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schepotkina [342]
3 years ago
12

A water wave traveling in a straight line on a lake is described by the equation

Physics
1 answer:
ivanzaharov [21]3 years ago
6 0

Answer:

A) The wave equation is given as

y(x,t) = A\cos(kx + \omega t)=(3.3\times 10^{-2})\cos(0.004x + 5.05t)\\

According to the above equation, k = 0.004 and ω = 5.05.

Using the following identities, we can find the period of the wave.

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

T = 1.25 s.

For the horizontal distance travelled by one period of time, x = λ.

\lambda = 2\pi / k = 2\pi / 0.004 = 1.57\times 10^3~m

y(x = \lambda,t = T) = 0.033\cos(0.004*1.57*\10^3 + 5.05*1.25) = 0.033~m

B) The wave number, k = 0.004 . The number of waves per second is the frequency, so f = 0.8.

C) The propagation speed of the wave is

v = \lambda f = 1.57\times 10^3 * 0.8 = 1.256\times 10^3~m/s

The velocity of the wave is the derivative of the position function.

v(x,t) = \frac{dy(x,t)}{dt} = -(5.05\times 0.033)\sin(0.004x + 5.05t)

The maximum velocity is when the derivative of the velocity function is equal to zero.

\frac{dv_y(x,t)}{dt} = -(5.05)^2(0.033)\cos(0.004*1.57\times 10^3 + 5.05t) = 0

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

0.004*1.57\times 10^3 + 5.05t = 5\pi /2\\t = 0.31~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 = 0.31) = -(5.05\times0.033)\sin(0.004\times 1.57\times 10^3 + 5.05\times 0.31) = -0.166~m/s

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ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

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Answer:

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Explanation:

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2 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

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A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi
Oxana [17]

Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A

Therefore, for a  60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

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