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Archy [21]
1 year ago
10

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

Physics
1 answer:
Sergeu [11.5K]1 year 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

#SPJ4

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Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la
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Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93  ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

v=v_0-g \,t

for the object's vertical displacement we have

y-y_0=v_0\,t - \frac{g}{2} \,t^2

If the maximum height reached by the object is given in meters, we use the value for g in m/s^2 which is: 9.8\,\,m/s^2

If the maximum height of the object is given in feet, we use the value for g in  ft/s^2  which is : 32\,\,ft/s^2

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}

and now we use this to express the maximum height in the second equation we typed:

y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g})  - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}

Then if the max height is 1.1 meters, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s

If the max height is 1.1 feet, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s

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