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Ronch [10]
3 years ago
8

A stone is thrown up from the top of a tower. It reaches the ground in 5.64 seconds. A second stone is thrown down from the same

tower. They have the same initial velocity. This stone reaches the ground in 3.6 seconds. The height of the tower and the initial velocities aren't given. You need to find them.
Physics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

add 5.64 to 3.6 where you'll find your awnser

Explanation:

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1.2

Explanation:

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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

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3 years ago
A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
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Answer:

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Net friction force = mass of hockey puck × acceleration

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s = 30 m

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a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

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