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3 years ago

9

A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a velocity of 8 m/s. What is the velocity of t

he skater immediately after?

1 answer:

KengaRu [80]3 years ago

5 0

Answer:

The velocity will be v1 = 0.58[m/s]

Explanation:

This problem can be solved by the law of conservation of the moment, which explains that the moment of a system remains constant because there are no external forces acting on it.

We have the following initial data:

m1 = mass of the skater = 55 [kg]

m2 = mass of the ball = 3 [kg]

v2 = velocity of the ball = 8 [m/s]

Therefore:

$m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}+m_{2}*v_{2}\\(50*0)+(3*0)=(50*v_{1})+(3*8)\\50+3-24=50*v_{1}\\v_{1}= 0.58[m/s]$

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3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

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Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

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Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

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b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

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2 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

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Part d)

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Explanation:

As we know that the speed of wave in string is given by

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so we have

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now we have

$v = \sqrt{\frac{17.5}{0.54}}$

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now we have

Part a)

$y_m$ = amplitude of wave

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part b)

$k = \frac{\omega}{v}$

here we know that

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so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

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