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1 year ago

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A sinusoidal wave in a string is described by the wave functiony=0.150 sin (0.800x - 50.0t)where x and y are in meters and t is

in seconds. The mass per length of the string is 12.0 g/m(a) Find the maximum transverse acceleration of an element of this string

1 answer:

AleksAgata [21]1 year ago

6 0

The maximum transverse acceleration of an element of this string is 375 m/s

<h3>What is the maximum transverse acceleration?</h3>

For a transverse wave, the velocity is known to be perpendicular to the direction of the propagation of the particular wave. To get the acceleration, what is necessary is to take the partial derivative as it compares to time and the velocity.

With this in mind, we would have

transverse acceleration

a(t) = d²y/dt² = -0.150*50.02sin(0.8x - 50t)

= 0.150*502 = 375 m/s2

Hence we can say that the maximum acceleration of the element on the string is given as 375 m/s²

Read more on acceleration here

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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

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The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

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The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

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The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

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The force exerted on the third charge by the first is

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Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

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$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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