Anyone know what type of circuit this is?

How large is the area of water found on Mars?

ehidna [41]

More than five million cubic kilometers of ice have been identified.

8 0

3 years ago

An AC generator consists of 20 circular loops of wire with an area of 75 cm2. It has a maximum induced voltage of 24 V. If its a

Monica [59]

Faraday's law allows us to find the magnetic field that produces the emf in the rotating system is:

The magnetic field is: B = 0.424 T

Faraday's law of induction states that when the magnetic flux changes in time, an induced electromotive force is produced.

fem = $- \frac{d \Phi_B }{dt}$

where fem is the induced electromotive force and Ф the flux,

The magnetic flux is the scalar product of the field and the area.

$\Phi_B = B . A = B A \ cos \theta$

In this case we have several turns, so the expression remains.

fem = $- N B A \ \frac{d cos \theta}{dt}$

Indicate that the turns rotate at a constant frequency, therefore we can use the uniform rotational motion ratio.

θ = w t

We substitute

$fem = - N B A \ \frac{d \ cos \ wt}{dt}\\fem = N B A w sin \ wt$

the maximum induced electromotive force occurs when the sine function is ±1

fem = N B A w

They indicate that the fem = 24 V, the number of the turn is N = 20, the area is A = 75 cm² = 75 10⁻⁴ m² and the frequency f = 60 Hz

Frequency and angular velocity are related.

w = 2π f

We substitute.

fem = N B A 2π f

$B = \frac{fem }{2 \pi \ NA \ f}$

Let's calculate.

$B= \frac{24 }{2\pi \ 20 \ 75 \ 10^{-4} 60}$ B = 24 / 2pi 20 75 10-4 60

B = 0.424 T

In conclusion, using Faraday's law we can find the magnetic field that produces the emf in the rotating system is:

The magnetic field is; B = 0.424 T

Learn more about Faraday's law here: brainly.com/question/24617581

8 0

2 years ago

A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

Ann [662]

Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = $n \times \frac{1}{4} \lambda$ ...............1

so

total length will be here

$L = \frac{\lambda}{2} + \frac{\lambda}{4}\\$

$L = \frac{3 \lambda }{4}$

so $\lambda$ will be

$\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}$

$\lambda$ = 40 cm

5 0

3 years ago

How does a bimetallic strip work in a sinple alarm system

tatiyna

The bimetallic strip in a fire alarm is made of two metals with different expansion rates bonded together to form one piece of metal. Typically, the low-expansion side is made of a nickel-iron alloy called Invar, while the high-expansion side is an alloy of copper or nickel. The strip is electrically energized with a low-voltage current. When the strip is heated by fire, the high-expansion side bends the strip toward an electrical contact. When the strip touches that contact, it completes a circuit that triggers the alarm to sound. The width of the gap between the contacts determines the temperature that will set off the alarm.

7 0

2 years ago

Read 2 more answers

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

2 years ago