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3 years ago

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An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to

109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =

1 answer:

mafiozo [28]3 years ago

8 0

Answer:

wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

w_f = w₀ - a t

w₀ = w_f + a t

w₀ = 109 + 4.7 1.87

w₀ = 117.8 rad / s

let's reduce to revolutions / s

w₀ = 117.8 rad / s (1 rev / 2pi rad)

w₀ = 18.75 rev / s

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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

geniusboy [140]

Answer:

$E=\frac{K}{ed}$

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done= $Force\times displacement$

Using the formula

Work done= $qE\times d=-eEd$

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

$E=\frac{K}{ed}$

Hence, the magnitude of the electric field= $E=\frac{K}{ed}$

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3 years ago

A 420-turn circular coil with an area of 0.0650 m2 is mounted on a rotating frame, which turns at a rate of 22.3 rad/s in the pr

Ivenika [448]

Answer:

33.48 V

Explanation:

Parameters given:

Number of turns, N = 420

Magnetic field strength, B = 0.055 T

Area, A = 0.065 m²

Angular velocity, ω = 22.3 rad/s

EMF induced in a coil is given as:

EMF = -dΦ/dt

where Φ = magnetic flux

Magnetic flux, Φ, is given as:

Φ = B * N * A * cosωt

EMF = -d( B * N * A * cosωt) / dt

EMF = B * N * A * ω * sinωt

where ωt = 90°

Therefore:

EMF = 0.055 * 420 * 0.065 * 22.3 * sin90°

EMF = 33.48 V

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3 years ago

Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in

Radda [10]

Answer:

The value is $t = 14.129 \ minutes$

Explanation:

From the question we are told that

The distance of planet Tatoone is $d = 1.7 \ AU = 1.7 *1.496* 10^{11}=2.543*10^{11} \ m$

The speed of light is $c = 3.0*10^{8} \ m/ s$

Generally the time taken is mathematically represented as

$t = \frac{d}{c}$

=> $t = \frac{2.543*10^{11}}{3.0*10^{8} }$

=> $t = 847.7 \ s$

Now converting to minutes

$t = \frac{847.7}{60}$

=> $t = 14.129 \ minutes$

8 0

3 years ago

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

$v=0+(9.8)(0.75)=7.35 m/s$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

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Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

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$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

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$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

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3 years ago