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N76 [4]
3 years ago
11

What is the Lorentz force law used for?

Physics
2 answers:
alexira [117]3 years ago
8 0

Answer:

Lorentz force, the force exerted on a charged particle q moving with velocity v through an electric E and magnetic field B. The entire electromagnetic force F on the charged particle is called the Lorentz force (after the Dutch physicist Hendrik A. Lorentz) and is given by F = qE + qv × B.

Explanation:

N/A

Mekhanik [1.2K]3 years ago
7 0

Answer:

To find the magnitude of a magnetic force.

Explanation:

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What part of the electromagnetic spectrum do you think is most important and why?
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Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th
andrew11 [14]

Answer:

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

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3 years ago
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

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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