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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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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
ddd [48]

Answer:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force exerted by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
2 years ago
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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T
atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.

<em>where</em>,

k is the Coulomb's constant, having value = \dfrac{1}{4\pi \epsilon_o}

\epsilon_o is the electrical permittivity of free space.

Also, r>>s, therefore, \rm r^2+s^2\approx r^2.

\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

\rm F=qE

Therefore, the electric force on the charge Q due to the dipole is given by

\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole is given by

\rm \tau = Fs\ \sin\theta

Since the charge Q is placed in the plane that bisects the dipole, therefore, \theta = 90^\circ.

\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.

4 0
2 years ago
A grasshopper jumps at a 65.0 degree angle at 5.42m/s. At what time does it reach its maximum height?
crimeas [40]
When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.
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2 years ago
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Nanotechnology and microtechnology devices are both very tiny. However, one property of nanotechnology distinguishes it from mic
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An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces a
marshall27 [118]
Refer to the free body diagram shown melow.

F =  applied force
R =  frictional force
m = 15 kg, the mass of the object

The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N

The normal reaction is
N = mg = (15 kg)*(9.8 m/s²)  = 147 N
The frictional force is
R = μN = 147μ N,  where μ =  coefficient of kinetic friction.

Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N,   (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)

Answer:
The most reasonable answer is R = 5.5 N

8 0
2 years ago
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