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1 year ago

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A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the ve

ry small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)Assume that the tension of the string is constant and equal to W. (a) How much time does it take a pulse to travel the fulllength of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

1 answer:

Natalka [10]1 year ago

5 0

Answer:

Explanation:

mass of string = .0125 / 9.8

= 1.275 x 10⁻³ kg

Length of string l = 1.5 m .

m = mass per unit length

= ( .1.275 / 1.5) x 10⁻³ kg/m

m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = $\sqrt{\frac{W}{m } }$

28.08 = $\sqrt{\frac{W}{.85\times10^{-3} } }$

788.48 = W / .85 X 10⁻³

W = 670 x 10⁻³ N .

c ) wave length

wave length =2π / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

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A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate a

qaws [65]

Answer:

impedance Z = 416.66 ohm

voltage across inducance V = 346.99 V

power factor = 0.720

Explanation:

given data

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voltage amplitude = 500 V

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to find out

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we apply here average power formula that is

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I = $\frac{Vrms}{Z}$

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impedance Z = 416.66 ohm

and

we know voltage across inductor is here express as

V = I × X .............2

so here X will be by inductance

Z² = R² + (X)²

(X)² = 416.66² - 300²

X = 289.15 ohm

and I = $\frac{V}{Z}$

I = $\frac{500}{416.66}$

I = 1.20 A

so from equation 2

V = 1.20 × 289.15

voltage across inducance V = 346.99 V

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1 year ago

HELP!!! HURRY DUE BY 4:30!!!

Ipatiy [6.2K]

Work = force * distance.
<span>You must produce twice as much energy as we are lifting the weight twice as high. </span>
<span>But because you aren't increasing the force, you need to increase the length of the ramp instead. </span>
<span>The new length will be twice as great as the previous length. </span>
<span>So 8 metres is required. </span>

Hope this helps.

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