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3 years ago

Alex has to pick up a 20 N stack of documents and raise them 5 meters from the floor.

Physics

1 answer:

Serjik [45]3 years ago

6 0

Answer:

1.Work done=100 joules

2.potential energy

3.kinetic energy

Explanation:

W=Force x distance

w= 20 N x 5 M

W = 100 JOULES

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A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h

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2 years ago

Which best describes most covalent compounds? O soft O brittle 3 O cold O warm

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3 years ago

A summary of a source is usually

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Answer:

A shorter than the original source and in the researcher's words

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The summary is an abridged version of the original source and in the researcher's own very words.

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A summary should not be detailed and must avoid overt illustrations. They must capture the true essence of piece leaving out flowery details.

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3 years ago

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2 years ago

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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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2 years ago