All categories

1 year ago

Which statements correctly describe the relationship between a solenoid and an electromagnet? Check all that apply.

Physics

2 answers:

prohojiy [21]1 year ago

8 0

Answer:

B. A solenoid carries an electric current.

C. An electromagnet is a solenoid with a metal core.

E. A solenoid induces a magnetic field in a metal core.

Explanation:

An electroiman is a large number of wire turns (solenoid) that are wrapped around a magnetic core made of a ferromagnetic or ferrimagnetic material, such as iron. The metal core is composed of "tiny imans", when a current is passed through the solenoid, its magnetic field penetrates the iron, and causes the "tiny imans" to aligning parallel to the magnetic field, creating a large magnetic field.

valkas [14]1 year ago

7 0

Answer:

B, C, E on edgeenuity

Explanation:

You might be interested in

In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat

Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

$\Delta T =$ The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

$Q_2 = 2*Q_1$

Replacing,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solvinf for $T_o$ ,

$T_o = -13.18$

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C

3 0

2 years ago

Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There

mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π) ΔV² / r⁴

Explanation:

a) For this exercise we can use Coulomb's law

F = - k Q² / r²

where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

Capacitance is defined by

C = Q / ΔV

Q = C ΔV

also the capacitance for a parallel plate capacitor is related to its shape

C = ε₀ A / r

we substitute

Q = ε₀ A ΔV / r

we substitute in the force equation

F = k (ε₀ A ΔV / r)² / r²

k = 1 / 4πε₀

F = ε₀ /4π L² ΔV² / r⁴4

F = L² ΔV² ε₀/ (4π r⁴)

F = (L² ε₀/4π) ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

F = q E

where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface

Ф = ∫E .dA = $q_{int}$ / ε₀

the plate have two side

2E A = q_{int} / ε₀

E = σ / 2ε₀

σ = q_{int} / A

substituting in force

F = q σ / 2ε₀

the charge total on the other plate is

q = σ A

q = σ L²

F= σ² L² /2ε₀

4 0

2 years ago

568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim

AlladinOne [14]

The answer to this question is:

C-"That moving clocks run slower"

Your Welcome :)

6 0

2 years ago

Read 2 more answers

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

A freight train has a mass of $1.83\times 10^{7}\,kg$ . The wheels of the locomotive push back on the tracks with a constant net force of $7.50\times 10^{5}\,N$ , so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is: