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Natalka [10]
3 years ago
15

Yung hangs a mass from a rod, sets the pendulum in simple harmonic motion, and determines the period of the pendulum. How could

Yung decrease the period of the pendulum?
He could use a longer rod.
He could use a shorter rod.
He could use a larger mass.
He could use a smaller mass.
Physics
2 answers:
Doss [256]3 years ago
4 0
He could use a shorter rod.

Also, if the rod is NOT massless, then using a smaller mass on the end
would also decrease the period (make the pendulum wiggle faster).
sashaice [31]3 years ago
3 0

Answer:

He could use a shorter rod.  

Explanation:

Period of a pendulum is given by:

T = 2 \pi \sqrt \frac{L}{g}

Where, L is the length of the rod and g is the acceleration due to gravity.

Period of a pendulum does not depend on the mass that is hung. It just depends upon the length of the rod by which the mass hangs. Period can be decreased if the length of the rod is decreased. Thus, Yung should use a shorter rod.

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NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}

With the following condition :

  • v = final velocity (m/s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • \sf{h_1} = initial height = 45.0 m
  • \sf{h_2} = final height = 10.0 m
  • g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

  • v = final velocity = ... m/s

Step by Step

\sf{v = \sqrt{2 \times g \times \Delta h}}

\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}

\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}

\sf{v = \sqrt{19.6 \times 35}}

\sf{v = \sqrt{686}}

\boxed{\sf{v \approx 26.2 \: m/s}}

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

<h3>See More :</h3>
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