Ask question

Ask question

All categories

2 years ago

14

A wave has a wavelength of 4. 9 m and a velocity of 9. 8 m/s. The medium through which this wave is traveling is then heated so

that the velocity is doubled. If the frequency remains constant, what is the wavelength of the heated wave? 2. 5 m 9. 8 m 14. 7 m 19. 6 m.

1 answer:

tatiyna2 years ago

3 0

The wavelength of a wave is obtained by taking the ratio of wave speed and frequency.

The wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

What is frequency of a wave?

The number of oscillations completed by a wave in one second is known as the frequency of a wave. It is expressed as the ratio of the velocity of the wave to its wavelength.

Given data-

The wavelength of the wave is, $\lambda = 4.9 \;\rm m$ .

The velocity of the wave is, v = 9.8 m/s.

The mathematical expression for the frequency of the wave is,

$f = \dfrac{v}{\lambda}$

Solving as,

$f = \dfrac{9.9}{4.9}\\\\f =2 \;\rm Hz$

Now, with constant frequency and the double magnitude of velocity (v' = 2 × 9.8 = 19.6 m/s). The wavelength of the heated wave is calculated as,

$f = \dfrac{v'}{\lambda'}$

here,

$\lambda'$ is the wavelength of the heated wave.

Solving as,

$2 = \dfrac{19.6}{\lambda'}\\\\\lambda' = \dfrac{19.6}{2}\\\\\lambda' = 9.8 \;\rm m$

Thus, we can conclude that the wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

Learn more about the frequency of wave here:

brainly.com/question/1324797

You might be interested in

A substance that does not react with metal, has a bitter taste, and turns litmus paper blue is a(n) ___.

Pavlova-9 [17]

The answer to the question is b because acid is a substance that tastes sour, reacts with metal and turns litmus paper blue

5 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

2 years ago

Which of these statements best describes an effect of changes in Earth's magnetic field?

Sveta_85 [38]

The only statement that makes sense as to which one describes an effect of changes in Earth's magnetic field is the first one - <span>Southern lights are seen at high altitude places like Alaska.
Normally, Southern lights are seen around the South Pole, which means that it is quite odd for them to be seen in Alaska, which is near the North Pole.</span>

6 0

3 years ago

Read 2 more answers

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$

So now, we see that $v_{m} < 0$ and $v_{M} > 0$ . So therefore, the smaller mass recoils out.

Hope this helps you!

Bye!

7 0

3 years ago

The relationship between frequency wavelength and speed holds for light waves because

tamaranim1 [39]

Answer:

all forms of electromagnetic radiation travel at a single speed in a vacuum.

Explanation:

5 0

2 years ago