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Luda [366]
3 years ago
6

The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in

centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string
Physics
1 answer:
zhannawk [14.2K]3 years ago
6 0

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

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kobusy [5.1K]

Answer:

(a) Spring constant =  1850 N/m.

(b) Energy stored in the spring = 9.25 J.

(c) Muzzle velocity of the dart =  43.01 m/s.

Explanation:

(a) Spring constant

From hook's law,

F = ke ...................... Equation 1

Where F = force on the gun, k = spring constant of the gun, e = extension of the gun's spring.

Making k  subject of the equation,

k = F/e ............... Equation 2

Given: F = 185 N, e = 10 cm = 0.1 m.

Substitute into equation 2

k = 185/0.1

k = 185/0.1

k = 1850 N/m.

Hence the spring constant = 1850 N/m.

(b) Energy stored in the spring,

E = 1/2ke²............... Equation 3

Where E = Energy stored in the spring, k = spring constant, e = extension.

Given: k = 1850 N/m, e = 0.1 m.

Substitute into equation 3

E = 1/2(1850)(0.1)²

E = 925(0.01)

E = 9.25 J.

(c) Muzzle velocity of the dart.

Kinetic energy of the dart = 1/2mv²

Note: The kinetic energy of the dart is equal to the energy stored in the sprig.

E = 1/2mv²

Where m = mass of the dart, v = velocity of the dart.

Making v the subject of the equation,

v = √(2E/m)............... Equation 4

Given: E = 9.25 J, m = 10 g = 0.01 kg

Substituting these values into equation 4

v = √(2×9.25/0.01)

v = √(18.5/0.01)

v = √1850

v = 43.01 m/s.

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Answer:

Everything

Explanation:

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Answer:

The perceived frequency is higher than the actual emitted sound frequency. that means that the received sound waves are of shorter wavelength.

Explanation:

When the source of a sound wave is moving toward the observer, the perceived frequency of the wave changes in relation to the observer producing a change in pitch. The effect is called Doppler effect in honor of the physicist who formulated the physical explanation.

In the case of the sound source approaching the observer, the perceived frequency is higher than that actually emitted by the sound source.

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Answer:

Option (c) Acceleration is 1/2x the original acceleration

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Case 1:

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Divide both side by 2m

a₂ = 20 / 2m

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Acceleration 1 (a₁) = 20 / m

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a₂ / a₁ = (10 / m) × (m / 20)

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a₂ = ½a₁

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