Find the distance the ball traveled if it is 20.0n and it took 780 j to move it

Explanation:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

here hope it helps

Explanation:

200 N.

Explanation:

Hello,

In this case, since the ball with a defined mass is moving at 15 m/s and the bat with a greater mass moves at 20 m/s, by recalling the newton's third law, we can infer that even when they have different velocities and masses, in order to successfully hit the ball with the bat, they must do the same force, that is 200 N.

Best regards.

200 N.

Explanation:

Hello,

In this case, since the ball with a defined mass is moving at 15 m/s and the bat with a greater mass moves at 20 m/s, by recalling the newton's third law, we can infer that even when they have different velocities and masses, in order to successfully hit the ball with the bat, they must do the same force, that is 200 N.

Best regards.

115 m/s

Explanation:

Momentum before = momentum after

mu = (M + m) v

u = (M + m) v / m

u = (20.0 + 0.900) v / (0.900)

u = 23.2 v

Sum of the forces in the radial direction:

∑F = ma

T − (M + m) g = (M + m) v² / r

600 − (20.0 + 0.900) (9.8) = (20.0 + 0.900) v² / 1.30

v = 4.96

u = 23.2 (4.96)

u = 115 m/s

You correctly found the speed of the ball/projectile after the collision.  You just have to use conservation of momentum to find the projectile's speed before the collision.

115 m/s

Explanation:

Momentum before = momentum after

mu = (M + m) v

u = (M + m) v / m

u = (20.0 + 0.900) v / (0.900)

u = 23.2 v

Sum of the forces in the radial direction:

∑F = ma

T − (M + m) g = (M + m) v² / r

600 − (20.0 + 0.900) (9.8) = (20.0 + 0.900) v² / 1.30

v = 4.96

u = 23.2 (4.96)

u = 115 m/s

You correctly found the speed of the ball/projectile after the collision.  You just have to use conservation of momentum to find the projectile's speed before the collision.

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

The height of the cliff is 20 m.

Explanation:

Given that,

Velocity of the ball v = 20 m/s

Gravitational field strength g= 10 N/Kg

It reaches the ground, all of its gravitational potential energy has been transferred into kinetic energy.

According to the conservation of energy,

Where,

m = mass of the ball

g = gravitational field strength

v= velocity of the ball

h = height

We need to calculate the height of the cliff

Using conservation of energy

Put the value into the formula

Hence, the height of the cliff is 20 m.

Reference :

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