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klio [65]
3 years ago
10

A child holds one end of a 33.0-meter long rope in her hand and moves it up-and-down to produce a sinusoidal wave by moving her

hand from 8.00 cm above her shoulder to 8.00 cm below her shoulder at a frequency of 2.00 Hz and a wavelength of 75.0 cm. If the child doubles the amplitude of her hand's motion on that same rope with the same tension in the rope, then what will the wavelength of the wave now be
Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

Wavelength=75 cm.

The wavelength well remain unchanged which is 75 cm.

Explanation:

The formula which will help us to answer the question is:

V=f*λ

Where:

V is the velocity

f is the frequency of wave

λ is the wave length

Now:

λ=V/f    Eq (1)

The equation show's that wavelength is independent of the amplitude but it depends on the frequency and the velocity with which wave is moving.

The wavelength well remain unchanged which is 75 cm.

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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

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3 years ago
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Explanation:

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Answer:

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Explanation:

Apply the formula:

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W = 5000 . 400.000

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I think it is that, can be wrong.

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Answer:

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So, the blanks should be filled with:

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