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ololo11 [35]
3 years ago
5

A 16-kg scooter is moving at a speed of 7 m/s. The scooter’s speed doubles. What is the scooter’s kinetic energy when its speed

doubles?
a. 784 J
b. 1568 J
c. 392 J
d. 112 J
Physics
1 answer:
lbvjy [14]3 years ago
5 0
The formula for kinetic energy is:

KE = 0.5mv^2

where:
m = mass
v = speed

Given this formula, the original KE of the scooter is calculated as 392 J. Since the speed is doubled, it can be observed in the formula that the change would actually affect the kinetic energy by quadrupling it. This is because 2^2 is 4. So, 392(4) = 1568 J or B.
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It seems like the question is asking for the frequency.
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The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec​
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Answer:

4 m/s^2

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3 years ago
Fnet = fa + ff or fnet = fa - ff
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5 0
2 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

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v = 21. 34 m/s

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