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wariber [46]
3 years ago
6

A horizontal wire is stretched with a tension of 98.4 N, and the speed of transverse waves for the wire is 406 m/s. What must th

e amplitude of a traveling wave of frequency 60.5 Hz be for the average power carried by the wave to be 0.391 W? Give an answer in mm. Pay attention to the number of significant figures.
Physics
1 answer:
Anika [276]3 years ago
4 0

Answer:

Amplitude = 4.725 mm

Explanation:

We have got the following values :

T = 98.4 N

T is the wire tension

v = 406 m/s

v is the transverse waves speed

f = 60.5 Hz

f is the frequency

ΔU = 0.391 W

ΔU is the average power carried by the wave

If μ is  the mass per unit length of the wire ⇒ μ = \frac{T}{v^{2}}

If ω is the angular frequency of the wire ⇒ ω = 2π.f

So ω = 2π(60.5 Hz) and  ω units are s^{-1}

The equation that relates all is :

ΔU = (1/2).(T/v^2).(ω^2).(A^2).v

Where A is the wave amplitude

0.391 W=\frac{1}{2}.\frac{98.4 N}{(406\frac{m}{s})^{2}}.(2.pi.(60.5Hz))^{2}.A^{2}.406\frac{m}{s}

This will give us A in meters

A=4.7253(10^{-3})m\\A=4.725 mm

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Answer:

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Explanation:

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HOPE THIS HELPS

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3 years ago
What can the magnetic field be used for​
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Answer:

the magnetic field can be used to make electricity

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3 years ago
A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0
3 years ago
Read 2 more answers
Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm
Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

q_{1} = charge on one sphere

q_{2} = charge on other sphere

q = Total charge on the two spheres = 40 μC

q_{1}+ q_{2} = q

q_{1}+ q_{2} = 40 x 10⁻⁶

q_{1} = (40 x 10⁻⁶) - q_{2}                                   eq-1

r = distance between the two spheres = 50 cm = 0.50 m

F = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

F = \frac{k q_{1} q_{2}}{r^{2}}

2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}

q_{2} = 1.44 x 10⁻⁶ C

7 0
3 years ago
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