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MissTica
3 years ago
8

What is the force on an electron in a CRT when it’s moving at 2.5 × 105 meters/second perpendicular to a magnetic field of 1.5 t

eslas? The charge for an electron is -1.6 × 10-19 coulombs.

Physics
1 answer:
Firdavs [7]3 years ago
4 0
F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

f = |q| \times v \times b \times sin \: \theta

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons
Newton (N) = C x m/s x T = (C x m x T)/s

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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th
Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0
3 years ago
Please do all of i will give you brainlest and thanks to best answer
vlabodo [156]

Answer:the answer is A

Explanation:

3 0
3 years ago
Which statement is correct
daser333 [38]

C Weight is the gravitational pull on an object

8 0
3 years ago
a feather is dropped on the moon from a height of 1.40meters. the acceleration of gravity on the moon is 1.67m/s^2. determine th
kykrilka [37]

Answer:

1min since there is no gravity on the moon so it will take time to drop on the moon.

Explanation:

6 0
3 years ago
In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev
Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

6 0
3 years ago
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