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3 years ago

What is the distance from the crest to the equilibrium of a wave called?

Physics

2 answers:

diamong [38]3 years ago

5 0

The answer would be amplitude. Amplitude is the largest displacement of a wave particle from its equilibrium position.

ss7ja [257]3 years ago

4 0

Answer:

Amplitude

Explanation:

The amplitude of a wave is a measure of the maximum variation of a physical quantity that changes periodically. It is the distance between the wave crest to its equilibrium point. For a mechanical wave the amplitude is a displacement and has units of length.

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Ions form bonds by __________ electrons.

goldenfox [79]

I would say it’s “donating” so giving up...

3 0

3 years ago

Read 2 more answers

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Anastaziya [24]

Answer:

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7 0

3 years ago

An object is located 10.0 cm from a convex mirror. The magnitude of the

Sunny_sXe [5.5K]

Answer:

B. 6.00 cm

Explanation:

6 0

3 years ago

In what way would a digital thermometer be preferable to those from a liquid-based thermometer?

garik1379 [7]

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5 0

3 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

4 years ago