The average speed of a plane was 600 km/hr. How long did it take

What affects the amount of potential energy stored in the magnetic field when a magnet is moved against a magnetic force?

LekaFEV [45]

Answer: Strength of magnet and distance from magnetic material

Explanation:

The potential energy of a magnet is determined by the strength of the magnet and the distance between a magnet and another magnet or a magnetic material. Magnetic materials are materials that would be attracted when brought close to a magnet, example of magnetic materials are most metals.

8 0

3 years ago

True or false atoms may be held together by an exchange of electrons

diamong [38]

The answer is true...............

8 0

3 years ago

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HELP BRAINLYEST TO CORRECT ANSWER

kow [346]

Answer:

physical

Explanation:

how thick something is is a physical trait, so it's a physical property!

3 0

3 years ago

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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

$\tau = I\alpha$

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