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zloy xaker [14]

2 years ago

5

A ball is thrown upwards. Neglecting air resistance, what initial upward speed does the ball need to remain in the air for a tot

al time of 10 seconds

1 answer:

melamori03 [73]2 years ago

5 0

Answer:

H = 1/2 g t^2 time to reach top of trajectory

v = g t time to reach top of trajectory when v is initial speed upwards

v = 5 g = 49 m/s 5 sec upwards and 5 sec downwards

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Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, $q_1=9\ \mu C=9\times 10^{-6}\ C$

Charge on sphere 2, $q_1=4\ \mu C=4\times 10^{-6}\ C$

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Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

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Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

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initial speed of the second ball, u₂ = 4 m/s

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The initial kinetic energy of the balls is calculated as;

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K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

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