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geniusboy [140]

1 year ago

5

It requires 55N force to ring the bell at a hammer swing carnival game. Donald Duck can generate 250W of power while swinging th

e hammer down in .85s, doe he produce enough force to ring the bell at a height of 7m? Donald doubled power to what happens to the force he generates ?

1 answer:

WARRIOR [948]1 year ago

5 0

Given

F = 55 N to ring the bell

P = 250 W

t = 0.85 s

Procedure

First let's calculate the force for 250W of power.

$\begin{gathered} P=\frac{Fd}{t} \\ F=\frac{Pt}{d} \\ F=\frac{250W\cdot0.85s}{7m} \\ F=30.35\text{ N} \end{gathered}$

With 250W it is not possible to reach the bell to ring it.

Now let's calculate the value for 500W

$\begin{gathered} F=\frac{500W\cdot\text{0}.85s}{7m} \\ F=60.71\text{ N} \end{gathered}$

Force is doubled when power is doubled

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A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are

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1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

1)

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

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F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

2)

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 225 N

Therefore, the coefficient of kinetic friction is

$\mu_k = \frac{F}{mg}=\frac{225}{(30)(9.8)}=0.765$

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3 years ago

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

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Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

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(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

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