Explanation:
Let acceleration due to Gravity for a planet is given by:
![g_X=GM/R^2](https://tex.z-dn.net/?f=g_X%3DGM%2FR%5E2)
Here,![g_X = 2g](https://tex.z-dn.net/?f=g_X%20%3D%202g)
Escape velocity is given by:
![v =\sqrt{ \frac{2GM} {R}} = \sqrt{2aR}](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B%20%5Cfrac%7B2GM%7D%20%7BR%7D%7D%20%3D%20%5Csqrt%7B2aR%7D)
Here,
and g_X = 2g
Therefore,![v=\sqrt(2(2g)(R/2))=v_0](https://tex.z-dn.net/?f=v%3D%5Csqrt%282%282g%29%28R%2F2%29%29%3Dv_0)
Answer:
1.The Sun is located at one of the foci of the planets' elliptical orbits.
2.The path of the planets around the Sun is elliptical in shape.
Explanation:
As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path
So all planets are in elliptical path always
Position of sun is always at one of the focus
so correct answer will be
1.The Sun is located at one of the foci of the planets' elliptical orbits.
2.The path of the planets around the Sun is elliptical in shape.
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
![W = F*d](https://tex.z-dn.net/?f=%20W%20%3D%20F%2Ad%20)
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
![W = F*d = 108.915 N*2.38 m = 259.22 J](https://tex.z-dn.net/?f=%20W%20%3D%20F%2Ad%20%3D%20108.915%20N%2A2.38%20m%20%3D%20259.22%20J%20)
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
7.D
8C
9A
10B
11C
For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N