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shutvik [7]
3 years ago
11

A spacecraft returning from a lunar mission approaches earth on a hyperbolic trajectory. At its closest approach A it is at an a

ltitude of 5000 km, traveling at 10 km/s. At A retrorockets are red to lower the spacecraft into a 500 km altitude circular orbit, where it is to rendezvous with a space station. Find the location of the space station at retrore so that rendezvous will occur at B.
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

Explanation:

rA= 5000km+6378km=11378 km

rB= 500km+6378km=6878 km

Eccentricity, e=[11378-6878] / [11378 +6878] =0.24649

Evaluating the orbital equation at perigee yields the angular momentum

rB= h22/u * [1/[1+e]]----------------[1]

Pluging the values in equation[1]

6878 km= h22/398600* [1/[1+0.24649]]

we get h2= 58458 km2/s

Period of transfer ellipse, T_{2}=\frac{2\pi }{\mu ^{2}}( \frac{h_{2}}{\sqrt{1-e^{2}}})^{3}

Pugging the values we get , T_{2}=\frac{2\pi }{398600 ^{2}}( \frac{58458}{\sqrt{1-0.24649^{2}}})^{3}=\mathbf{8679.1s}

The period of circular orbit 3 is ,T_{3}=\frac{2\pi }{\mu }*r_{B}^{\frac{3}{2}}

T_{3}=\frac{2\pi }{398600 }*6878^{\frac{3}{2}}=\mathbf{5676.8s}

The time of flight from C to B on orbit 3 must be equal to  time of flight from A to B on orbit 2

\bigtriangleup t_{CB}=\frac{1}{2}T_{2}=\frac{1}{2}*8679.1=4339.5s

here orbit 3 is a circle, hence

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What is the mass of a person running at 2.5 m/s with momentum of<br> 200 kg m/S
kondor19780726 [428]

Answer:

m = 80 kg

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p = m × v

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3 years ago
If a sound wave travels 680 m in 2 seconds and has a frequency of 400 hz, what is the wavelength of the wave?
Kipish [7]

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3 years ago
Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has
Shalnov [3]

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

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m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

3 0
3 years ago
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