Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 72.0° between them. Wha

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

kicyunya [3294]

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

$V_h = \frac{iB}{neL}$

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

As a result, the charge carrier density is $6.6*10^{27}e/m^3$

5 0

2 months ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

2 months ago

A positive charge moves in the direction of an electric field. Which of the following statements are true?

kicyunya [3294]

Answer:

The potential energy tied to the charge diminishes.

The electric field performs negative work on the charge.

Explanation:

3 0

2 months ago

Read 2 more answers

If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,

Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0

1 month ago