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velikii [3]
2 years ago
10

How do you find the amplitude of a wave?

Physics
2 answers:
Sergio039 [100]2 years ago
6 0

Answer:

To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form y(x,t)=Asin(kx−ωt+ϕ).

The amplitude can be read straight from the equation and is equal to A.

The period of the wave can be derived from the angular frequency (T=2πω).

MatroZZZ [7]2 years ago
4 0

Answer: The person above me is absolutely correct!

Explanation: The formula for calculating amplitude of a wave is:

x = displacement of wave (meter)

A = amplitude

ω = Angular frequency (rad/s)

t = Time period

ϕ = Phase angle

Happy to help, if I did!

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3 years ago
When does sea breeze occur
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3 0
4 years ago
Read 2 more answers
Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?
UNO [17]

Answer:

1.55 m

Explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

V =\frac{k*q}{r}

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m

r2 =\frac{k*q}{V2} = \frac{9e9 N*m2/C2*1.7e-8C}{54V} = 2.83 m

The distance between both points, is just the difference between the radius of both spheres, as follows:

r₂ - r₁ = 1.55 m

5 0
3 years ago
Ohm’s law is I = V ÷ R. What is the current of a circuit with a 3.0-volt battery and 1.0 ohm of resistance? The current of a cir
Angelina_Jolie [31]
Current = voltage \ resistance
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7 0
3 years ago
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liq [111]

Answer:

Option C

Explanation:

Adiabatic cooling systems function similarly to dry cooling systems, but with the incorporation of pre-cooling pads; running water over pre-cooling pads and drawing air through the pads depresses the ambient dry bulb of the incoming air. The depressed dry bulb allows for greater system heat rejection.

he result of this is that adiabatic systems are highly effective in hot, dry environments, while using less water than traditional evaporative units. Adiabatic units also deliver the required cooling capacity in a smaller footprint and/or lower fan motor horsepower than a completely dry cooler/condenser.

5 0
4 years ago
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