All categories

3 years ago

A crest in a transverse wave corresponds to a in a longitudinal wave

Physics

2 answers:

Lerok [7]3 years ago

7 0

Answer: A crest in a transverse wave corresponds to a Compression in a longitudinal wave.

Explanation:

A transverse wave has crests and troughs. Crest and trough are the points on the wave to which maximum displacement of medium particles in upward and downward direction occurs respectively. The medium particles vibrate perpendicular to the direction of propagation of wave.

A longitudinal wave has compression and rarefaction. The medium particles vibrate parallel to the direction of propagation of wave. A compression is high density region and rarefaction is a low density region.

A crest is the point on the transverse wave to which the medium particle rises maximum. Correspondingly, in a longitudinal wave, the medium particles come closer to each other and form a denser region. This is known as compression.

Katena32 [7]3 years ago

6 0

Compression..............

You might be interested in

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w

FromTheMoon [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

$\lambda = \frac{1}{1523611.1112}$

$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

$\lambda = 434nm$

For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

7 0

3 years ago

20. Q: How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?

saul85 [17]

Answer:

2.45s

Explanation:

is explanation needed too?

7 0

1 year ago

In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every

Mademuasel [1]

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

Also, we know that an extra solar planet which is also called as an exo planet is that planet which is outside our solar system and orbits any star other than our sun. The system in consideration is extra solar system with a single planet.

Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

4 0

3 years ago

A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The

lesya692 [45]

I think the answer is C

7 0

3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago