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3 years ago

0. A 1700-W machine operates at 120 V. What i

Physics

1 answer:

kap26 [50]3 years ago

4 0

Option D

The resistance in machine is 8.5 ohms

Solution:

Given that,

Power = 1700 W

Voltage = 120 V

We have to find the resistance in machine

The formula used is:

$P = \frac{V^2}{R}$

Where,

"P" is the electric power and "V" is the voltage and "R" is the resistance

Substituting the given values we get,

$1700 = \frac{120^2}{R}\\\\1700 = \frac{14400}{R}\\\\R = \frac{14400}{1700}\\\\R = \frac{144}{17}\\\\R = 8.47 \approx 8.5$

Thus resistance in machine is 8.5 ohms

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Rufina [12.5K]

Answer:

The angular speed is $w = 5.89 \ rad/s$

Explanation:

From the question we are told that

The time taken is $t = 1.6 s$

The number of somersaults is n = 1.5

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substituting values

$\theta = 1.5 * 2 * 3.142$

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substituting values

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3 0

3 years ago

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

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$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

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3 years ago

A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and ampl

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Answer:

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3 years ago

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The correct option to the question is Matter.

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Moreover, Matter can be described as,

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8 0

3 years ago

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
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$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
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You can also use the equation of linear motion,
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$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
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$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
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4 0

3 years ago