A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the velocity with which the camera hits the ground.

42.05 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 239 m

Acceleration due to gravity (g) = 3.7 m/s²

Final velocity (v) =?

The velocity with which the camera hits the ground can be obtained as follow:

v² = u² + 2gh

v² = 0² + 2 × 3.7 × 239

v² = 0 + 1768.6

v² = 1768.6

Take the square root of both side

v = √(1768.6)

v = 42.05 m/s

Therefore, the velocity with which the camera hits the ground is 42.05 m/s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards