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Genrish500 [490]
3 years ago
5

A. The distance to a star is approximately 8.58 ✕ 10^18 m. If this star were to burn out today, in how many years would we see i

t disappear?
b. How long does it take sunlight to reach Earth?
c. How long does it take for a microwave radar signal to travel from Earth to the Moon and back?
Physics
1 answer:
Tamiku [17]3 years ago
7 0

Answer:

906.27931 years

8.34 minutes

2.56266 seconds

Explanation:

v = Velocity of light = 3\times 10^8\ m/s

s = Distance

Time is given by

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{8.58\times 10^{18}}{3\times 10^8\times 365.25\times 24\times 3600}\\\Rightarrow t=906.27931\ years

We would see the star disappear after 906.27931 years after it actually disappeared.

Distance between Earth and Sun = 150.18 billion m

t=\dfrac{150.18\times 10^9}{3\times 10^8\times 60}\\\Rightarrow t=8.34\ minutes

It takes 8.34 minutes for sunlight to reach Earth.

Distance between Earth and Moon = 384.4 million m

t=\dfrac{2\times 384.4\times 10^6}{3\times 10^8}\\\Rightarrow t=2.56266\ s

It takes 2.56266 seconds for a microwave radar signal to travel from Earth to the Moon and back.

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Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

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We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

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We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

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Hence, The voltage across the capacitor is 1.57 V.

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