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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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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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2 years ago
A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
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Answer:

a)R = 171μΩ

b)E = 1.7 *10^{-4} V/m

c)R_{2} = 1.16 *10^{-4}Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder  =  1.1 cm

resistivity depends on the distance x from the left

p(x)=a+bx^2 ............(i)

using equation

R = \frac{pl}{a}

let dR is the resistance of thickness dx

dR =\frac{p(x)dx}{a}

where p(x) is resistivity  l is length

a is area

\int\limits^R_0 {dR}  =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx  \\.........................(2)

after integration

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}  ...............(3)

it is given p(0) = a = 2.25 * 10 ^{-8}Ωm

p(L) = a + b(L)^2  = 8.5 * 10 ^{-8} Ωm

8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\

(here * stand for multiplication )

on solving we get

b = 2.78* 10^{-8} Ωm

put each value of a  and b and r value in equation 3rd we get

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}

R = 1.71 * 10^{-4}Ω

R = 171μΩ

FOR (b)

for mid point  x = L/2

E = p(x)L

for x = L/2

p(L/2) = a+b(L/2)^2

for given current  I = 1.75 A

so electric field

 

E = \frac{[a+b(L/2)^2]I }{\pi  r^2}

by substitute the values

we get;

E = 1.7 *10^{-4} V/m

(here * stand for multiplication )

c ).

75 cm means length will be half

 that is   x =  L/2

integrate  the second equation with upper limit  L/2  

Let resistance is R_{1}

so after integration we get

R_{1}  =  \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}

substitute the value of a , b and L we get

R_{1} = 5.47 * 10 ^{-5}Ω

for second half resistance

R_{2} =  R- R_{1}

R_{2}  = 1.7 *10^{-4} -5.47 *10^{-5}

R_{2} = 1.16 *10^{-4}Ω

(here * stand for multiplication )

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The question is missing the diagram. Also, the choices must have pictorial representation. So, I have attached the missing diagram and the pictorial representation of the vectors.

Answer:

The correct representation is attached below. Force and acceleration will be towards the center of rotation while the velocity will be along the tangent to the circular motion. <u>Option (D).</u>

Explanation:

From the figure, we can conclude the following points:

1. The cylinder is under a uniform circular motion as the circular table is moving at constant speed.

2. For a circular motion, velocity acts along the tangent to the circular path.

3. For a circular motion, centripetal force acts on the body that causes it move around a circular path.

4 The direction of the centripetal force is radially inward towards the center of rotation.

5. The centripetal force causes a centripetal acceleration acting on the body.

6. From Newton's second law, the net acceleration of a body is in the same direction as that of the net force acting on it. So, centripetal acceleration also acts in the radially inward direction.

Therefore, from the above conclusions, it is clear that velocity will act in the horizontal direction at the given instance of time and force and acceleration will act vertically down for the given instance.

This is shown in the picture below. The option (D).

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