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Gnoma [55]
3 years ago
5

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

the wave is to transfer energy at a rate of 75 W, what should the amplitude of the wave be?
Physics
1 answer:
matrenka [14]3 years ago
8 0

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

V = \sqrt{\frac{T}{\mu}}

Our values are given as:

f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w

Replacing we have that the velocity is

V = \sqrt{\frac{T}{\mu}}

V = \sqrt{\frac{65}{0.230}}

V = 16.81m/s

From the theory of wave propagation the average power wave is given as

P =\frac{1}{2} \mu \omega^2 A^2 V

Where,

A = Amplitude

\omega = 2\pi f \rightarrow Angular velocity

A^2 = \frac{2P}{\mu \omega^2 V}

A^2 = \frac{2P}{\mu (2\pi f)^2 V}

Replacing,

A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}

A = 0.0165m

Therefore the amplitude of the wave should be 0.0165m

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Answer:

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3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

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8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

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b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

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