Ask question

Ask question

All categories

3 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:

tatiyna3 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 100.0 mL sample of 1.020 M HCl is mixed with a 50.0 mL sample of 2.040 M NaOH in a Styrofoam cup. If both solutions were initi

I am Lyosha [343]

Answer:

$t_2 = 33.793$

Explanation:

Given data:

0.1 L HCl × 1.020 mol/L = 0.102 mol HCL

0.05 L NaOH × 2.040 mol/L = 0.102 mol NaOH

NaOH + HCl \rightarrow H_2O + NaCl

0.102 mol HCl /1 mol HCl × 1 mol H2O = 0.102 mol H2O

0.102 mol H2O / 1 mol H2O × 57000 J = 5814 J

$V(t ) = V_1+V_2 = 100 + 50 = 150 mL$

M(t)=150*1=150 g

we know that heat energy is given as

$Q = mc(t_2 - t_1)$

where c is specific heat

total energy released is = 57 \times 0.102 = 5.814 kJ

$5814 = 150 \times 4.184 \times (t_2 - 24.53)$

$t_2 = 33.793$

8 0

3 years ago

Examples of positive effects of fluid forces<br><br><br> note: and pls I don't want the definition

densk [106]

Answer:

slider,topspin,hook,curveballs

4 0

3 years ago

Read 2 more answers

When a bow is stretched back and an arrow is shot, what type of energy conversion has occurred?

Lelechka [254]

D. mechanical to elastic

6 0

3 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

5 0

3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago