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givi [52]
2 years ago
12

Which phenomenon occurs when an object absorbs wave energy that matches the object's natural frequency?

Physics
2 answers:
zheka24 [161]2 years ago
4 0

When object absorbs energy then due to this energy it will have increase in it natural frequency

This increase in natural frequency will lead it to the increase in the amplitude of the object

Now when natural frequency of the object will match with the frequency of the wave that is absorbed by the object then this phenomenon is known as Resonance.

In resonance the amplitude of object will rise high and it will reach to its maximum limit.

so at condition

f_n = f_{wave}

then in this condition it is known as resonance

uysha [10]2 years ago
3 0
The answer to the question above is resonance. This is a phenomenon in which the absorbed waves match the frequency the object's natural frequency which is also sometimes referred to as resonant frequency. This would cause an increase in amplitude of oscillation or vibration of the system. 
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7 0
2 years ago
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A horizontal pipe of diameter 0.81 m has a smooth constriction to a section of diameter 0.486 m . The density of oil flowing in
Lerok [7]

Answer:

2.06 m³/s

Explanation:

diameter of pipe, d = 0.81 m

diameter of constriction, d' = 0.486 m

radius, r = 0.405 m

r' = 0.243 m

density of oil, ρ = 821 kg/m³

Pressure in the pipe, P = 7970 N/m²

Pressure at the constriction, P' = 5977.5 N/m²

Let v and v' is the velocity of fluid in the pipe and at the constriction.

By use of the equation of continuity

A x v = A' x v'

r² x v = r'² x v'

0.405 x 0.405 x v = 0.243 x 0.243 x v'

v = 0.36 v' .... (1)

Use of Bernoulli's theorem

P+\frac{1}{2} \rho v^{2}=P' +\frac{1}{2}\rho'v'^{2}

7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'²    from (i)

1992.5 = 357.3 v'²

v' = 5.58 m/s

v = 0.36 x 5.58

v = 2 m/s

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2 years ago
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Burka [1]

Answer:

m = 1.82E+23 kg

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r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

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r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

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v = √(Gm/r)

v² = [√(Gm/r)]²

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m = rv²/G

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k = spring constant of the spring = 85 N/m

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the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

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inserting the values

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8 0
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