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LiRa [457]
4 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]4 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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Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface
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We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = \frac{1}{2} m*v^2
GPE (Gravitational Potential Energy) = m*g*h

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
TE =  \frac{1}{2} m*v^2 + m*g*h

Solving:
TE = \frac{1}{2} m*v^2 + m*g*h
TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50
TE =  \frac{2.0*100}{2} + 1000
TE =  \frac{200}{2} + 1000
TE = 100 + 1000
\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark


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Answer:

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Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

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Dividing eq-1 by eq-2

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