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Ksivusya [100]
4 years ago
14

What quantity measures the number of complete cycles an oscillation makes per second? A. period B. amplitude C. frequency D. for

ce
Physics
1 answer:
Free_Kalibri [48]4 years ago
8 0

Hello, there Jcparris

Your answer is going to be C. Frequency

If my answer helped please leave a thank rate 5 stars and the most important rank me braiinliest thank you and have a great day!

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Evaporation. The glass of water was less full after 10 minuets because the water got evaporated




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A certain radio wave has a frequency of 2.0 x 10^6 Hz. What is its wavelength?
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<span>1.    </span>Electromagnetic radiation are represented in waves. Each type of wave has a certain shape and length. The distance between two peaks in a wave is called the wavelength. This value is equal to the speed of light divided by the frequency.

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Wavelength = 3x10^8 / </span><span>2.0 x 10^6</span>

Wavelength = 150 m

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4 years ago
One star has a mass slightly more than the mass of the sun. Which of these graphs would best model and predict the lifetime of t
Oksanka [162]

Answer:

b

Explanation:

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4 years ago
Weight is measured in: grams, kilograms, metric tons, or pounds
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8 0
3 years ago
A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,
kolbaska11 [484]

Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

3 0
3 years ago
Read 2 more answers
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