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MArishka [77]
2 years ago
8

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

produced by the pipe, in SI units, is closest to:
Physics
1 answer:
NemiM [27]2 years ago
8 0

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

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Answer:

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Explanation:

you be in serious injury.

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
3241004551 [841]

Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

  • gravitational field strength of moon at a distance R from its center, g__R
  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0
2 years ago
The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?
AnnZ [28]
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
3 0
2 years ago
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On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

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W_2 = 4*5cos180

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Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
2 years ago
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