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VladimirAG [237]
3 years ago
15

What are the first three harmonics of a note produced on a 0.31 m long violin string if the waves on this string have a speed of

274.4 M/s
Physics
1 answer:
Gelneren [198K]3 years ago
5 0

Answer:

442 Hz, 882 Hz, 1330 Hz

Explanation:

Given data:  length of the string= 0.31 m = 31 cm

speed of the waves = 274.4 M/s or 2740 cm/s

Wavelengths are

62 cm

31 cm and

62 / 3  =  20.67 cm

The frequencies associated with these wavelengths ( which are the 1st, 2nd and 3rd harmonics )

are

27440/ 62 =  442.6 = 442 Hz

27440/31  = 885.2  = 882 Hz

27440 / 20.67  = 1328 = 1330 Hz

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1 year ago
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A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

4 0
3 years ago
A figure skater glides along a circular path of radius 6.70 m. If she coasts around one half of the circle, find the following.
adoni [48]

(a) Her distance from the starting location is 21.05 m.

(b) The length of the path she skated is 21.05 m.

<h3>Distance of the skater from the starting position</h3>

The distance around a complete circular path is calculated as 2πr.

The distance for a half circle is calculated as ¹/₂ x 2πr = πr

Distance from the starting location = π x 6.7 m = 21.05 m

The length of the path she skated is the same as her distance from the starting location = 21.05 m.

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Who is known as the father of India​
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With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.
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Answer:

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v_{1e} = velocity of object 1 relative to earth = 0.80 c

v_{21} = velocity of object 2 relative to object 1 = 0.80 c

v_{2e} = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}

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