Any one their to help

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

You are retrieving your boat from the water. as a courtesy to others, when should you pull your boat into a launch lane?

OLga [1]

The best way to pull the boat in the launch lane is when the towing vehicle has reached the boat ramp as well as the trailer. It is best to retrieve the boat in this manner for it is more appropriate, convenient and it is a way of showing courtesy towards others because it is a proper manner.

7 0

3 years ago

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1. The statement about Newton's reflecting telescope contains 1 error. Find the error, write it, and

Jlenok [28]

Answer:

i. The error is the rough convex mirror.

ii. This should be replaced with a smooth convex morror.

Explanation:

Reflection is dependent on the surface involved and has two types; diffuse and specular. When the surface is rough, diffused reflection is observed. The surface causes a distortion of the incident light (the rays would be reflected at different angles to one another) after reflection. This makes some rays to interfere with one another. While specular reflection is observed with a smooth surface.

In the statement, the rough convex mirror would produce a distorted reflection which would produce diffused reflection. The effect is that few or no rays (depending on the degree of how rough the surfce is) would be reflected to the other smooth, flat diagonal mirror.

8 0

2 years ago

Which of the following describes the effect of a convex mirror on light rays?

natulia [17]

B. It reflects light rays outward.

A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.

3 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

SIZIF [17.4K]

Answer:

GFCI outlets are found in wet areas.

GFCI outlets prevent electrocution if you are touching a wet appliance.

5 0

3 years ago

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