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1 year ago

A transverse wave on a string is described by the wave function

Physics

1 answer:

svp [43]1 year ago

5 0

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

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Keith_Richards [23]

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

$y=\frac{m\lambda D}{d}$

m: order of the fringes

lambda: wavelength 500nm

D: distance to the screen 5 m

d: separation of the slits 1mm=1*10^{-3}m

With the formula you can calculate the separation of two adjacent slits:

$\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm$

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3 years ago

A potter’s wheel moves from rest to an angular speed of 0.50 rev/s in 28.9 s. Assuming constant angular acceleration, what is it

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Answer:

0.108 rad/s².

Explanation:

Given that

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 0.5 rev/s

We know that

1 rev/s = 6.28 rad/s

ωf= 3.14 rad/s

t= 28.9 s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

3.14 = 0 + α x 28.9

$\alpha=\dfrac{3.14}{28.9}\ rad/s^2\\\alpha=0.108\ rad/s^2$

Therefore the acceleration will be 0.108 rad/s².

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