Answer:
Condition

Explanation:
Basically black hole is an object from which light rays can not escape it means to go out from gravitational field , that body should thrown with speed greater then light.
Let's do some calculation
Gravitational potential at surface =
If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.
=
⇒
It will be more better for black hole if above ratio (analogous to density ) is more then above calculated
If the object is moving at a constant speed, acceleration is 0. So:

The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so
Friction =
90 N
Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ =
L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Acceleration is found if we have the force and mass.
With the following equation: F = ma, we can find the missing values.
F = 25n
M = 0.5 kg
a = ?
a = f/m
a = 25/0.5
a = 50
a = 50 m/s
So, the acceleration is 50 m/s^2
The electric current passing through the bulb would be 3.3A
<u>Explanation:</u>
Given:
Electric charge, q = 800C
Time, t = 4 min
= 4 X 60 sec
= 240 sec
Electric current, I = ?
We know,

On substituting the value we get:

Thus, the electric current passing through the bulb would be 3.3A