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8 months ago

How long did it take for a message sent from earth to reach the first astronauts?

Physics

1 answer:

Sergeu [11.5K]8 months ago

8 0

Time taken for a message to sent from earth to reach the first astronauts is 140 second.

It takes about three days for a spacecraft to reach the Moon. in the course of that time a spacecraft travels at the least 240,000 miles 386,400 km that is the space between Earth and the Moon. The specific distance depends at the specific path selected.

The distance from earth to moon is approximately3.84*108m

So, the distance from earth to the astronauts is3.84*108m - 10500*103m =3.735*108m

The speed of light c = 2.99792458*108m/s

So, divided the distance from earth to astronauts by the speed off light,

3.735*108m/2.99792458*108m/s ˜1.25seconds

Divided the Pluto-earth distance by speed of light:

Hence, 4.2*1010m/2.99792458*108m/s = 140seconds

Learn more about astronauts here:-brainly.com/question/24496270

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3 years ago

A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the

svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength is given by

$\lambda = \dfrac{2L}{n}$

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

$f = n\dfrac{v}{2L}$

now, wavelength calculation

n = 1

$\lambda_1 = \dfrac{2\times 2}{1}$

λ₁ = 4 m

n = 2

$\lambda_2 = \dfrac{2\times 2}{2}$

λ₂ = 2 m

n =3

$\lambda_3 = \dfrac{2\times 2}{3}$

λ₃ = 1.333 m

now, frequency calculation

n = 1

$f = n\dfrac{v}{2L}$

$f_1 =1\times \dfrac{50}{2\times 2}$

f₁ = 12.5 Hz

n = 2

$f = n\dfrac{v}{2L}$

$f_2 =2\times \dfrac{50}{2\times 2}$

f₂= 25 Hz

n = 3

$f = n\dfrac{v}{2L}$

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3 years ago

What effect would this have on the momentum?

Mariana [72]

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2 years ago

A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

Keith_Richards [23]

Answer

given,

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mass of rod = 0.39 Kg

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mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

$I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}$