Newton's first law of motion states that an object at rest tends to stay at rest, while an object in motion tends to stay in motion unless an external force acts upon it. This law appears in basketball when the player is shooting the ball. When the player is holding the ball, the ball is at rest but when a player shoots the ball, they use force to throw the ball in the hoop.
Answer:
Average acceleration on first part of the chunk is given as
![a_1 = 13.125 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%2013.125%20m%2Fs%5E2)
Average acceleration on second part of the chunk is given as
![a_2 = -13.125 m/s^2](https://tex.z-dn.net/?f=a_2%20%3D%20-13.125%20m%2Fs%5E2)
Explanation:
By momentum conservation along x direction we will have
![mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2](https://tex.z-dn.net/?f=mv_i%20%3D%20%5Cfrac%7Bm%7D%7B2%7Dv_1%20%2B%20%5Cfrac%7Bm%7D%7B2%7Dv_2)
so we have
![v_1 + v_2 = 2v](https://tex.z-dn.net/?f=v_1%20%2B%20v_2%20%3D%202v)
![v_1 + v_2 = 4.68](https://tex.z-dn.net/?f=v_1%20%2B%20v_2%20%3D%204.68)
also by energy conservation
![\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm%7D%7B2%7D%29v_1%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm%7D%7B2%7D%29v_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%2017%20J)
![\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7Dm%28v_1%5E2%20%2B%20v_2%5E2%29%20-%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%2017%20)
![(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)](https://tex.z-dn.net/?f=%28v_1%5E2%20%2B%20v_2%5E2%29%20-%202v%5E2%20%3D%20%5Cfrac%7B4%7D%7B7.7%7D%2817%29)
![(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83](https://tex.z-dn.net/?f=%284.68%20-%20v_2%29%5E2%20%2B%20v_2%5E2%20-%202v%5E2%20%3D%208.83)
![21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83](https://tex.z-dn.net/?f=21.9%20%2B%202v_2%5E2%20-%209.36%20v_2%20-%2010.95%20%3D%208.83)
![2v_2^2 - 9.36v_2 + 2.12 = 0](https://tex.z-dn.net/?f=2v_2%5E2%20-%209.36v_2%20%2B%202.12%20%3D%200)
by solving above equation we will have
![v_1 = 4.44 m/s](https://tex.z-dn.net/?f=v_1%20%3D%204.44%20m%2Fs)
![v_2 = 0.24 m/s](https://tex.z-dn.net/?f=v_2%20%3D%200.24%20m%2Fs)
Average acceleration on first part of the chunk is given as
![a_1 = \frac{4.44 - 2.34}{0.16}](https://tex.z-dn.net/?f=a_1%20%3D%20%5Cfrac%7B4.44%20-%202.34%7D%7B0.16%7D)
![a_1 = 13.125 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%2013.125%20m%2Fs%5E2)
Average acceleration on second part of the chunk is given as
![a_2 = \frac{0.24 - 2.34}{0.16}](https://tex.z-dn.net/?f=a_2%20%3D%20%5Cfrac%7B0.24%20-%202.34%7D%7B0.16%7D)
![a_2 = -13.125 m/s^2](https://tex.z-dn.net/?f=a_2%20%3D%20-13.125%20m%2Fs%5E2)
Answer:
![E = 2.9775\times10^5 J](https://tex.z-dn.net/?f=E%20%3D%202.9775%5Ctimes10%5E5%20J)
Explanation:
Given: The latent heat of fusion for Aluminum is ![L = 3.97\times10^5 J/Kg](https://tex.z-dn.net/?f=L%20%3D%203.97%5Ctimes10%5E5%20%20J%2FKg)
mass to be malted m = 0.75 Kg
Energy require to melt E = mL
![E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J](https://tex.z-dn.net/?f=E%20%3D%203.97%5Ctimes10%5E5%5Ctimes0.75%20%3D%202.9775%5Ctimes10%5E5%20J)
Therefore, energy required to melt 0.75 Kg aluminum
![E = 2.9775\times10^5 J](https://tex.z-dn.net/?f=E%20%3D%202.9775%5Ctimes10%5E5%20J)
A). Convection is heating the soup in the pot.
When you stick the spoon into the hot soup,
conduction heats the spoon all the way up to the end.
b). Water conducts heat a little bit.
But convection is much more responsible for the
uniform distribution of temperature in the kiddie pool.
c). The heat from the metal bench conducts directly
to the buttus epidermis when you sit on it.
d). You feel the heat on your face ... but not on the back of your
neck ... on account of radiation from the fire and the hot grill.