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2 years ago

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A tuning fork is sounded above a resonating tube (one end closed), which resonates at a length of 0.20 m and again at 0.60 m. If

the tube length were extended further, at what point will the tuning fork again create a resonance condition?

1 answer:

Nina [5.8K]2 years ago

6 0

To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then

$n_1 = \frac{V}{4L_1}$

Third resonance occurs at $5n_2$

$n_2 = \frac{V}{4L_2}$

At resonance $5n_2=n_1$

$5\frac{V}{4L_2} = \frac{V}{4L_1}$

$L_2 = 5L_1$

$L_2 = 5*0.2$

$L_2 = 1m$

Therefore at 1m will the tuning fork again create a resonance condition

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A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused

mojhsa [17]

Answer:

Explanation:

If I assume that the wind did not cause the plane to chage its velocity.

The plane will have a velocity of vp = (0*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (0*i + 100*j) + (0*i - 20*j) = (0*i + 80*j) km/h relative to the ground.

If I assume that the wind move the plane:

The plane will have a velocity of vp = (-40*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (-40*i + 100*j) + (0*i - 20*j) = (-40*i + 80*j) km/h relative to the ground.

In reality the wind would move the plane a little, not to the full speed of the wind, somewhere between these two values, but without more data it cannot be calculated.

6 0

2 years ago

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

2 years ago

Read 2 more answers

Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going

DerKrebs [107]

Answer:

The impact force will be same for both the cases.

Explanation:

The rate of change of momentum is known as the Impulse and is given by:

$I = \frac{\Delta p}{\Delta t}$

where

I = Impulse

$\Delta p = change in momentum$

$\Delta t = time interval$

Now,

In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.

Also, in both the cases the car does not bounces off the things it hit.

Thus

$\Delta p = 0 - m\times v = - mv$

Thus

Impact force, $F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}$

Therefore, impact force is same for both the cases.

5 0

2 years ago

Read 2 more answers

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere

Nataliya [291]

Answer:

The correct answer is D $I_{30}$ / $I_{50}$ = 1.5

Explanation:

In this exercise the moment of inertia equation should be used

I = ∫ r² dm

In addition to the parallel axis theorem

I = $I_{cm}$ + M D²

Where $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

L = 100.00 cm (1m / 100 cm) = 1.0000 m

$x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

λ = M / L = dm / dx

dm = λ dx

Let's replace in the moment of inertia equation

I = ∫ x² ( λ dx)

We integrate

I = λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]

I = λ/3 L³/4

I = 1/12 λ L³

Let's replace the linear density with its value

I = 1/12 (M/L) L³

I = 1/12 M L²

Let's calculate with the given values

I = 1/12 M 1²

I = 1/12 M

This point is the center of mass of the rod

Icm = I = 1/12 M = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

D = $x_{cm}$ - x

D = 0.50 - 0.30

D = 0.20 m

Let's calculate

$I_{30}$ = $I_{cm}$ + M D²

$I_{30}$ = 1/12 M + M 0.202

$I_{30}$ = M (1/12 + 0.04)

$I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

$I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

$I_{30}$ / $I_{50}$ = 1.48

The correct answer is d 1.5

6 0

2 years ago

A typical adult human has a mass of about 70 kg.

Misha Larkins [42]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N

b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>

3 0

2 years ago