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3 years ago

When a car's velocity is positive and its acceleration is negative, what is happening to the car's motion?

Physics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

Assuming rightward is positive, the velocity is positive whenever the car is moving to the right, and the velocity is negative whenever the car is moving to the left. The acceleration points in the same direction as the velocity if the car is speeding up, and in the opposite direction if the car is slowing down.

Explanation:

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A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo

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The fundamental frequency on a vibrating string is given by:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension

$\mu$ is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}$

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor $\sqrt{2}$

In this case, the mass per unit length is doubled:

$\mu'=2\mu$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}$

So, the fundamental frequency will decrease by a factor $\sqrt{2}$ .

(c) the fundamental frequency will increase by a factor $\sqrt{2}$

In this case, the tension is doubled:

$T'=2T$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f$

So, the fundamental frequency will increase by a factor $\sqrt{2}$ .

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