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8090 [49]
2 years ago
13

Two sounds are played for you. The first has a frequency of 196 Hz. The second has a frequency of 392 Hz. Which sound will be "h

igher" to your ear?
A) The First
B) The Second
C) They will sound the same
Physics
1 answer:
kykrilka [37]2 years ago
5 0

Answer:

The second one, the higher the hertz the higher the frequency we hear

Explanation:

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Counter clockwise torque is 360Nm. 
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If frequency is kept constant, how are the velocity and wavelength of a wave related?
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<h3><u>Answer;</u></h3>

Velocity and wavelength are directly proportional when frequency is kept constant.

<h3><u>Explanation;</u></h3>
  • <em><u>Frequency of a wave is the number of complete oscillations made by a given wave in one second. </u></em>
  • <em><u>Wavelength on the other hand, is the distance between two successful crests or troughs in a transverse wave or two successful rarefactions or compressions in a longitudinal waves.</u></em>
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Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

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