I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle? 2. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of woof is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the woof block move together as one object. How fast are they traveling?

3. A big league hitter attacks a fastball. The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?

Answers

Answer 1

Answer:

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

$v_1 = (m_1 u_1 - m_2 v_2)/(m_1)=((0.4 kg)(18m/s)-(0.2 kg)(25 m/s))/(0.4 kg)=5.5 m/s$

2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2) v$

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

$(m_1 u_1)/(m_1 +m_2)=((0.04 kg)(300 m/s))/(0.04 kg+0.5 kg)=22.2 m/s$

3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

$I=\Delta p$ (1)

The impulse can be rewritten as product between force and time of collision:

$I=F \Delta t$

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

$\Delta p = m\Delta v = m(v_f -v_i)$

So, eq.(1) becomes

$F \Delta t = m(v_f -v_i)$

where:

F = ? is the unknown force

$\Delta t = 0.002 s$ is the duration of the impact

m = 0.16 kg is the mass of the ball

$v_f = 44 m/s$ is the final velocity of the ball

$v_i = -38 m/s$ is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

$F=(m (v_f -v_i))/(\Delta t)=((0.16 kg)(44 m/s-(-38 m/s)))/(0.002 s)=6560 N$

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a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s^2. what is the velocity of the stroller after it has traveled 4.75m?

Answers

You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)

Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN

(V)final= 2.179m/s

Hello!

A person pushing a stroller start from rest uniformly accelerating at a rate of 0.500 m/s². What is the velocity of the stroller after it traveled 4.75 m ?

We have the following data:

a (acceleration) = 0.500 m/s²

Vf (final velocity) = ? (in m/s)

Vi (initial velocity) = 0 m/s

Δx (displacement) = 4.75 m

Solving:

Let's apply the Torricelli Equation, to find the velocity of the stroller, let's see:

$V_f^2 = V_i^2 + 2*a*\Delta{x}$

$V_f^2 = 0^2 + 2*0.500*4.75$

$V_f^2 = 0 + 4.75$

$V_f^2 = 4.75$

$V_f = √(4.75)$

$\boxed{\boxed{V_f \approx 2.18\:m/s}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

The velocity of the stroller is approximately 2.18 m/s

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the three small spheres shown in (figure 1) carry charges q1 = 3.50 nc , q2 = -7.50 nc , and q3 = 2.05 nc .

Answers

Answer:

It seems you're describing a scenario with three small spheres, each carrying a different electric charge:

1. Sphere 1: Charge $q_1 = 3.50 \, \text{nC}$

2. Sphere 2: Charge $q_2 = -7.50 \, \text{nC}$

3. Sphere 3: Charge $q_3 = 2.05 \, \text{nC}$

If you have a specific question or if there's a particular calculation or scenario you'd like to explore with these charges, please provide more details, and I'll be happy to assist you further.

PLEASE help me with these questions as soon as possible i need to turn it in!!!!A 60 watt light bulb uses 60 joules of electrical energy every second. However, only 6 joules of electrical energy is converted into light energy each second.
a. What is the efficiency of the light bulb? Give your answer as a percentage.

b. What do you think happens to the “lost” energy?

19. The work output is 300 joules for a machine that is 50% efficient. What is the work input?

20. A machine is 75% efficient. If 200 joules of work are put into the machine, how much work output does it produce?

Answers

a. I believe this might be the answer but
6 J/60 J= .1
.1 x 100= 10%

b. I'm not sure of this question

19. I'm not sure if I did this correct way or if there is a better way (I'm rusty)
300 J/.50 %= 600
600/100= 6 (answer)

20. (200 J)(.75%) = 150 (answer)
Check:
Efficiency: output/input x 100
150 J/200 J= .75
.75 x 100= 75%

Who speaks the line "Lord, what fools these mortals be"? A. Oberon
B. Cobweb
C. Mustardseed
D. Puck

Answers

The answer is D.Puck.

♡♡Hope I helped!!! :)♡♡

How much physical activity do adults need?

Answers

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30? Physical activity

If a group of workers can apply a force of 1000 newtons to move a crate 20.0 meters on a frictionless ramp: How high up a ramp will the crate travel?

Answers

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