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Katarina [22]
2 years ago
12

Which refers to the change of pitch heard when the source of the sound approaches and then moves away from a stationary person?

Doppler effect volume resonance effect vibration
Physics
2 answers:
babunello [35]2 years ago
5 0
<h2><em>The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound.</em></h2><h2><em>HOP</em>E IT HELPS </h2><h2>THANK YOU </h2>
goldfiish [28.3K]2 years ago
4 0

Answer: Doppler effect

Explanation:

I hoped I helped

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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

   = 0.55 \ \frac{m}{sec^2}\\\\

F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

6 0
3 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a
anyanavicka [17]

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

F_2-F_1-F_f=Ma          (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero

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3 years ago
g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the qu
Travka [436]

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

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artcher [175]

Answer:

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3 years ago
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laila [671]

3b. No

4a. 50N


I hope this helps


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