I need someone toguide me on these questions, pls.
Itry my best but little bit not good in physics

1. A 0.25 kg ball is attached to a string of 1.50 m and swung in a horizontal circle with a velocity of 2.4 m/s. Find the centripetal force of the ball. Then find the centripetal acceleration.
F= m v^2/r = (0.25 kg) (2.4)^2/1.5m is this right?
2. A 300 N box is placed on an incline plane that makes a 30 degree angle from the horizontal. Find the component of the weight force parallel to the planes surface.
F =
3. A student lifts a box that weighs 120N. The box is lifted 10m. How much work does the student do?
W = fd (120N) (10m) = 1200 J - correct?
4.A sailor pulls a boat 20 m along a dock using a rope that makes a 25 degree angle with the horizontal.
a. How much work does the sailor do on the boat?
b. if the boat’s mass is 2000kg, find velocity of boat.
Help!! On this one
5. A stone is thrown horizontally from the top of a 80.50m cliff. The stone lands at a distance of 10.00 m from the edge of the cliff. What is the initial horizontal velocity of the stone?
Im not good with projectiles.
6.A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net force on the probe is zero, what can be said about the relative masses of the moons?
a.Moon A is four times as massive as moon B.
b.Moon A is half as massive as moon B.
c.Moon A has the same mass as moon B.
d.Moon A is twice as massive as moon B.


Hope I get help. Grazie

1. A 0.25 kg ball is attached to a string of 1.50 m and swung in a horizontal circle with a velocity of 2.4 m/s. Find the centripetal force of the ball. Then find the centripetal acceleration.
F= m v^2/r = (0.25 kg) (2.4)^2/1.5m is this right?

The force part is correct. The centripetal acceleration = v^2/r

2. A 300 N box is placed on an incline plane that makes a 30 degree angle from the horizontal. Find the component of the weight force parallel to the planes surface.

F = 300 sin 30.

The component of the weight force perpendicular to the surface

F = 300 cos 30.

Treat the weight force as the hypotenuse of a right triangle and drop a perpendicular to the surface. You will be able to figure out how from the definition of sine and cos in a right triangle.

3. A student lifts a box that weighs 120N. The box is lifted 10m. How much work does the student do?
W = fd (120N) (10m) = 1200 J - correct?

Correct.
A trick question would be how much work is done by the box.
The answer would be -1200J

5. A stone is thrown horizontally from the top of a 80.50m cliff. The stone lands at a distance of 10.00 m from the edge of the cliff. What is the initial horizontal velocity of the stone?
Im not good with projectiles.

There is no force on it in the horizontal direction ( we neglect air resistance) so the horz vel does not change.
Since it was thrown horizontally, its initial velocity in the vertical dierection is zero. Use s = ut + 0.5gt^2 withu, the initial vel = 0.
From this, find t.
The horizontal distance is simply Vhorz x t. So

Vhorz x t = 10.00 .Solve for Vhorz.

In projectiles always resolve the velocity into its vertical and horizontal components. Then treat them independently remembering that the horz vel doesn't change in problems where air resistance is neglected.
It is quite simple actually one you get the hang of it.

6.A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net force on the probe is zero, what can be said about the relative masses of the moons?
a.Moon A is four times as massive as moon B.
b.Moon A is half as massive as moon B.
c.Moon A has the same mass as moon B.
d.Moon A is twice as massive as moon B.

Let the distance of the probe from moon B = R.
Then distance from moon A = 2R

Let Ma = mass of moon A, Mb = mass of moon B, m= mass of probe
The forces exerted by both the moons on the probe are equal and opposite. Equating the magnitude of the forces on the probe, we have,
from Newton's law of Gravitation,

G Ma m / (2 R)^2 = G Mb m / R^2 from which we get,

Ma/Mb = 4 or Ma = 4 Mb. So the answer is a.

4.A sailor pulls a boat 20 m along a dock using a rope that makes a 25 degree angle with the horizontal.
a. How much work does the sailor do on the boat?
b. if the boat’s mass is 2000kg, find velocity of boat.
Help!! On this one

Resolve the compnents of the force F that the sailor exerts into its horz and vertical components.

Fh = F cos 25 and Fv = F sin 25.

Work is done only by the horz component as work is defined as a dot product F.s = Fs cos theta where theta is the angle between the force and displacement. For the vertical cos 90 = 0.

Work = F x 20 x cos 25.


The work done on the boat goes only into increasing its K.E. as no friction is being considered. Thus,

F x 20 x cos 25 = 0.5 x 2000 x v^2. Solve for v