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Kay [80]
3 years ago
13

A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot

ion. Hz
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

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Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res
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Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

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Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through
Norma-Jean [14]

Answer:

1.\theta=29.84^{0}

2.\theta=60.15^{0}

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of  first polarizer

which is given as

I=\frac{I_{0} }{2}

I= \frac{655\frac{W}{M^{2} } }{2}

I=327.5\frac{W}{m^{2} }

For a smaller angle for the first polarizer:

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I_{2} =I_{1} Cos^{2}(90^{0} - \theta)

I_{2} =I_{1} sin^{2}\theta

\frac{I_{2} }{I_{1} }=Sin^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = sin\theta

\theta=Sin^{-1}(0.4977)

\theta=29.84^{0}

For a larger angle for the first polarizer:

According to Malus Law

I_{2} =I_{1} cos^{2}\theta

\frac{I_{2} }{I_{1} }=Cos^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = cos\theta

\theta=Cos^{-1}(0.4977)

\theta=60.15^{0}

7 0
3 years ago
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