Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
![v=gt](https://tex.z-dn.net/?f=v%3Dgt)
And the distance traveled downwards is:
![\displaystyle y=\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7Bgt%5E2%7D%7B2%7D)
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
![\displaystyle t=\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Replacing into the first equation:
![\displaystyle v=g\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3Dg%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Rationalizing:
![\displaystyle v=\sqrt{2gy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Csqrt%7B2gy%7D)
Let's call v1 the final speed of the package dropped from a height H. Thus:
![\displaystyle v_1=\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_1%3D%5Csqrt%7B2gH%7D)
Let v2 be the final speed of the package dropped from a height 4H. Thus:
![\displaystyle v_2=\sqrt{2g(4H)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Csqrt%7B2g%284H%29%7D)
Taking out the square root of 4:
![\displaystyle v_2=2\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D2%5Csqrt%7B2gH%7D)
Dividing v2/v1 we can compare the final speeds:
![\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D%5Cfrac%7B2%5Csqrt%7B2gH%7D%7D%7B%5Csqrt%7B2gH%7D%7D)
Simplifying:
![\displaystyle v_2/v_1=2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D2)
The final speed of the second package is twice as much as the final speed of the first package.
Part a)
Magnitude of electric field is given by force per unit charge
![E = \frac{F}{q}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BF%7D%7Bq%7D)
![E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B4.3%20%2A%2010%5E%7B-6%7D%7D%7B2%20%2A%2010%5E%7B-9%7D%7D)
![E = 2150 N/C](https://tex.z-dn.net/?f=E%20%3D%202150%20N%2FC)
Part b)
Electrostatic force on the proton is given as
F = qE
![F = 1.6 * 10^{-19} * 2150](https://tex.z-dn.net/?f=F%20%3D%201.6%20%2A%2010%5E%7B-19%7D%20%2A%202150)
![F = 3.44 * 10^{-16} N](https://tex.z-dn.net/?f=F%20%3D%203.44%20%2A%2010%5E%7B-16%7D%20N)
PART C)
Gravitational force is given by
![F_g = mg](https://tex.z-dn.net/?f=F_g%20%3D%20mg)
![F_g = 1.6 * 10^{-27}*9.8](https://tex.z-dn.net/?f=F_g%20%3D%201.6%20%2A%2010%5E%7B-27%7D%2A9.8)
![F_g = 1.57 * 10^{-26} N](https://tex.z-dn.net/?f=F_g%20%3D%201.57%20%2A%2010%5E%7B-26%7D%20N)
PART d)
Ratio of electric force to weight
![\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%20%5Cfrac%7B3.44%20%2A%2010%5E%7B-16%7D%7D%7B1.57%2A10%5E%7B-26%7D%7D)
![\frac{F_e}{F_g} = 2.2 * 10^{10}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%202.2%20%2A%2010%5E%7B10%7D)
The answer is A.) igneous rocks
Okay, first off, the formula for Kinetic Energy is:
<em>KE = 1/2(m)(v)^2</em>
<em>m = mass</em>
<em>v = velcoity (m/s)</em>
Using this formula, we can then calculate the kinetic energy in each scenario:
1) KE = 1/2(100)(5)^2 = 1,250 J
2) KE = 1/2(1000)(5)^2 = 12,500 J
3) KE = 1/2(10)(5)^2 = 125 J
4) KE = 1/2(100)(5)^2 = 1,250 J