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beks73 [17]
3 years ago
15

A puck of mass 0.5100.510kg is attached to the end of a cord 0.8270.827m long. The puck moves in a horizontal circle without fri

ction. If the cord can withstand a maximum tension of 126126N, what is the highest frequency at which the puck can go around the circle without the cord breaking?
Physics
1 answer:
yanalaym [24]3 years ago
6 0

Answer: 2.75 1/sec

Explanation:

The only external force (neglecting gravity) acting on the puck, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r (1)

Our unknown, is the frequency at which the puck can go around the circle, which is the inverse of the period Tp.

By definition, a period is the time needed by the puck to complete one entire circle.

By definition also , angular velocity is the rate of change of the angle advanced, so we can express this way:

ω = ∆θ / ∆t  

The angle advanced during one period, is exactly (by angle definition) 2 π radians.

So, we can always write the angular velocity, ω, as follows:

ω = 2π / Tp = 2πf

Now, there is a relationship between linear and angular velocity, that can be found applying simply the definition of velocity and of an angle too, as follows:

v = ∆s / ∆t = r ∆θ/∆t = ω r

Replacing in (1), we have:

T = mω2 r2 / r = m ω2r (2)

We have just found that ω= 2πf, so, replacing in (2) :

T = m (2π)2 f2 r  

Solving for f:

f = 1/2π√(T/mr) = 1/2π 17.28 1/sec = 2.75 1/sec

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Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
e-lub [12.9K]

Answer:

f = 421.8 Hz

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}

so it will be

\Delta L = 12.5 - 12.093

\Delta L = 0.4066

now we know that

\frac{\lambda}{2} = 0.4066

\lambda = 0.813

now frequency of sound is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.813}

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3 years ago
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What type of plate boundary decreases the amount of the Earth's crust?
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