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geniusboy [140]
11 months 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 ?

Physics
1 answer:
WARRIOR [948]11 months 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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let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

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now similarly the gravitational force on the man if he is at the surface of Saturn

F_2 = \frac{G*100M*m}{(10R)^2}

F_2 = \frac{GMm}{R^2}

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\frac{F_1}{F_2} = 1

so here we can say

F1 = F2

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7 0
3 years ago
When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b
Nesterboy [21]

Answer:

Explanation:

Given

Displacement is \frac{1}{3} of Amplitude

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Potential Energy is given by

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U=\frac{1}{2}k(\frac{A}{3})^2

U=\frac{1}{18}kA^2

Total Energy of SHM is given by

T.E.=\frac{1}{2}kA^2

Total Energy=kinetic Energy+Potential Energy

K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2

K.E.=\frac{8}{18}kA^2

Potential Energy is \frac{1}{8} th of Total Energy

Kinetic Energy is \frac{8}{9} of Total Energy

(c)Kinetic Energy is 0.5\times \frac{1}{2}kA^2

P.E.=\frac{1}{4}kA^2

\frac{1}{2}kx^2=\frac{1}{4}kA^2

x=\frac{A}{\sqrt{2}}                  

7 0
3 years ago
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Answer:

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As the mass of a soccer ball is more than the mass of a tennis ball, so

m_2 > m_1

Let d_1 be the distance between the centers of both the balls near each other and d_2 be the distance between the centers of both the balls touching each other.

So, d_2 > d_1

The gravitational force, F, between the two objects having masses M and m and separated by distance d is

F=\frac{GMm}{d^2}

Where G is the universal gravitational constant.

As, the gravitational force is directly proportional to the product of both the masses and inversely proportional to the square of the distance between them,  so selecting the larger mass (m_2, soccer ball) separated by a lesser distance (d_2, touching) to get more gravitational force.

Therefore, there will be a larger gravitational force between them when two soccer balls touching each other.

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3 0
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Answer:increases

Explanation:

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If the radial distance is increased then the tangential velocity of the object must be increased because the time period is the same.

This can be best explained by taking an example of a car moving in a circle of radius r. If radial is increased for the same period then the car has to travel at a higher velocity to make in time.              

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How much time does it take if an object is traveling at 55 m/s and going for a distance of 6.5 meters?
wlad13 [49]

Answer:

0.12 seconds

Explanation:

7 0
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