5%20m%2Fs%5E2)
Explanation:
The motion of the bullet is a uniformly accelerated motion, therefore we can find its acceleration by using the following suvat equation:
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the bullet in this problem:
u = 350 m/s is the initial velocity of the bullet
v = 0 is the final velocity (the bullet comes to a stop)
s = 0.125 m is the stopping distance of the bullet
Therefore, by solving the equation for a, we find its acceleration:
![a=\frac{v^2-u^2}{2s}=\frac{0^2-350^2}{2(0.125)}=-4.9\cdot 10^5 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2s%7D%3D%5Cfrac%7B0%5E2-350%5E2%7D%7B2%280.125%29%7D%3D-4.9%5Ccdot%2010%5E5%20m%2Fs%5E2)
And the negative sign tells that the direction of the acceleration is opposite to that of the velocity.
The answer to this problem can be given through energy conservation as well as Newton's first law.
Newton's first law states that as long as there is no force exerted on a body, its movement will be constant or at rest. In this way the amount of linear movement of the satellite before the explosion will be equal to the sum of the movement generated in the pieces after the explosion. In the absence of external forces but the preservation of the internal force as the only one to act, these will not change the total momentum of the system.
The ramp does 480J of useful work with an efficiency of 80% .
<h3>What is efficiency of work done ?</h3>
- Efficiency is the ratio of the useful energy released by a system to the input energy .
- Mathematically, efficiency of energy = out put energy/ input energy
<h3>
What is the useful work done by the ramp having efficiency 80% and an input work done 600J?</h3>
- The efficiency =output work done/ input work done
- 80% =output work done/ 600J
- output work done =( 80×600)/100
=480J
Thus, we can conclude that the useful work done by the ramp is 480J.
Learn more about efficiency of energy here:
brainly.com/question/14280607
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In physics we refer to how heavy the object is as the mass.
Fg = mass x acceleration due to gravity
Force is directly proportional to mass, thus if force of gravity increases mass also increases.