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Sedaia [141]
2 years ago
8

Is this object showing acceleration for the first 2 seconds? explain your answer.

Physics
1 answer:
iVinArrow [24]2 years ago
3 0
Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.
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8 0
3 years ago
What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?
joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

3 0
2 years ago
What is a scalar?
Tcecarenko [31]

a quqntity that doea not indict a direction

3 0
3 years ago
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What type of electromagnetic wave can be used to transmit information?
JulijaS [17]
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5 0
3 years ago
Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find
True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0
3 years ago
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