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PSYCHO15rus [73]
3 years ago
11

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

ers/second in 3 seconds. What’s the acceleration of the motorcycle over this time? Use . 9 m/s2 8 m/s2 6 m/s2 5 m/s2 3 m/s2
Physics
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer: 3 m/s^2

Explanation: The acceleration is defined as the rate of change of the velocity:

We know that at the time t = 0 the velocity was 15 m/s

and at the time t = 3s the velocity was 24 m/s

Then the average acceleration may be calculated as the slope of the linear relationship between these two points; this is if the points are (0, 15) and (3, 24)

The slope is : S = (23 - 15)/(3 - 0)m/s^2 = 3m/s^2

So the average acceleration of the motorcycle in this period of time is 3 meters per second squared.

andrew11 [14]3 years ago
4 0

Answer: 3 m/s^2

Explanation:

The acceleration of the motorcycle is given by

a=\frac{v-u}{t}

where

v=24 m/s is the final velocity of the motorcycle

u=15 m/s is the initial velocity

t=3 s is the time taken

Substituting these numbers into the equation, we find

a=\frac{24 m/s-15 m/s}{3 s}=\frac{9 m/s}{3s}=3 m/s^2

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Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

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        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

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b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

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                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

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