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icang [17]
3 years ago
10

In the lab activity, you will examine sound waves as they are emitted from a moving source. Predict what will happen to the soun

d waves when the sound source is in motion. Record your prediction (hypothesis) as an “if then” statement. (For example: If you select the GO button, then the train will move)
Physics
2 answers:
irina1246 [14]3 years ago
6 0
<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
 Hope this helps</span>
Serjik [45]3 years ago
5 0

Answer:

Explanation:

If the source is the same, the characteristic sound waves when they are emitted do not change (wave speed), but if the source is in relative motion with the observer, a change in the wavelength must be seen by this relative movement, such as the speed is the same this will be reflected in a change in the frequency heard by the observer

This effect is called the Doppler effect for a moving source is described by the expression, and

     f ’= [(v / (v± vs)] f

With v the speed of sound, vs the speed of the source, f the frequency emitted and f ’the frequency heard, where the sign corresponds to the source away (-) or the source approaching (-)

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How are ocean ridges formed
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5 0
3 years ago
Read 2 more answers
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
A sphere of volume 1.20×10−3m3 hangs from a cable. When the sphere is completely submerged in water, the tension in the cable is
KATRIN_1 [288]

Answer:

B = 62.9 N

Explanation:

This is an exercise on Archimedes' principle, where the thrust force equals the weight of the  liquid

         B = ρ g V

write the equilibrium equation

         T + B -W = 0

         B = W- T               (1)

use the density to write the weight

         ρ = m / V

        m = ρ V

         W = ρ g V

substitute in  1

         B = m g -T

         B = \rho_{body} g V - T

To finish the calculation, the density of the material must be known, suppose it is steel  \rho_{body} = 7850 kg / m³

calculate

         B = 7850 9.8 1.20 10⁻³ - 29.4

          B = 92.3 - 29.4

          B = 62.9 N

4 0
3 years ago
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