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N76 [4]
3 years ago
7

The objective lens in a DVD player must move quickly to keep the laser beam focused on the disc's reflective layer. Why must thi

s lens have a very small mass?
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Explanation:

The small mass of lens in a DVD player accounts for small inertia of the lens. Hence, even on application of force of small magnitude, the lens is rapidly accelerated. Moreover, the small mass of lens also accounts for lower resistance to change in its motion and therefore, again can be rapidly accelerated back and forth.

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Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
jek_recluse [69]

Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

cos\theta=\dfrac{0}{50}

\theta=90^{\circ}

6 0
3 years ago
Why is measuring the duration of a number of swings a better way to determine the period of a pendulum than by measuring a singl
Alekssandra [29.7K]

Answer:

It is to reduce the expected relative error of the measurement.

Explanation:

If there was a way to measure without error, this method would be unnecessary. In practice, the pesky error is always there. The sources are varied: inexact instrument, small inaccuracies in starting/stopping the timer, etc. But, it is reasonable to assume that such an error is random and has an expected spread that is <em>independent</em> of the actual duration of measurement. Under such assumptions, the methods offers a great advantage:

Let ε denote an additive measurement error. Let the error be random, symmetric (negative/positive), distributed in some fixed range independent of the actual measured value. The error represents an additive component in our measurement, i.e., (measurement) = (true value) + (error). In the case of one period T, we get to measure the duration T':

T' = T + \epsilon

so the relative error is

\frac{|T'-T|}{T}=\frac{|T+\epsilon-T|}{T}=\frac{|\epsilon|}{T}

In a separate experiment, suppose you measure n periods. Same error applies:

T_n'=n\cdot T+\epsilon

we can get a single period by dividing the measured value by n:

\frac{T_n'}{n}=\frac{n\cdot T +\epsilon}{n}=T+\frac{\epsilon}{n}

and the relative error of such a result will be:

\frac{|T+\frac{\epsilon}{n}-T|}{T}=\frac{|\epsilon|}{n\cdot T}

which is n times smaller than the relative error of the single measurement above. The more periods are included in the measurement, the smaller the expected error!

5 0
3 years ago
Ajdaifsgodtistizhxtsgoachoach
Norma-Jean [14]

Hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm maybe

8 0
3 years ago
Help??? Phases of the moon. Am I right
Ksenya-84 [330]

Answer:

Yes. You are correct. Great job!

Explanation:

4 0
4 years ago
4. As a 500 N woman sits on the floor, the floor exerts a force on her of
monitta

Answer: 1000

Explanation:

Its double her weight

7 0
3 years ago
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