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madreJ [45]
3 years ago
12

A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

rcular motion is a constant 50.00 cm, and the time it takes the stopper to make one full revolution is 0.2500 seconds, what is the tangential speed of the stopper?
Physics
1 answer:
alexandr402 [8]3 years ago
3 0

Answer:

v = 12.57 m/s

Explanation:

As we know that the radius of the circular motion is given as

R = 50.00 cm

time period of the motion is given as

T = 0.2500 s

now we know that it is moving with uniform speed

so it is given as

v = \frac{2\pi R}{T}

now plug in all data

v = \frac{2\pi(0.50)}{0.25}

v = 12.57 m/s

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Answer:

ωi = 15.4 rev/sec

Explanation:

Since the movement of the fan is rotating, we are thus dealing with Rotational motion. In rotational motion, for angular speed to take place also means angular acceleration is also occurring.

angular acceleration = α = (change in speed)/(change in time)

angular acceleration = α = Δw/Δt = (ω - ωi) /(t- t₀) ..........(equation 1)

                                      α =  (ω -ωi) /(t- 0)

                                      α =  (ω-ωi) /t

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We replace the values for ω, t and α

ωi = 105 rad/sec - ( 4.4 rad/sec² )(1.85s) = 96.86 rad/s = 15.415747788 rev/sec

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The main component of all computer memory is
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Hi!

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3 years ago
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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

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q_1 = 47.0\mu C

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by relation for electric field we have following relation

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FROM FIGURE

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E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

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