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qaws [65]
2 years ago
11

A water wave traveling in a straight line on a lake is described by the equation:y(x,t)=(2.75cm)cos(0.410rad/cm x+6.20rad/s t)Wh

ere y is the displacement perpendicular to the undisturbed surface of the lake. a. How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? b. What are the wave number and the number of waves per second that pass the fisherman? c. How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?
Physics
1 answer:
Georgia [21]2 years ago
6 0

Answer:

A) The wave equation is defined as

y(x,t) = A\cos(kx + \omega t)=0.0275\cos(0.0041x + 6.2t)\\

Using the wave equation we can deduce the wave number and the angular velocity. k = 0.0041 and ω = 6.2.

The time it takes for one complete wave pattern to go past a fisherman is period.

\omega = 2\pi f\\ f = 1/ T

T = 1.01 s.

The horizontal distance the wave crest traveled in one period is

\lambda = 2\pi / k = 2\pi / 0.0041 = 1.53\times 10^3~m

y(x = \lambda,t = T) = 0.0275\cos(0.0041*1.53*\10^3 + 6.2*1.01) = 0.0275~m

B) The wave number, k = 0.0041 . The number of waves per second is the frequency, so f = 0.987.

C) A wave crest travels past the fisherman with the following speed

v = \lambda f = 1.53\times 10^3 * 0.987 = 1.51\times 10^3~m/s

The maximum speed of the cork floater can be calculated as follows.

The velocity of the wave crest is the derivative of the position with respect to time.

v(x,t) = \frac{dy(x,t)}{dt} = -(6.2\times 0.0275)\sin(0.0041x + 6.2t)

The maximum velocity can be found by setting the derivative of the velocity to zero.

\frac{dv_y(x,t)}{dt} = -(6.2)^2(0.0275)\cos(0.0041*1.53\times 10^3 + 6.2t) = 0

In order this to be zero, cosine term must be equal to zero.

0.0041*1.53\times 10^3 + 6.2t = 5\pi /2\\t = 0.255~s

The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.

v(x=\lambda, t = 0.255) = -(6.2\times0.0275)\sin(0.0041\times 1.53\times 10^3 + 6.2\times 0.255) = -0.17~m/s

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KE = (0.5) times the mass times the velocity^2
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square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
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height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
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5 0
2 years ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

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1 year ago
An elevator held by a single cable is ascending but slowing down. Is the work done by tension positive, negative, or zero? What
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Explanation:

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2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.

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Following are the correct options:

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A substance gain energy and changes from solid to gas.

vaporization

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2 years ago
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