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3 years ago

If the amplitude of a wave is 4 feet, what is the Wave Height?

2 answers:

5 0

A wave is a disturbance that moves along a medium from one end to the other. If one watches an ocean wave moving along the medium (the ocean water), one can observe that the crest of the wave is moving from one location to another over a given interval of time. The crest is observed to cover distance. The speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled per time of travel. In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form,

Vesnalui [34]3 years ago

3 0

Explanation :

The height of the wave is defined as the elevation of the crest minus the elevation of the trough.

The wave height is defined as the twice of the amplitude.

It is given that the amplitude of the wave is 4 feet. The height of the wave is given as :

$H=2a$

Where a is the amplitude of the wave.

So, H = 8 feet.

Hence, this is the required solution.

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A wave produced by a doorbell

Vlad [161]

A wave produced by a door bell?
There is a sound wave yes.

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3 years ago

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3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t

grigory [225]

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2 is 10000K .

4 0

3 years ago

As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.09 mm apart and position

kow [346]

Answer:

pretty dark

Explanation:

i dont know

4 0

3 years ago