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Dennis_Churaev [7]
3 years ago
9

An object moves in simple harmonic motion described by the equation d equals one fifth sine 2 t where t is measured in seconds a

nd d in inches. Find the maximum​ displacement, the​ frequency, and the time required for one cycle.
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

The maximum​ displacement, the​ frequency, and the time required for one cycle are 1/5 inch, 1/π Hz and π sec.

Explanation:

Given that,

The equation of simple harmonic motion

d = \dfrac{1}{5}\sin 2t.....(I)

We need to calculate the maximum amplitude

Using equation of simple harmonic motion

y = a \sin\omega t

Where, a = amplitude

\omega =frequency

t = time

On comparing equation (I) and general equation

The amplitude is a maximum displacement traveled by a wave.

a = \dfrac{1}{5}

So, the maximum displacement is

d= \dfrac{1}{5}\ inch

We need to calculate the frequency

\omega=2\pi f

f = \dfrac{1}{\pi}\ Hz

We need to calculate the time required for one cycle

t =\dfrac{1}{f}

t =\pi\ sec

Hence, The maximum​ displacement, the​ frequency, and the time required for one cycle are 1/5 inch, 1/π\ Hz and π\ sec.

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If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

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in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th
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R = 73.25 m

Explanation:

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Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.
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Answer: FR=2.330kN

Explanation:

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Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

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Answer:

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In the given case electron is given and the magnitude of charge on electron is e=1.6\times 10^{-19}C

Electric field can be represented as,

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As we know that electric field start from the positive charge and vanish in the negative charge.

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