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devlian [24]
3 years ago
5

PLEASE HELPPPP!!!!

Physics
2 answers:
ivanzaharov [21]3 years ago
5 0

Answer:

The right option is higher pitch due to change to a higher frequency

Explanation:

Pitch can be defined as the characteristics of a note which enables us to differentiate a high note from a low one.

Frequency: this is the number of oscillation per seconds expressed in Hertz (Hz)

Pitch of a note depends on frequency of a sound wave

I.e as the frequency become high, The higher will be the pitch.

From the question above, when the sound source is moving towards the observer,

Let the frequency of the sound source = f

Let the apparent frequency (frequency which the observer will hear) = f¹

Let the velocity of the source sound = Vₓ

Let the velocity of the observer = V

∴ The apparent frequency (f¹) when the source sound is moving towards the observer = {V/(V-Vₓ)}f

f¹ = {V/(V-Vₓ)}f

since (V-Vₓ ) is less than V, therefore f¹ will be increased.

If f¹ is increased, the pitch is also increased.

The right option is higher pitch due to change to a higher frequency

Tresset [83]3 years ago
3 0

you will hear a higher pitch due to a higher frequency.

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We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
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Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

                q_C = 3/4 Q

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