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AURORKA [14]
3 years ago
9

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?

Physics
1 answer:
Nina [5.8K]3 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 boy is standing 4 meter from a plane mirror how far and in what distance must te move so that he will be 4 meter from his imag
Alona [7]

Answer:

2 meters towards the mirror.

Explanation:

In a plane mirror the image distance is equal to the object distance. Therefore, by moving 2 meters towards the mirror, the boy reduces the distance between him and the mirror to two meters which is the object distance. The image distance is also 2 meters. add the two distances you will get four meters.

6 0
3 years ago
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?​
Ksju [112]

Answer:

Explanation:

T = 2π\sqrt{L/g}

(T / 2π)² = L/g

g = 4π²L/T²

g = 4π²(0.75000)/(1.7357)²

g = 9.82814766...

g = 9.8281 m/s²

6 0
2 years ago
A 1-kg mass at the earth's surface weighs about? a. 1 n. b. 5 n. c. 10 n. d. 12 n. e. none of these
ValentinkaMS [17]

A 1-kg mass at the earth's surface weighs about C. 10N

The third planet from the Sun is the Earth. It is the seventh largest in terms of size and weighs roughly 5.98 1024 kg. The inherent quality of mass is unaffected by the environment of the object or the technique employed to quantify it.

Newton's law of gravitation can be used to estimate the mass of the Earth. This is set to the fundamental formula, which reads: force (F) = mass (m) times acceleration. Gravitational acceleration (G) is equal to 9.8 m/s2, the Earth's radius is 6.37 106 m, and the gravitational constant (G) is 6.673 1011 Nm2/kg2. The Earth has a mass of 5.96 1024 kg after rearranging the equation and entering all the numbers.

To learn more about earth please visit-
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3 0
1 year ago
A cook had a jar containing a sweet food and a jar containing a sour food. The sweet food has a strong attraction between its mo
patriot [66]

Answer:

The sweet food changed because the molecules were able to move fast enough to overcome the attraction between them with its molecules now moving away from each other.

Explanation:

We are told that the sweet food has a strong attraction between its molecules, and the sour food has a weak attraction between its molecules.

This means that the molecules in the sweet food would be moving at a faster rate than in the sour food because of the strong forces of attraction. Therefore, the molecules in the sweet food would be moving far away from each other hence the change of phase.

7 0
3 years ago
A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi
motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

3 0
3 years ago
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