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IrinaK [193]
3 years ago
14

Calculate the magnitude of the average gravitational force between earth and the moon

Physics
1 answer:
zepelin [54]3 years ago
5 0
F = G mM / r^2, where 
<span>F = gravitational force between the earth and the moon, </span>
<span>G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2, </span>
<span>m = mass of the moon = 7.36 × 10^(22) kg </span>
<span>M = mass of the earth = 5.9742 × 10^(24) and </span>
<span>r = distance between the earth and the moon = 384,402 km </span>

<span>F </span>
<span>= 6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 </span>
<span>= 1.985 x 10^(26) N</span>
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The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

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In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

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Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

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y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

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By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

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b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

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In this problem,

L = 5.00 m is the length of the string

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d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

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In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

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