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Karo-lina-s [1.5K]

3 years ago

15

Newtons second law states that Force=Mass × Acceleration.The acceleration of gravity in 9.8 m/s.How much force is gravity exerti

ng on the 100g ball

1 answer:

Naddika [18.5K]3 years ago

5 0

.98 Newton’s because you convert 100 g to kg which is .1 kg them you multiply.1 kg by 9.8 and get .98 and the units of the force are in Newton’s

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Answer:

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Explanation:

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2 years ago

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The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

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b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

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c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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Salsk061 [2.6K]

Answer:

Option e) 320 s

Explanation:

Here, distance = 3.0 km = 3000 m

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Upstream velocity = velocity of boat in still water - velocity of river flow

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Option E is the correct answer.

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3 years ago

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Straight
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