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mamaluj [8]
3 years ago
15

A violinist is tuning the A string on her violin by listening for beats when this note is played simultaneously with a tuning fo

rk of frequency 440 Hz. She hears a beat frequency of 4 Hz. She notices that, when she increases the tension in the string slightly, the beat frequency decreases. What was the frequency of the mistuned A string?
Physics
1 answer:
omeli [17]3 years ago
4 0

Answer:

f_2  = 436 Hz

Explanation:

Given:

- Frequency of the note played f_1 = 440 Hz

- Number of beats heard per second b = 4 Hz

Find:

What was the frequency of the mis-tuned A string?

Solution:

- The beat is the difference in frequency of the notes played and is given by the expression:

                                   | f_1 - f_2 | = beats

- we heard 4 beats, so

                                   | f_1 - f_2 | = 4

- Plug in the value:     | 440 - f_2 | = 4

Hence,                         f_2  = 436 Hz

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A see-saw is balanced on a pivot with two children on it. One child is sitting 1.5 m to the left of the pivot and has a mass of
Lorico [155]

Answer:

Explanation:

Remark

This is a second class lever. It is much more efficient than the fishing pole problem. All distances are measured from the pivot in these kinds of questions.

Givens

d1 = 1.5

d2 = ?

m1 = 50 kg

m2 = 30 kg

The lighter child will have to sit further away from the pivot to make the two conditions equal.

Formula

d1*m1 = d2*m2

1.5*50 = d2 * 30

75 = 30 * d2

75/30 = d2

d2 = 2.5

Remark

Notice that the distance is longer for the lighter child. The fact that these are masses and not forces does not matter, but you should take note of it. There is a difference between masses and forces. See the fishing pole problem.

Answer to the multiple Choice question. No motion on this kind of problem means equal moments. The answer is D

Problem 2

1) The wheels are further apart making B more stable. The wider the distance the wheels are apart, the harder it would be to tip the concrete mixer over

2) The center of gravity is lower. The higher the force is the more chance you have of exerting an external force to tip the mixer over.

7 0
3 years ago
What is the final speed of a 60 kg boulder dropped from a 111 meter cliff
saveliy_v [14]

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

       Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.



3 0
3 years ago
Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.
MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>
  • It is given that, the infinite well having a width of 0.050 mm.
  • We have the expression for energy of electron in an infinite well as,

                  E_n=\frac{n^2h^2}{8mL^2}

  • where;

                m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1

  • Thus, the lowest energy of electron in an infinite well is,

                E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

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7 0
1 year ago
A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.
Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

F=mg

\dfrac{mv^2}{r}=mg

\dfrac{v^2}{r}=g

v=\sqrt{rg}

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

v=\sqrt{250\times9.8}

v=49.49\ m/s

Hence, The speed of space station floor is 49.49 m/s.

6 0
3 years ago
Which best describes why Keplers observation of planetary motion is a law instead of a theory
svet-max [94.6K]

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5 0
3 years ago
Read 2 more answers
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