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nirvana33 [79]
2 years ago
14

For a sine wave depicting simple harmonic motion, the smaller the amplitude of the wave, the smaller the of the pendulum from th

e equilibrium position. The shorter the period, the the pendulum’s rod.
Physics
2 answers:
stiks02 [169]2 years ago
7 0
Displacement   , shorter 
OLga [1]2 years ago
7 0

Answer:

<em>The smaller the amplitude of the wave, the smaller </em><em>the displacement</em><em> of the pendulum from the equilibrium position. The shorter the period, </em><em>the shorte</em><em>r the pendulum’s rod.</em>

<em />

To find the right answer, we could look the analysis of a simple pendulum movement. The relation that defines this movement is:

T=2 \pi \sqrt{\frac{l}{g} }

T: \ period\\l: \ pendulum's \ rod \ length\\g: \ gravity

From the equation, we can see the relation between the period and the length of the pendulum's rod. They are directly proportional, this means that if the period increases, the length increases, or vice versa. So, the shorter the period, the shorter the pendulum's rod, because they are directly proportional.

On the other hand, the amplitude of the harmonic motion refers to the displacement of the object from the equilibrium point which is in the center of the periodic movement. This allow us to deduct that if the amplitude is small, the displacement of the object is small, if the amplitude is big, the displacement from the equilibrium point is bid.

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For circular motion.

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Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m

m = mass = 175g = 0.175kg.

Angular speed, ω = Angle covered / time

                         = 2 revolutions / 1 second

                         = 2 * 2π  radians / 1 second

                         = 4π  radians / second

Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5    Use a calculator

                                                         ≈13.817  m/s²

The magnitude of acceleration ≈13.817  m/s² and it is directed towards the center of rotation.

Tension in the string = m*a

                                   = 0.175*13.817

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5 0
2 years ago
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If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c
Sloan [31]

Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2  into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

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A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
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Answer:

592.92 x 10³ Pa

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Mole of ammonia required = 10 g / 17 =0 .588 moles

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From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

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3 0
2 years ago
A wildebeest and chicken participate in a race over a 2.00km long course. the wildebeest travels at a speed of 16.0m/s and chick
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Answer:

(a)  The distance of the chicken from the finish line is 62.5 m

(b) The stationary time of the wildebeest is 675 s

Explanation:

Given;

total distance traveled by wildebeest and chicken, d = 2 km = 2000 m

speed of the wildebeest, v_w = 16 m/s

speed of the chicken, v_c = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

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t(stationary) = t(chicken) - t(wildebeest)

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t = 675 s

(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?

The time taken for the wildebeest to run 1.6 km (1600 m) is given by;

t = 1600 / 16 = 100 s

The total time spent by the wildebeest before it resumed the race = stationary time + 100s

t (total) = 675 s + 100 s = 775 s

Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m

Thus, the distance of the chicken from the finish line = 2000 m -  1937.5 m

the distance of the chicken from the finish line = 62.5 m

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