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Ksju [112]
3 years ago
11

A body weighing 250 grams was dropped from a helicopter flying at an altitude of 100 meters. Determine the potential energy of t

his body. (g = 10 m/s² ). PLEASE HURRY ITS A TEST​
Physics
1 answer:
kogti [31]3 years ago
7 0

Answer:

the potential energy of this body is 245 J.

Explanation:

Given;

mass of the body, m = 250 g = 0.25 kg

height from which the body was dropped, h = 100 m

acceleration due to gravity, g = 9.8 m/s²

The potential energy of this body is calculated as;

P.E = mgh

substitute the given values and solve for the potential energy of this body;

P.E = 0.25 x 9.8 x 100

P.E = 245 J.

Therefore, the potential energy of this body is 245 J.

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Explanation:

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2 years ago
Which best summarizes a concept related to the work-energy theorem?
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Answer:

When work is positive, the environment does work on an object.

Explanation:

According to the work-energy theorem, the net work done by the forces on a body or an object is equal to the change produced in the kinetic energy of the body or an object.

The concept that summarizes a concept related to the work-energy theorem is that ''When work is positive, the environment does work on an object.''

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3 years ago
A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
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