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Xelga [282]
2 years ago
8

A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's

Physics
1 answer:
9966 [12]2 years ago
6 0

By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is

<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²

where

<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant

• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet

Solve for <em>M</em> :

<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)

<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))

<em>M</em> ≈ 2.8 × 10¹⁴ kg

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3 years ago
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

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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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