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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:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

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Explanation:

first we have to state the general form of the equation

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comparing the general form with the given equation

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The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the
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Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

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If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

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The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

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I have made a sketch for you, and added it as attachment.  

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