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Masteriza [31]
3 years ago
9

A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net for

ce is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Physics
2 answers:
LenKa [72]3 years ago
7 0

Answer:

The acceleration is half of its original value

Explanation:

Dmitriy789 [7]3 years ago
6 0

Answer:

The answer  is The acceleration is double its original value.

Explanation:

<h2><u>It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.</u></h2><h2><u></u></h2>

Hope this helps....

Have a nice day!!!!

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Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10
NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

     2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v

           -4.12 \times 10^{3} kg m/s = (10^{4} kg) v

                   v = \frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

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3 years ago
Cardiovascular exercise can
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Can <span>get your heart rate up and increases blood circulation throughout the body.</span>
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3 years ago
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A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i
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2 years ago
calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
bulgar [2K]

Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

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g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

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<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

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\rule[225]{225}{2}

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