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agasfer [191]
3 years ago
6

Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .

Physics
1 answer:
Vlad [161]3 years ago
5 0
<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

               = 4 - (-5)

               = 9 cm

Thus;

work done = 95 × 9/100

                  <u>= 8.55 Joules </u>

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
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What is the velocity of a wave with a frequency of 45 Hertz and a wavelength of 3 meters?
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The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
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On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
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<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

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<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
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