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AysviL [449]
3 years ago
8

Use the following information for question 26 and 27. A 550-g ball traveling at 8.0 m/s undergoes a sudden head-on perfectly ela

stic collision with a 250-g ball traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass just after the collision
Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

The speed of 250 g ball after collision is 14 m/s.

Explanation:

mass of first ball, m = 550 g = 0.55 kg

initial velocity of first ball,  u = 8 m/s

mass of second ball, m' = 250 g = 0.25 kg

initial velocity of second ball, u' = - 8 m/s

Let the speed of the balls is v and v' after the collision.

Use conservation of momentum

m u + m' u' = m v + m' v'

0.55 x 8 - 0.25 x 8 = 0.55 v + 0.25 v'

0.55 v + 0.25 v' = 2.4 ..... (1)

Use the formula  of coefficient of restitution,

For elastic collision, e = 1

e =\frac{v'-v}{u - u'}\\\\1 =\frac{v'-v}{8+8}\\\\16 =v' - v...... (2)

By solving (1) and (2)

v = - 2 m/s and v' = 14 m/s

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Answer:

2,8

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3 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
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The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

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2 years ago
!!! 100 POINTS !!! PLEASE HELP !!!
hram777 [196]

Q1. Option 2: basketball


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<span>Q3. A basketball sitting on the floor stays there and a basketball rolling on court keeps on rolling.</span>

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<span>Q4 Second law says acceleration is dependent upon net force and mass of the object.</span>


Q5. Basketball accelerates when a player tries to dunk it with both hands.


<span>Q6. Third law says f<span>or every action, there is an equal and opposite reaction.</span></span>

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<span><span>Q7. As a player dribbles, the force the basketball hits the floor with is the same as the force from the floor on the ball. That is why the ball bounces back up in air.</span></span>

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8 0
3 years ago
Read 2 more answers
A block is thrown with an initial velocity of 30.0 m/s at an angle of 25.0o above the horizontal. What is the highest elevation
Serga [27]

The highest elevation reached by the ball in its trajectory is 16.4 m.

To find the answer, we need to know about the maximum height reached in a projectile.

What's the mathematical expression of the maximum height reached in a projectile motion?

  • The maximum height= U²× sin²(θ)/g
  • U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity

What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?

  • Here, U = 30.0 m/s and θ= 25°
  • Maximum height= 30²× sin²(25)/9.8

= 16.4m

Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.

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2 years ago
1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being
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