Answer:1. Roche limit
2.hydrogen
3.atmosphere
4.mercury
5.venus
6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet
7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets
Explanation:
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration, 
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,



Frictional force is given by :


F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
The initial volume of the gas is

while its final volume is

so its variation of volume is

The pressure is constant, and it is

Therefore the work done by the gas is

where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is

And the change in internal energy of the gas can be found by using the first law of thermodynamics:

where the positive sign means the internal energy of the gas has increased.
Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s
Answer:

Explanation:
We are given that
Linear charge density of wire=
Radius of hollow cylinder=R
Net linear charge density of cylinder=
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem



Where surface area of cylinder=
