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maria [59]
2 years ago
6

When a beam of charged particles moves through a magnetic field, what is the evidence that particles in the beam have momentum g

reater than the value mv?
Physics
1 answer:
Vsevolod [243]2 years ago
4 0

The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.

A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.

Thus, we can see that the charged particles are often deflated in a magnetic field.

Learn more about magnetic field:brainly.com/question/23096032

#SPJ1

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7 0
3 years ago
What is heat?
OLEGan [10]
Well you need to have lots of heat
7 0
4 years ago
If a vehicle accelerating at 2.7 m/s2what is it's velocity at 20 meters 12.63m/s,1.03m/s,10.39m/s,6.39m/s
netineya [11]

Answer:

The final velocity of the vehicle is 10.39 m/s.

Explanation:

Given;

acceleration of the vehicle, a = 2.7 m/s²

distance moved by the vehicle, d = 20 m

The final velocity of the vehicle is calculated using the following kinematic equation;

v² = u² + 2ah

v² = 0 + 2 x 2.7 x 20

v² = 108

v = √108

v = 10.39 m/s

Therefore, the final velocity of the vehicle is 10.39 m/s.

5 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

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Firlakuza [10]

Answer:

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Explanation:

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7 0
3 years ago
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