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Zanzabum
3 years ago
12

B. A guitar string has a length of 0.90 m. When you pluck it, it plays a “C” that has a frequency of 256 Hz. How fast is the wav

e moving back and forth along the string? (Choose one of the equations from the box at the top of the page to solve this problem.)
Equations to use: v= λ ∙ f v=d/t
Physics
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

461  m/s

Explanation:

Given the string length of guitar, l = 0.90 m

Frequency , f = 256 Hz

Using frequency, wavelength and velocity relation as

v=f λ

Here λ is wavelength.

Now for guitar string the length must be equal to half of wavelength, so

λ = 2l

Therefor above formula becomes as  

v = f×2l

Substituting the given values, we get

v =256×2×0.90

v = 460.8 m/s = 461 m/s

Thus, the wave moving back and forth along the string with velocity 461 m/s.

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2 years ago
In Einstein’s analysis, a photon striking the surface of a conductor is absorbed by an electron. Which statement describes the s
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Answer:

the correct answer is A

Explanation:

In an Einstein-type analysis, the photon is absorbed, it loses all its energy, therefore the electron must receive all or none of the energy of the incident photon. In a type of inelastic shock.

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A) true. In photon it is completely absorbed or passes without interaction

B) False. The photon must change energy, but in this case there is no absorption of the photon

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2 years ago
Explain why frog will not look green under the red light?
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A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.

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8 0
3 years ago
When planning an oral presentation, you want to make sure to do all of the following except _____.
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3 years ago
Read 2 more answers
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
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