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3 years ago

A wave with a short wave length will also have...?

1 answer:

3 0

It'll have a higher frequency.

The product of (wavelength) times (frequency) for a wave
is always the same number ... it's the wave speed.
So if one of them is small, the other one has to be big.

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A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago

This chemical equation represents a ______________ reaction.

Ede4ka [16]

Answer:

you didnt put anything

Explanation:

7 0

3 years ago

Read 2 more answers

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.