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3 years ago

6

A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at

a constant vertical acceleration of 6.89 m/s2 the initial plane horizontal speed was 86.9 m/s, how far away from the survivor will it hit the waves

1 answer:

borishaifa [10]3 years ago

7 0

Answer: 19.15 meters on the waves away from the survivor.

Explanation:

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2 years ago

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The best activity for her to do to improve her range of motion is flexibility.

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1 year ago

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

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The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

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<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

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2 years ago