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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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<h3>What can be the challenges of wave power?</h3>

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The major challenges that face the use of wave power in electricity generation is the unreliability of the waves which leads to uncertainty in the quantity of power generated Also, the wave direction and direction of ocean currents all limit the amount of power generated by this method. However, in spite of challenges, it can be regarded as a reliable source of energy because the ocean currents are always moving.

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6 0
1 year ago
Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p
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Answer:

Energy density will be 14.73 J/m^3

Explanation:

We have given capacitance C=225\mu F=225\times 10^{-6}F

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm 0.2\times 10^{-3}m

We know that there is relation between electric field and potential

E=\frac{V}{d}, here E is electric field, V is potential and d is separation between the plates

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3 years ago
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl
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Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

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F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )

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The buoyant force of copper block

F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219  \ N

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