Ask question

Ask question

All categories

3 years ago

13

The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?

1 answer:

ICE Princess25 [194]3 years ago

3 0

Answer:

660V

Explanation:

V=IR

V=?,I=11A,R=60w

V=60×11

=660V

You might be interested in

Just asking how everyone is doing cause of the virus

Bess [88]

Answer:

We are all fiine

Explanation:

How are you doing also

3 0

3 years ago

Read 2 more answers

IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0

3 years ago

Read 2 more answers

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

<h3>b. </h3>

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

A motor does 8000j of work in 20 seconds. What is the power of the motor

lozanna [386]

Power = work / time = 8000J / 20s = 400W

5 0

3 years ago

A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0

a_sh-v [17]

Answer:

a. $-12.7 Nm$

b. $-7.9 rad/s^2$

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

Mass of solid disk, $M = 24.3kg$

and radius of solid disk, $r = 0.364m$

a.) The formula for determining torque ( $T$ ), is $T = r * F$

Hence the net torque produced by the two forces is given as a summation of both forces:

$T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm$

b.) The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

We then relate the torque and angular acceleration ( $\alpha$ ) with the formula:

$T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2$

4 0

4 years ago