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3 years ago

7

A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fas

t is it moving after being struck?

2 answers:

Likurg_2 [28]3 years ago

8 0

Answer:

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

o-na [289]3 years ago

8 0

Answer:

4.25 m/s

Explanation:

An impulse is defined as a change in momentum, and can be found my multiplying an applied force by its duration. In this case,

that's 22×0.029 = 0.638 kg-m/s. Momentum is also mass times velocity, and your velocity is unknown.

Momentum P= mv

0.15 v= 0.638

v= 4.25 m/s

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3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

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3 years ago

What current is used to power the United States power grid?

alexdok [17]

The answer is Alternating Current

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3 years ago

Read 2 more answers

I need help with this

fredd [130]

We have here what is known as parallel combination of resistors.

Using the relation:

$\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\$
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
$\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}$

To ger effective resistance take the inverse:
we get,
$\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm$

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.

7 0

3 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

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3 years ago