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zmey [24]
3 years ago
9

In the oscillating spring ball system, where is the velocity of the ball the greatest?

Physics
1 answer:
nlexa [21]3 years ago
6 0

The speed of an object in a mass-spring system is given under the function

v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}

Here,

m = mass

k = Spring constant

A = Amplitude

x = Position

When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as

v_{max}= A\sqrt{\frac{x}{m}}

So the correct answer is B.

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Explanation:

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A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change
lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg

Therefore, the mass that must be added is

\Delta m = 1.128 - 0.500 = 0.628 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

5 0
4 years ago
While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega
Paraphin [41]

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

 

   Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

   In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

   If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

4 0
3 years ago
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