Answer:
Explanation:
From the given information:
Distance 
Speed of the comet 
At distance 
where;
mass of the sun = 

To find the speed
:
Using the formula:

![E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}](https://tex.z-dn.net/?f=E_f%20%3D%20E_i%20%2B%200%20%5C%5C%20%5C%5C%20%20K_f%20%2B%20U_f%20%3D%20K_i%20%2B%20U_i%20%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7B1%7D%7B2%7DmV_f%5E2%20%2B%20%20%5Cdfrac%7B-GMm%7D%7Bd%5E2%7D%20%3D%20%20%5Cdfrac%7B1%7D%7B2%7DmV_i%5E2%2B%20%5Cdfrac%7B-GMm%7D%7Bd_i%7D%20%5C%5C%20%5C%5C%20V_f%20%3D%20%5Csqrt%7BV_i%5E2%20%2B%202%20GM%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7Bd_2%7D-%20%5Cdfrac%7B1%7D%7Bd_i%7D%5CBig%20%5D%7D)
![V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}](https://tex.z-dn.net/?f=V_f%20%3D%20%5Csqrt%7B%289.1%20%5Ctimes%2010%5E%7B4%7D%29%5E2%20%2B%202%20%286.67%5Ctimes%2010%5E%7B-11%7D%29%20%2A%281.98%20%2A%2010%5E%7B30%7D%20%29%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7B6%2A10%5E%7B12%7D%7D-%20%5Cdfrac%7B1%7D%7B4.8%2A10%5E%7B10%7D%7D%5CBig%20%5D%7D)

She ran for 3s
Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
<span>two filters with vertical polarization</span>