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serg [7]
4 years ago
6

Some passengers on an ocean cruise may suffer from motion sickness as the ship rocks back and forth on the waves. At one positio

n on the ship, passengers experience a vertical motion of amplitude 1.0 m with a period of 13 s .
What fraction is this of g?
What is the maximum acceleration of the passengers during this motion?

Physics
2 answers:
Eddi Din [679]4 years ago
8 0

Explanation:

Below is an attachment containing the solution to the question.

Vera_Pavlovna [14]4 years ago
5 0

Answer: 1936/8281 m/ssquare

Maximum acceleration = 0.23m/ssquare

Explanation: using the formular f=1/T, frequency is calculated as 1/13hertz.

But g=(2Πf)square × A

Where Π=pie=22/7, f is the frequency=1/13, and A is the amplitude =1 m .

By insputing the values in the formula, we have g= (2/13×22/7)square × 1

This will give 4/169×22/7×22/7

Which is equal to 1936/8281 m/ssquare

And when divided gives 0.23m/ssquare as the maximum acceleration.

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Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
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Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
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