Find the distance the ball traveled if it is, №18011165, 11.09.2021 13:42

Physics

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

Step-by-step answer

P Answered by Master

39m

Explanation:

$w = f \times s$

$780 = 20 \times s$

$s = 780 \div 20$

s = 39m

Physics

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s) hit the ball with? O 200 O 15 O 20 O 20

Step-by-step answer

P Answered by PhD

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).

Chemistry

a baseball player throws a ball with a force of 10N. The ball travels 20 meters. What is the total work done on baseball?

Step-by-step answer

P Answered by Master

here hope it helps

Explanation:

a baseball player throws a ball with a force of 10N. The ball travels 20 meters. What is the total w

Physics

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s) hit the ball with?

Step-by-step answer

P Answered by Master

1

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.

Physics

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s) hit the ball with?

Step-by-step answer

P Answered by Master

1

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.

Physics

A20.0 kg wood ball hangs from a 1.30 m -long wire. the maximum tension the wire can withstand without breaking is 600 n . a 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball. what is the largest speed this projectile can have without causing the cable to break? original work: m = 20 kg wood ball + .9 kg projectile = 20.9 kg ft max = 600 n 1.3 m radius f = mar = ft - fg m(v^2)/r = ft - fg v = sqrt((r(ft - fg))/m) v = sqrt((1.3m(600n - (20.9 kg)(9./20.9 kg) v = 4.96 m/s

Step-by-step answer

P Answered by PhD

2

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.

Physics

A20.0 kg wood ball hangs from a 1.30 m -long wire. the maximum tension the wire can withstand without breaking is 600 n . a 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball. what is the largest speed this projectile can have without causing the cable to break? original work: m = 20 kg wood ball + .9 kg projectile = 20.9 kg ft max = 600 n 1.3 m radius f = mar = ft - fg m(v^2)/r = ft - fg v = sqrt((r(ft - fg))/m) v = sqrt((1.3m(600n - (20.9 kg)(9./20.9 kg) v = 4.96 m/s

Step-by-step answer

P Answered by PhD

2

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.

Physics

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposite direction. If the ballis in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is a. 230 N. b. 400 N. c. 360 N. d. 320N,

Step-by-step answer

P Answered by Specialist

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

Physics

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposite direction. If the ballis in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is a. 230 N. b. 400 N. c. 360 N. d. 320N,

Step-by-step answer

P Answered by Master

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

Physics

Aball is dropped from the top of a cliff. by the time it reaches the ground, all of its gravitational potential energy has been transferred into kinetic energy. if the ball is travelling at 20m/s when it hits the ground, what height was it dropped from? (assume that the gravitational field strength is 10n/kg.)

Step-by-step answer

P Answered by PhD