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ArbitrLikvidat [17]
3 years ago
12

describe an experiment to show how the frequency of a note emitted by a vibrating string depends on the tension of the string

Physics
1 answer:
mart [117]3 years ago
3 0
Easy ! 

Take any musical instrument with strings ... a violin, a guitar, etc.

The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.

Pluck any string.  Then, slightly twist the tuning peg for that string,
and pluck the string again.

Twisting the peg only changed the string's tension; the length
couldn't change.

-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.

-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.
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Plate tectonics is unique to which terrestrial planet?
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Answer:  [B]:  Earth .
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6 0
3 years ago
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In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc
Strike441 [17]

The solution would be like this for this specific problem:

 

Given:  

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

 

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m <span>

Whenas n = 1, maxima (grating equation) sinθ = λ/d 
Grant distance of each maxima from centre = y .. 
<span>As sinθ ≈ y/D  y/D = λ/d λ = yd / D </span>

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D 
∆λ = (d/D) [y2 -y1] 

<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>

6 0
3 years ago
A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?
Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

Explanation:

g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2.   I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0
3 years ago
1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15
kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity=\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 m/s^2

Distance covered by the car,s = 75 m

Initial velocity of the car ,v_i = 6 m/s

Final velocity of thre car ,v_f=?

Using third equation of motion:

v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2

v_{f}=14.69 m/s

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0
3 years ago
How does the carbon rod of a cell beomes a positive<br>terminal?​
ivann1987 [24]

Answer:

In the middle of a dry cell, is a rod made of carbon. Around the carbon rod is a chemical paste. At the same time, the carbon rod becomes positively charged. When this happens, electrical current flows out of the cell when a conductor is attached between the cell's positive and negative terminals.

7 0
3 years ago
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