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DiKsa [7]
3 years ago
7

1. How would the motion of a pendulum change at high altitude like a high mountain top? How would the motion change under weight

less conditions?
Physics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

A pendulum moves with the principle of small angles that show that the frequency and period of the pendulum are not dependent on the initial angular displacement of the mass it carries.

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

Explanation:

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

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A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p
MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) f=1617.9\ Hz

(b) T=0.618\ ms

(c) A=20 \ Volts

(d) \varphi=60^o

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

V(t)&=A\cdot \sin(2\pi ft+\varphi )

Where

A=Amplitude

f=frequency

\varphi=Phase\  angle

The voltage of the question is the sinusoid expression  

V(t)=20cos(5\pi\times 103t+60^o)

(a) By comparing with the general formula we have

f=5\pi\times 103=1617.9\ Hz

\boxed{f=1617.9\ Hz}

(b) The period is the reciprocal of the frequency:

\displaystyle T=\frac{1}{f}

\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec

Converting to milliseconds

\boxed{T=0.618\ ms}

(c) The amplitude is

\boxed{A=20 \ Volts}

(d) Phase angle:

\boxed{\varphi=60^o}

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We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

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From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

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