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alexandr1967 [171]
4 years ago
8

A sprinter accelerates from rest to 10.0 m/s in 1.35 seconds. What is their acceleration?

Physics
2 answers:
riadik2000 [5.3K]4 years ago
5 0

Answer:

7.407 recurring m/s^2

Explanation:

Initial velocity= 0 m/s [as it starts from rest]

Final velocity= 10 m/s

Time taken= 1.35 s

Hence,

Acceleration = <u>Final velocity - Initial velocity</u>

                                 Time taken

a=<u>10</u>

   1.35

a=7.407 recurring m/s^2

ipn [44]4 years ago
4 0

Answer:

0.135

Explanation:

You need to divide

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A charge of +0.001 C is 1 m to your right and another charge of +1000 C is 1 m to your left. You are holding a charge of −1 C. W
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Answer:

C and D

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Charge on right = +0.001 C

Charge on left = 1000 C

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Which shows charge on left exerts 1,000,000 times more force then charge on right

Charge on left is 1000 times greater in magnitude than charge held between and charge held between is 1000 times greater in magnitude than charge on right. So the magnitude of the force on the charge you are holding would be the same if it were +1 C instead of −1 C.

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A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou
kupik [55]

Answer:

The acceleration of the satellite is 0.87 m/s^{2}

Explanation:

The acceleration in a circular motion is defined as:

a = \frac{v^{2}}{r}  (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

v = \frac{2 \pi r}{T} (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite (1.50x10^{7} m) and the Earth radius (6.38x10^{6} m) :

r = 1.50x10^{7} m+6.38x10^{6}m

r = 21.38x10^{6}m

Then, equation 2 can be used:

T = 8.65 hrs \cdot \frac{3600 s}{1hrs} ⇒ 31140 s

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Hence, the acceleration of the satellite is 0.87 m/s^{2}

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