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irina [24]
3 years ago
13

A 14 kg tennis ball moves at a velocity of 11 m/s. The ball is struck by a racket, causing it to rebound in the opposite directi

on at a speed of 5 m/s. What is the change in the ball’s momentum? Report the number of kg‧m/s with the appropriate sign.
Physics
1 answer:
Charra [1.4K]3 years ago
7 0

To solve this problem we will apply the concepts related to the momentum.

This is defined as the product between the change in velocity and the mass of the object, that is

p = m\Delta v

p = m (v_f-v_i)

Where,

m = mass

v_f = Final velocity

v_i = Initial velocity

Our values are given as,

m = 14kg

v_i= 11m/s

v_f= - 5m/s \rightarrow <em>the negative Symbol implies that the direction is opposite to the initial one and therefore there is also a change in the sense of magnitude</em>

p = (14)(-5-11)

p = 14(-16)

p = -224 kg \cdot m/s

The negative symbol indicates that the momentum has a direction opposite to that of the initial velocity. Or failing that, it has the same direction of the final speed

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa
Alborosie
<h2>The work done = - 2 x 10⁴ J</h2>

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process  , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure  and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

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3 years ago
Science is made up of two things, what is it
rodikova [14]

Answer:

Natural science is concerned with the description, prediction, and understanding of natural phenomena based on empirical evidence from observation and experimentation. It can be divided into two main branches: life science (or biological science) and physical science.

Explanation:

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3 years ago
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ipn [44]
The set of all sets that are not members of themselves. This contradiction is Russell's paradox.
5 0
3 years ago
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