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Oksana_A [137]
3 years ago
14

The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth

er with a tuning fork that is oscillating accurately at a concert A (400Hz). What is the period of the violin string oscillations
Physics
1 answer:
Basile [38]3 years ago
3 0

Answer:

T=2.5*10^{-3}s

Explanation:

From the question we are told that:

Beat frequency F_b=4

Frequency F=400Hz

Generally the equation for Frequency of the violin is mathematically given by

 f_v=F_b+F

 f_v=4+400Hz

 f_v=404Hz

Therefore the period of the violin string oscillations is

 T=\frac{1}{f_v}

 T=\frac{1}{404}

 T=2.5*10^{-3}s

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Image 4 ?

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4 years ago
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

W=0.31 J

4 0
3 years ago
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When scientists observed the light from stars and galaxies, they noticed that their color shifted toward the end of the visible
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Color shifted towards the end is C red
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3 years ago
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A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
I need to know the right answer to that question
ad-work [718]
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation

v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.
8 0
3 years ago
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