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AlladinOne [14]
3 years ago
5

A 90.0kg person stands 1.25m from a 60.0kg person sitting on a bench nearby. What is the magnitude of the gravitational force be

tween them ?
Physics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer: 23(10)^{-8} N

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2} (1)

Where:

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

m1=90 kg is the mass of the first person

m2=60 kg is the mass of the second person

r=1.25 m  is the distance between both persons

Solving:

F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(90 kg)(60 kg)}{(1.25 m)^2} (2)

Finally:

F=23(10)^{-8} N  This is the gravitational force between both persons. As you can see is too small.

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Answer:

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Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with
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If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     v=\sqrt{\frac{T}{(m/l)}}

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

(m/l)=\frac{10}{V^2}

Explanation:

From the question we are told that

Speed of a transverse wave given by

v=\sqrt{\frac{T}{(m/l)}}

Maximum Tension is T=10.0N

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v=\sqrt{\frac{T}{(m/l)}}

v^2=\frac{T}{(m/l)}

(m/l)=\frac{T}{V^2}

(m/l)=\frac{10}{V^2}

Therefore the Linear mass in terms of Velocity is given by

(m/l)=\frac{10}{V^2}

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I hope I helped:)
  
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