Answer:
190.3759 pounds
Explanation:
If the astronaut weighs 210 pounds when he is at a distance of 3978 miles from the Earth's center, then he will weigh
210 [3978 / (3978 + 200) ]² = 190.3759 pounds
at an altitude of 200 km.
<span>The motion of the medium is parallel to a longitudinal wave
and perpendicular to a transverse wave.</span>
When water vaporizes into the air, it becomes humid out.
Answer:
B) 4/5
Explanation:
The magnitude of the electric force between the two spheres is given by
![F=k\frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D)
where
k is the Coulombs' constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
Initially, we have
![q_1 = 5.0\mu C=5.0\cdot 10^{-6}C\\q_2 = 1.0 \mu C=1.0 \cdot 10^{-6}C\\r = L](https://tex.z-dn.net/?f=q_1%20%3D%205.0%5Cmu%20C%3D5.0%5Ccdot%2010%5E%7B-6%7DC%5C%5Cq_2%20%3D%201.0%20%5Cmu%20C%3D1.0%20%5Ccdot%2010%5E%7B-6%7DC%5C%5Cr%20%3D%20L)
So the force is
![F_1=k\frac{(5.0\cdot 10^{-6}C)(1.0\cdot 10^{-6}C)}{L^2}=(5.0 \cdot 10^{-12} C^2)\frac{k}{L^2}](https://tex.z-dn.net/?f=F_1%3Dk%5Cfrac%7B%285.0%5Ccdot%2010%5E%7B-6%7DC%29%281.0%5Ccdot%2010%5E%7B-6%7DC%29%7D%7BL%5E2%7D%3D%285.0%20%5Ccdot%2010%5E%7B-12%7D%20C%5E2%29%5Cfrac%7Bk%7D%7BL%5E2%7D)
Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).
The total charge is
![Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^{-6}C](https://tex.z-dn.net/?f=Q%3Dq_1%20%2B%20q_2%20%3D%20%2B4.0%20%5Cmu%20C%3D4.0%5Ccdot%2010%5E%7B-6%7DC)
So each sphere will have a charge of
![q=\frac{Q}{2}=2.0\cdot 10^{-6} C](https://tex.z-dn.net/?f=q%3D%5Cfrac%7BQ%7D%7B2%7D%3D2.0%5Ccdot%2010%5E%7B-6%7D%20C)
So, the new force will be
![F_2=k\frac{(2.0\cdot 10^{-6}C)(120\cdot 10^{-6}C)}{L^2}=(4.0 \cdot 10^{-12} C^2)\frac{k}{L^2}](https://tex.z-dn.net/?f=F_2%3Dk%5Cfrac%7B%282.0%5Ccdot%2010%5E%7B-6%7DC%29%28120%5Ccdot%2010%5E%7B-6%7DC%29%7D%7BL%5E2%7D%3D%284.0%20%5Ccdot%2010%5E%7B-12%7D%20C%5E2%29%5Cfrac%7Bk%7D%7BL%5E2%7D)
And so the ratio of the two forces is
![\frac{F_2}{F_1}=\frac{4.0\cdot 10^{-12} C}{5.0\cdot 10^{-12} C}=\frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7BF_2%7D%7BF_1%7D%3D%5Cfrac%7B4.0%5Ccdot%2010%5E%7B-12%7D%20C%7D%7B5.0%5Ccdot%2010%5E%7B-12%7D%20C%7D%3D%5Cfrac%7B4%7D%7B5%7D)