The object with the greater mass has a greater gravitational force and that determines what satellites orbit around it. An object with more mass will never orbit an object with less.
Answer:
<h2>
The potential difference increases </h2>
Explanation:
from the relation 
where E= electric field (force per coulomb)
V= voltage
d= distance
Hence the voltage is going to be V= E×d.
Therefore this means that increasing the distance increases the voltage.
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
</span>
Answer:
When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u> m.
When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u> m.
When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is <u>0.87</u> m.
When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is <u>1.25</u> m.
When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is <u>1.86</u> m.
Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!
