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Anettt [7]
3 years ago
6

Which word equation is used to calculate the acceleration of an object? A. Subtract the initial velocity from the final velocity

and multiply the result by the time. B. Subtract the initial velocity from the final velocity and divide the result by the time. C. Add the initial velocity and the final velocity and divide the result by the time. D. Add the initial velocity and the final velocity and multiply the result by the time.
Physics
2 answers:
saveliy_v [14]3 years ago
6 0

Answer:Subtract the initial velocity from the final velocity and divide the result by the time.

Explanation:

zlopas [31]3 years ago
3 0
Correct answer is: 
<span>B. Subtract the initial velocity from the final velocity and divide the result by the time

In fact, the formula to calculate the acceleration is
</span>a= \frac{v_f-v_i}{t}
<span>where vf is the final velocity, vi the initial velocity, and t the time.</span>
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Two movers are pushing a large crate with a force of 60.0 n each. one pushes north, the other east. what is the equilibrant forc
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Two precautions in images of a convex lens
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1 . rays should passs through correct center or points(i.e. optic center or focus)

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4 years ago
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Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a
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Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

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For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

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3 years ago
Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The
Korolek [52]

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f} (1)

Where:

M - Mass of the merry-go-round, in kilograms.

m_{1}, m_{2}, m_{3} - Masses of the three children, in kilograms.

R - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

\dot n_{o}, \dot n_{f} - Initial and final angular speed, in revolutions per minute.

If we know that M = 182\,kg, m_{1} = 17.4\,kg, m_{2} = 28.5\,kg, m_{3} = 32.8\,kg, R = 1.60\,m and \dot n_{o} = 15.3\,\frac{rev}{min}, then the final angular speed of the system is:

\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)

\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right]

\dot n_{f} = 18.388\,\frac{rev}{min}

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

7 0
3 years ago
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