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Black_prince [1.1K]
3 years ago
9

A rock from the top of a hill is falling from rest.

Physics
1 answer:
dsp733 years ago
3 0

Object name:- <u>R</u><u>o</u><u>c</u><u>k</u>

Starting energy:- <u>K</u><u>i</u><u>n</u><u>e</u><u>t</u><u>i</u><u>c</u><u> </u><u>e</u><u>n</u><u>e</u><u>r</u><u>g</u><u>y</u>

Conversion:- <u>Potential</u><u> </u><u>to</u><u> </u><u>kinetic</u><u> </u><u>energy</u>

Final energy form:- <u>P</u><u>o</u><u>t</u><u>e</u><u>n</u><u>t</u><u>i</u><u>a</u><u>l</u><u> </u><u>e</u><u>n</u><u>e</u><u>r</u><u>g</u><u>y</u>

Non usable form if energy:- <u>N</u><u>i</u><u>l</u><u> </u> [ As it has potential & kinetic both]

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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

6 0
3 years ago
Read 2 more answers
A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest
9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

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Solution:

Initial energy = Final energy

U_{i} =  (K.E) + U_{f}

As the person was at height initially so it has the potential energy only.

mg(h_{1} +h_{2}) = K.E + mgh_{2}

K.E = mgh_{1}

K.E = (61.0)(9.8)(10)\\K.E = 5978 J

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

F.h_{2} = 5978\\

F = \frac{5978}{3}

F = 1992.67 N

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