Answer:
![\Delta K=0.07969 J - 0.0849 J = -0.00521 J](https://tex.z-dn.net/?f=%5CDelta%20K%3D0.07969%20J%20-%200.0849%20J%20%3D%20-0.00521%20J)
Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being
and
the masses of pucks a and b respectively, the initial momentum of the system is
![M_1=m_av_a+m_bv_b](https://tex.z-dn.net/?f=M_1%3Dm_av_a%2Bm_bv_b)
Since b is initially at rest
![M_1=m_av_a](https://tex.z-dn.net/?f=M_1%3Dm_av_a)
After the collision and being
and
the respective velocities, the total momentum is
![M_2=m_av'_a+m_bv'_b](https://tex.z-dn.net/?f=M_2%3Dm_av%27_a%2Bm_bv%27_b)
Both momentums are equal, thus
Solving for ![v_a](https://tex.z-dn.net/?f=v_a)
![v_a=\frac{m_av'_a+m_bv'_b}{m_a}](https://tex.z-dn.net/?f=v_a%3D%5Cfrac%7Bm_av%27_a%2Bm_bv%27_b%7D%7Bm_a%7D)
![v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}](https://tex.z-dn.net/?f=v_a%3D%5Cfrac%7B0.254Kg%5Ctimes%20%28-0.123%20m%2Fs%29%2B0.367Kg%20%280.651m%2Fs%29%7D%7B0.254Kg%7D)
The initial kinetic energy can be found as (provided puck b is at rest)
![K_1=\frac{1}{2}m_av_a^2](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B1%7D%7B2%7Dm_av_a%5E2)
![K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B1%7D%7B2%7D%280.254Kg%29%20%280.8176m%2Fs%29%5E2%3D0.0849%20J)
The final kinetic energy is
![K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B1%7D%7B2%7Dm_av_a%27%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_bv_b%27%5E2)
![K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B1%7D%7B2%7D0.254Kg%20%28-0.123m%2Fs%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D0.367Kg%20%280.651m%2Fs%29%5E2%3D0.07969%20J)
The change of kinetic energy is
![\Delta K=0.07969 J - 0.0849 J = -0.00521 J](https://tex.z-dn.net/?f=%5CDelta%20K%3D0.07969%20J%20-%200.0849%20J%20%3D%20-0.00521%20J)
Answer:
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg.
Answer:
The weight of an object at the Earth's South Pole is slightly more than its weight at the Equator because the polar radius of the Earth is slightly less than the equatorial radius. Though the mass of an object remains constant, its weight varies according to its location.
Explanation:
The best and most correct answer among the choices provided by the question is drag<span>.
</span><span>Drag </span>kind of friction exists between solid objects moving in water.
Hope my answer would be a great help for you.
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