All categories

2 years ago

A transverse wave on a long horizontal rope with a

Physics

1 answer:

frosja888 [35]2 years ago

4 0

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and speed by the equation

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

where

f is the frequency

v is the speed of the wave

$\lambda$ is the wavelength

For the wave in this problem,

v = 2 m/s

$\lambda=8 m$

So the frequency is

$f=\frac{2}{8}=0.25 Hz$

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

$T=\frac{1}{f}=\frac{1}{0.25}=4 s$

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

$t=\frac{T}{2}=\frac{4}{2}=2 s$

You might be interested in

Heeeeeeeeeeeeeeeeeelp

RoseWind [281]

Im pretty sure the answer would be thermometer

7 0

2 years ago

Leverrier predicted that an invisible planet was pulling the planet Uranus off its predicted course around the sun. Likewise, hu

Alexandra [31]

Human beings are pulled off course as a result of the invisible forces of the <u>unconscious.</u>

According to Leverrier, he stated that an invisible planet was pulling the planet Uranus off its predicted course around the sun.

In such a way, human beings are pulled off course by the invisible forces of their unconscious minds. The unconscious minds of people control the thoughts of people.

Read related link on:

brainly.com/question/25588203

3 0

2 years ago

A student produces a wave in a long spring by

Sliva [168]

Answer: 2

Explanation:

4 0

2 years ago

You are operating a pwc in an area where swimmers are in the water. When must you slow your pwc to "slow, no wake speed"?.

Lostsunrise [7]

You should slow your pwc to "slow, no wake speed" when within 100 feet of anchored vessels or non-motorized craft.

This is the process of operating a personal watercraft at the slowest possible speed.

This helps to maintain steerage which prevents different forms of accident or risks when in motion in the water.

Read more about Slow-no-wake here brainly.com/question/10410716

#SPJ1

4 0

1 year ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago