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3 years ago

Which of the following could be units of speed? A. m B. m/s C. m north D. degrees north

2 answers:

34kurt3 years ago

7 0

Since velocity is vector quantity, it'll depend on both direction as well as magnitude hence options C and D are rejected.
Speed = distance/time
Therefore, unit of velocity is m/s.

patriot [66]3 years ago

4 0

B. M/s because north is to to with locating areas degrees north is temperature and speed is not measure in meters

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3 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Sergio039 [100]

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

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2 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

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3 years ago

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Answer:

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3 years ago

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