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anastassius [24]
3 years ago
10

A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

d and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?
Read the question and use the picture to help you with this one


use all 3 of the sentence starters
-The speed and direction of the satellite was
-The crash affected the motion of the satellite because . . .
-The _____ crash caused . . .

PLEASE HURRYYYY
Physics
1 answer:
irina1246 [14]3 years ago
7 0

Answer & Explanation:

Crashing into the asteroid would cause the satellite to slow down, stop, or reverse direction, because it is a force in the opposite direction to the satellite's motion. Whichever crash was a stronger force would cause it to change motion more. It takes a stronger force to change the velocity of a more massive object.

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The volume of an object as a function of time is V(t) = At³, where A is a constant. Let L and T denote dimensions of length and
Greeley [361]

Answer:

c) L³/T³

Explanation:

If t stands for time, the units are:

(V) = L³, (t) = T

The units for the equation:

V(t) = At³

must be:

L^{3} = \frac{L^{3} }{T^{3}} T^{3}

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What is the relationship between momentum and kinetic energy.
Margaret [11]

Answer:

constant object, momentum increases directly with speed

Explanation:

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A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
( timed for this!! please help!!) Which best defines scientific question?
Viefleur [7K]
My opinion, the answer is b

5 0
3 years ago
A passenger jet travels from Los angles to Bombay, India, in 22 h. The return flight takes 17 h. The difference in flight times
Hoochie [10]

Answer:

70.5 mph

Explanation:

A passenger jet travels from Los Angeles to Bombay, India, in 22h.

The return flight takes 17 h.

The difference in flight times is caused by winds over the Pacific Ocean that

blow primarily from west to east.

If the jet's average speed in still air is 550 mi/h what is the average speed

of the wind during the round trip flight? Round to the nearest mile per hour.

Is your answer reasonable?

:

Let w = speed of the wind

:

Write a distance equation (dist is the same both ways

17(550+w) = 22(550-w)

9350 + 17w = 12100 - 22w

17w + 22w = 12100 - 9350

39w = 2750

W = 2750/39

w = 70.5 mph seems very reasonable

:

Confirming if the solution by finding the distances using these value

17(550+70.5) = 10549 mi

22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph

6 0
3 years ago
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