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rodikova [14]
2 years ago
7

A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at

1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
Physics
1 answer:
Jobisdone [24]2 years ago
8 0

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = <em>360 km-m/s </em>

The direction of the change is <em>up /\ </em>.

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Answer:

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The dust present in the clouds.

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2 years ago
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We will convert the 1dm3 in terms of cm3 as follows:

1dm^3 = (10 cm)^3
= 1000 cm^3

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So, to calculate the number of bars we will use the formula as follows;
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4 0
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Answer:

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Explanation:

Below is an attachment containing the solution

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