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Over [174]
3 years ago
5

A turntable spins up from rest with an angular acceleration of 1.0 rad/s2. A rubber eraser of mass 20 g with coefficientof stati

c friction 1.0 is at a distance of 0.1 m from the center of the turntable, when it begins to spin. How long does ittake for the eraser to begin to slide off the turntable? Note that the tangential acceleration of the eraser is very small.
Physics
1 answer:
atroni [7]3 years ago
5 0

Answer:

9.89 seconds

Explanation:

Angular acceleration, α = 1 rad/s^2

mass of eraser, m = 20 g

μ = 1

r = 0.1 m

here, the centripetal force is balanced by the frictional force

m x r x ω^2 = μ m x g

r x ω^2 = μ g

ω^2 = 1 x 9.8 / 0.1 = 98

ω = 9.89 rad/s

Here, ωo = 0

Use

ω = ωo + α x t

9.89 = 0 + 1 x t

t = 9.89 seconds

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Answer:

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Explanation:

if    V ⇒ voltage

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V = IR

220 = 10 x R

220 / 10 = R

22 = R

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14. It takes an airplane nearly ¾ of a mile to stop. Which law of motion is being used?
swat32
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A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

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d)through a railroad track

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