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tamaranim1 [39]
2 years ago
6

A coin is located 35.0 mm from the center of a turntable which accelerates from rest to a speed of 7.95 rad/s in 0.270 s. What i

s the minimum coefficient of friction to avoid slippage of the coin
Physics
1 answer:
beks73 [17]2 years ago
5 0

The minimum coefficient of friction to avoid slippage of the coin is 0.23.

<h3>Minimum coefficient of friction</h3>

The minimum coefficient of friction can determined from the maximum speed of the coin.

ma = μmg

mv²/r = μmg

v²/r = μg

v² = μgr

μ = v²/(gr)

μ = (ωr)²/(gr)

μ = ω²r/g

μ = (7.95² x 0.035)/(9.8)

μ = 0.23

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ11

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8. not sure sorry ; (

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4 0
3 years ago
A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli
vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

x=V_{o} cos\theta t (1)

y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2} (2)

Where:

x=0.25 m is the horizontal displacement of the pencil

V_{o} is the pencil's initial velocity

\theta=0\° since we are told the pencil rolls <u>horizontally</u> before falling

t is the time since the pencil falls until it hits the ground

y_{o}=1 m  is the initial height of the pencil

y=0  is the final height of the pencil (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity, always acting vertically downwards

Begining with (1):

x=V_{o} cos(0\°) t (3)

x=V_{o}t (4)

Finding t from (2):

0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2} (5)

t=\sqrt{\frac{-2y_{o}}{g}} (6)

Substituting (6) in (4):

x=V_{o}\sqrt{\frac{-2y_{o}}{g}} (7)

Isolating V_{o}:

V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}} (8)

V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}} (9)

Finally:

V_{o}=0.55 m/s

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