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kifflom [539]
3 years ago
15

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x directi

on. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?
Physics
1 answer:
Readme [11.4K]3 years ago
7 0

Answer:

25.9 m/s

Explanation:

mass of ball, m = 0.145 kg

initial velocity, u = + 32 m/s

It bounce back with the velocity but in opposite direction so final velocity,

v = - v

Impulse, I = - 8.4 Ns

Impulse is defined as the change in momentum

I = m v - mu = m (v - u)

- 8.4 = 0.145 x (- v - 3 2)

- 57.9 = - v - 32

v = 57.9 - 32 = 25.9 m/s

Thus, the final speed of the ball is 25.9 m/s

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We only see objects because they absorb light. <br> True or False?
Alenkasestr [34]

Answer:

true

Explanation:

i think it's true because I took a quiz on this

3 0
3 years ago
The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0
3 years ago
Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
Tasya [4]

Answer:

The rock's speed after 5 seconds is 98 m/s.

Explanation:

A rock is dropped off a cliff.

It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².

Speed after 5 seconds = V

We know that acceleration = average speed/time

In our case,

g = ((0+V)/2)/5

9.8*5 = V/2

=> V = 2*9.8*5

V = 98 m/s

3 0
3 years ago
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How much have jellyfish populated since 2015?
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4 0
3 years ago
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A person pushes a large 42.9 kg box at a constant velocity of 9 m/s across a horizontal floor for 3.8 s. Find the
Gwar [14]

Answer:

10 kJ

Explanation:

W = Fd

W = (μN)(vt)

W = μ(mg)vt

W = 0.7(42.9)(9.81)(9)(3.8)

W = 10,075.12506 J

W ≈ 10 kJ

6 0
3 years ago
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