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laila [671]
3 years ago
13

When the Voyager I and Voyager II spacecraft were exploring the outer planets, NASA flight controllers had to plan the crafts' m

oves well in advance. How many seconds elapse between the time a command is sent from Earth and the time the command is received by Voyager at Neptune? Assume the distance from Earth to Neptune is 4.76E+12 m.
Physics
1 answer:
ella [17]3 years ago
5 0

Answer:

1.59*10^4 s

Explanation:

The signals sent from the earth, travel at the speed of light, therefore the travel time of the signal between the earth and neptune, can be calculated with the following equation, which relates time, speed and distance:

t=\frac{x}{v}\\t=\frac{4.76*10^{12} m}{3*10^8 m/s}=1.59*10^4 s\\

The time elapsed is approximately 4.41 hours

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Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

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So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

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D because those are both concerning.

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