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1 year ago

8

We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e

arth-sun distances are 3. 8 x 105 km and 1. 5 x 108 km

1 answer:

Tju [1.3M]1 year ago

6 0

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>

So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

$d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m$

As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

$t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min$

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

brainly.com/question/11495758

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Explanation:

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3 years ago

In which of the fire plume structure zones do the gases accelerate upward?

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3 years ago

1. What happens when like charges come together?*

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When like charges come together, they repel each other. For instance, when the north and south poles of a magnet come together, they push each other apart. The like poles in the magnet repel each other and unlike poles attract each other much. The same reaction occurs in like and unlike charges. Also, the repulsion acts along the line between the two charges.

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3 years ago

A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

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Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

$EMF = 0.1 V/s$

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

n = number of turns per unit length

$n = \frac{N}{L}$

$n = \frac{350}{0.20}$

$n = 1750 turn/m$

now we know that magnetic field due to solenoid is

$B = (4\pi \times 10^{-7})(1750)(0.100)$

$B = 2.2 \times 10^{-4} T$

Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

$\phi = (2.2 \times 10^{-4})(\pi r^2)$

$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

$\phi_{total} = N\phi$

$\phi_{total} = 20(2.76 \times 10^{-4})$

$\phi_{total} = 5.52 \times 10^{-6}$

now we have

$M = \frac{5.52 \times 10^{-6}}{0.100}$

$M = 5.52 \times 10^{-5} H$

Part C)

As we know that induced EMF is given as

$EMF = M \frac{di}{dt}$

$EMF = 5.52 \times 10^{-5} (1800)$

$EMF = 0.1 V/s$

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3 years ago