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o-na [289]
3 years ago
5

Find the frequency of a wave produced by a generator that emits 100 pulses in 2.0 s.

Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

The frequency of the wave is 50 Hz

Explanation:

The frequency of the wave is defined as the number of cycles per second of the wave:

f=\frac{N}{t}

where

N is the number of cycles completed in a time t.

Frequency is measured in Hertz (Hz).

In this problems, the wave has

N = 100 pulses

in

t = 2.0 s

Therefore, its frequency is

f=\frac{100}{2.0}=50 Hz

Learn more about waves and frequency here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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Answer:

c. hot material must be rising from the Sun's hotter interior

Explanation:

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<u>So, the granulation pattern that astronomers have observed on the surface of the Sun tells us that hot material must be rising from the Sun's hotter interior.</u>

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Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r
Naddik [55]

Explanation:

Fgravity = G*(mass1*mass2)/D²

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A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
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