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Sunny_sXe [5.5K]
1 year ago
8

A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 re

volutions before coming to rest.
Find the constant angular acceleration (in rad/s2) of the centrifuge.


______rad/s2

Physics
1 answer:
eduard1 year ago
7 0

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

<h3> Constant angular acceleration</h3>

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

  • ωf is the final angular velocity when the centrifuge stops = 0
  • ωi is the initial angular velocity
  • θ is angular displacement
  • α is angular acceleration

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

#SPJ1

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Thus

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f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

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Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

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Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
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Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

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