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marin [14]
2 years ago
11

A circular disk is rotating about a central axis with an angular velocity which is increasing at a constant rate. Point 1 on the

disk is located at a distance r1 from the center of the disk and point 2 on the disk is located at a distance r2 from the center of the disk. If r2 = 2r1, which of the following ratios is the largest? 1 and 2 refer to being at points 1 and 2 in the subscripts below, all values occurring at the same time.
a. ω2/ω1
b. α2/α1
c. ac2/ac1
d. All the ratios are 1 so none is larger.
Physics
1 answer:
antiseptic1488 [7]2 years ago
8 0

The largest ratio of the circular disk at the two points is ac2/ac1.

The given parameters:

  • Location of point 1 = r1
  • Location of point 2 = r2 = 2r1

<h3>What is angular speed?</h3>
  • The angular speed of an object is the rate of change of angular displacement of the object.

The angular speed of each disk is given as;

\omega = 2\pi N

where;

  • N is the number of revolutions

thus, the angular speed is independent of radius of the circular disk

ω2/ω1 = 1

the centripetal acceleration of each disk is calculated as follows;

a_c = \frac{v^2}{r} \\\\a_c_1 = \frac{v^2}{r_1} \\\\a_c_2 = \frac{v^2}{2r_1} \\\\\frac{a_c_2}{a_c_1} = \frac{2r_1}{v^2} \times \frac{v^2}{r_1} = 2

the angular  acceleration of the disk is calculated as follows;

\alpha = \frac{\omega }{t}

the angular acceleration is independent of the radius of the disk.

α2/α1 = 1

Thus, the largest ratio of the circular disk at the two points is ac2/ac1.

Learn more about centripetal acceleration here: brainly.com/question/79801

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Sergio039 [100]
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  • Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
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It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>
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5 0
1 year ago
What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a
Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

==> T= 38.38 N

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Answer:

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Explanation:

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JulijaS [17]
The answer is definitely C.) their molecules move at the same average speed
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Yakvenalex [24]

Answer:

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3 0
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