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NISA [10]
3 years ago
10

A child is riding in a wagon. What reference frame might have been used if an observer said the child was not moving.

Physics
1 answer:
GarryVolchara [31]3 years ago
3 0

Answer:

the wagon should be used as frame of reference if an observer said the child was not moving.

Explanation:

The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.

<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>

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3 years ago
A tuba creates a 4th harmonic of
xxTIMURxx [149]

Answer:

93.54 Hz

Explanation:

✓From the question, Number of harmonic frequency is 4

✓ the frequency (f₄ )= 116.5 Hz

✓harmonic frequency can be calculated using below expresion

fₙ = [ (nv)/4L]..........eqn(1)

v = speed of sound= 343 m/s

n = number of given harmonic frequency

L = Length of the rope

Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope

fₙ = [ (nv)/4L]

f₄ = 4× 343 /4L

f₄ = 343 /L

L= 343 /f₄

But f₄= 116.5 Hz

L= 343/116.5= 2.944m

Hence, initial Lenght of the rope= 2.944m

We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)

= ( 2.944m + 0.721m )

= 3.667m

Hence, new length= 3.667m

To find the new frequency of the 4th harmonic we will use eqn(2)

f₄ = v/l ...............eqn(2)

From equation (2) If we substitute the values we have

f₄ = (343/3.667)

= 93.54 Hz

Hence, the the new frequency of the 4th harmonic is

93.54 Hz

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A chamber of volume 51 cm^3 is filled with 32.4 mol of Helium. It is intially at 459.38°C. (a) The gas undergoes isobaric heatin
krek1111 [17]

Answer:

Explanation:

The pressure of the gas can be found out as follows

Gas law formula is as follows

PV = nRT

P = nRT / V

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3866.45 X 10⁶ Pa.

The first change is isobaric therefore

V₁ / T₁ = V₂ / T₂

V₂ = V₁ X T₂/ T₁

= 51 X 10⁻⁶ X ( 855.6 +273) / (459.38 +273)

= 78 X 10⁻⁶

78 cm³

After the first operation , the pressure of the gas remains at

= 3866.45 X 10⁶ Pa.

Now volume of the gas changes from 78 cm³ to 25.3 cm³ isothermally so

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

= 3866.45 X 10⁶ X 78 / 25.3

= 11920 X 10⁶ . Pa

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