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3 years ago

Why does a stationary electromagnet attached to an AC source induce current in a wire coil?

Physics

1 answer:

lora16 [44]3 years ago

4 0

This is basically Michael Faraday's law and this is known as electromagnetic induction

That's all I know

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Question 1 Multiple Choice Worth 2 points)

Vinil7 [7]

Answer:

2 atoms of oxygen in carbon dioxide

Explanation:

4 0

3 years ago

Will give brainliest.... Physics with work please

marishachu [46]

Answer:

so first wrote down

Explanation:

that will be concluded as the answer

6 0

2 years ago

Which question could be tested in a scientific manner?

andrey2020 [161]

The questions from this lot which could be tested in a scientific manner would be "what causes some people to be color-blind"
and
"what are the best shoes to wear when exercising"
Both of these questions can be tested in a scientific way through an experiment.

5 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

8 0

3 years ago