Ask question

Ask question

All categories

3 years ago

12

A pendulum bob is made from a sphere filled with water. What would happen to the frequency of vibration of this pendulum if the

sphere had a hole in it that allowed the water to leak out slowly?

1 answer:

Nady [450]3 years ago

7 0

Answer:

The frequency would not change because the frequency of a simple pendulum does not depend on mass.

Explanation:

You might be interested in

Please help thank you

Valentin [98]

Diagram A is the correct diagram

5 0

3 years ago

b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept

olasank [31]

Answer:

fftgjjjuhgybcrgjjuhg ui

Explanation:

trggfyh

4 0

3 years ago

An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of

blondinia [14]

Answer:

v = 12.12 m/s

Explanation:

It is given that,

Radius of circle, r = 30 m

The coeﬃcient friction between tires and road is 0.5,

The centripetal force is balanced by the force of friction such that,

v = 12.12 m/s

So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.

5 0

3 years ago

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$ wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$ , then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

3 0

3 years ago

Which particle rarely interacts with matter?

Marysya12 [62]

D. Neutrino
Neutrinos are particles that rarely interact with matter.

4 0

3 years ago