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sergey [27]
3 years ago
13

A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving

it a speed of 0.14m/s directly away from the shuttle. Seven and a half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

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Papessa [141]

Answer:

\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is \dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )

Now, using the pulse spread formula:

\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda

\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ )  \times (45 \ nm)

Thus, the pulse spreading as a result of  material dispersion is:\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

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