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denis23 [38]
3 years ago
5

What is the frequency of a rubber band that vibrates back and forth 50 times in 1 second

Physics
1 answer:
ch4aika [34]3 years ago
4 0

If it makes 50 complete back-and-forth wiggles in 1 second, then its frequency is <em>50 Hz</em>.

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Electrical metallic tubing would be used in an electrical installation because it
maw [93]
Hey There!

Electrical metallic tubing would be used in an electrical installation because it <span>is less expensive.</span>
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4 years ago
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The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s
V125BC [204]

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

8 0
2 years ago
What is the wavelength in air of red light from a helium neon laser?
Elza [17]

Answer:

632.8 nm is the wavelength (in air) of red light from a helium neon laser.

4 0
2 years ago
High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th
murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s  

4 0
3 years ago
An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the
bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N   +  N--->N   +   N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda

The frequency is calculated as follows;

F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0
3 years ago
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