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3 years ago

(01.01 LC) Two boxes are 8 cm apart. Which of the following should Janet do to

Physics

2 answers:

Tanya [424]3 years ago

6 0

Hello,

The correct answer is....

B) Place the boxes 10 cm apart.

Hope this helps!!!!

**(Vanessa)**

Trava [24]3 years ago

4 0

Place the boxes 10cm apart, if they are closer that concentrates the mass of the boxes more

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Please help on this one?

dezoksy [38]

the answer is heat engine

3 0

3 years ago

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = $\frac{energy}{time}$

The energy needed to light one bulb= $power*time$

Given time = 1minute = 60 seconds

Energy = $32W*60sec$ =1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = $1920*24$ =46080J=46.08KJ

3 0

3 years ago

Read 2 more answers

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$