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LiRa [457]
3 years ago
5

Macy and Sam are trying to push a large box across a floor. Both girls push with an equal amount of force. The total amount they

exert on the box is 500 Newtons. What amount of force does each girl apply? *
Physics
1 answer:
klemol [59]3 years ago
7 0

Answer:

250 N

Explanation:

Assuming both girls are pushing in the same direction

Given

Total amount of force,they exert on box= 500 N

let <u>force exerted by Macy</u> = F_{M}

     <u>force exerted by Sam</u>= F_{S}

⇒ F_{M}+F_{S}=500 N

But also given <u>both girls push with equal amount of force</u>

⇒ F_{M}=F_{N}

Therefore

F_{M}+ F_{M}=500 N

2\times F_{M}=500 N

F_{M}=250 N

⇒  F_{M}=250 N , F_{N}= 250 N

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A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh
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Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

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time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

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3 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

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