Ask question

Ask question

All categories

3 years ago

5

Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di

stance between the.two consecutive crest on the string as 5 cm. What was the velocity of the wave?

1 answer:

Goshia [24]3 years ago

6 0

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v = 250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

You might be interested in

A box experiences a force of 2 N to the left and 3 N to the right. Which is true of the box's motion?

Nostrana [21]

Below are the choices:

<span> A) The box will slow down.
B) The box's velocity will be 1 m/s.
C) The box's velocity will not change.
D) The box will experience acceleration
</span>
The answer is D) The box will experience acceleration

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

5 0

3 years ago

Read 2 more answers

A sculpture is suspended in equilibrium by two cables, one from a wall and the other

aleksandrvk [35]

Answer:

$T_1=6655.295917 \approx 6655.3N$

Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

7 0

3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

7 0

4 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

Consider the following scenario: You live near the shoreline and hear a report that an earthquake hit the ocean floor near your

Harlamova29_29 [7]

A. a tsunami
(if the earthquake is hitting the ocean, the water will get effected)

5 0

4 years ago

Read 2 more answers