Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=![4.67\times 10^{-3} kg](https://tex.z-dn.net/?f=4.67%5Ctimes%2010%5E%7B-3%7D%20kg)
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,![m_1=1177g=1.177kg](https://tex.z-dn.net/?f=m_1%3D1177g%3D1.177kg)
Mass of block 2,![m_2=1626 g=1.626 kg](https://tex.z-dn.net/?f=m_2%3D1626%20g%3D1.626%20kg)
Velocity of block 1,![v_1=0.681m/s](https://tex.z-dn.net/?f=v_1%3D0.681m%2Fs)
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
![mv=m_1v_1+(m+m_2)v_2](https://tex.z-dn.net/?f=mv%3Dm_1v_1%2B%28m%2Bm_2%29v_2)
![4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2](https://tex.z-dn.net/?f=4.67%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20357%2B1.177%280%29%2B1.626%280%29%3D1.177%5Ctimes%200.681%2B%284.67%5Ctimes%2010%5E%7B-3%7D%2B1.626%29v_2)
![1.66719=0.801537+1.63067v_2](https://tex.z-dn.net/?f=1.66719%3D0.801537%2B1.63067v_2)
![1.66719-0.801537=1.63067v_2](https://tex.z-dn.net/?f=1.66719-0.801537%3D1.63067v_2)
![0.865653=1.63067v_2](https://tex.z-dn.net/?f=0.865653%3D1.63067v_2)
![v_2=\frac{0.865653}{1.63067}](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B0.865653%7D%7B1.63067%7D)
![v_2=0.531m/s](https://tex.z-dn.net/?f=v_2%3D0.531m%2Fs)
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
![K_i=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K_i%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)](https://tex.z-dn.net/?f=k_i%3D%5Cfrac%7B1%7D%7B2%7D%284.67%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20%28357%29%5E2%29)
![k_i=297.59 J](https://tex.z-dn.net/?f=k_i%3D297.59%20J)
Final kinetic energy after collision
![K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7B2%7Dm_1v%5E2_1%2B%5Cfrac%7B1%7D%7B2%7D%28m%2Bm_2%29v%5E2_2)
![K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7B2%7D%281.177%29%280.681%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%284.67%5Ctimes%2010%5E%7B-3%7D%2B1.626%29%280.531%29%5E2)
![K_f=0.5028 J](https://tex.z-dn.net/?f=K_f%3D0.5028%20J)
Now, he ratio of the total kinetic energy after the collision to that before the collision
=![\frac{k_f}{k_i}=\frac{0.5028}{297.59}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_f%7D%7Bk_i%7D%3D%5Cfrac%7B0.5028%7D%7B297.59%7D)
=0.00169
Answer:
NO
Explanation:
No, a machine cannot be 100% efficient. This is due to the movement of the moving parts siding against each other and causing friction. This friction is the one that creates heat and causes wear and tear between moving ports f the machine hence making the machine to decrease in efficiency with time
The source and the observer are moving towards each other. The observer is moving toward the source. The source is moving away from the observer
Upwarped mountain is created when rock layers are pushed up by forces inside the Earth.
Answer:A
Explanation: number that shows the total atomic mass of the substance