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morpeh [17]
3 years ago
5

What is a wave frequency? A. the number of waves that pass a fixed point in a fixed period of time B. the speed that a wave is p

assed from particle to particle through a medium C. the loudness of a wave as it travels though the air D. the height of a wave between peak and trough
Physics
2 answers:
Tanzania [10]3 years ago
8 0
Frequency - number of waves (vibrations, revolutions, of cycles) that pass a given point per unit time.

the correct answer should be A

hope this helps :)

Tema [17]3 years ago
7 0

Answer:

A. the number of waves that pass a fixed point in a fixed period of time

Explanation:

A wave is a disturbance that travels in a medium. Frequency of a wave can be defined as the number of waves passing through a fixed time per second. It is measured in Hertz. Wavelength is the distance between two crests or troughs. Amplitude is half height of a wave between peak and trough.

Thus, option A is correct.

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3 years ago
which statements describe oceanic crust? select the two correct answers. a. it is between 25 and 70 km thick. b. its age ranges
Lady_Fox [76]

C & D --------- APEX

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4 years ago
Read 2 more answers
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

C = \frac{855.6}{220^2 \times 2\pi \times 60}

C = 4..689 \times 10^{-5} Faraday

C = 46.891 \mu F

6 0
4 years ago
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6 0
3 years ago
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coe
Rainbow [258]

The desk is in equilbrium, so Newton's second law gives

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

==>   <em>n</em> = <em>mg</em>

==>   <em>p</em> = <em>f</em> = <em>µn</em> = <em>µmg</em> = 0.400 (80.0 kg) <em>g</em> = 313.6 N

The student pushes the desk 3.00 m, so she performs

<em>W</em> = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J

of work.

6 0
3 years ago
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