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2 years ago

How was the speed of light determined

2 answers:

5 0

Presently, the speed of light in a vacuum is defined to be exactly 299,792,458 m/s (approximately 186,282 miles per second). . An early experiment to measure the speed of light was conducted by Ole Romer, a Danish physicist, in 1676. Using a telescope, Ole observed the motions of Jupiter and one of its moons, Io

klemol [59]2 years ago

3 0

Answer:

Speed of light = 3 * 10 ^8 m / s. Speed is related to the distance and time and so its unit is in meters and seconds. The speed of light value mentioned above is the value calculated in vacuum or air with the help of lasers.

But when it comes to water, refractive index has to be taken into account.The speed of light is faster in vacuum than other mediums.

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Brums [2.3K]

A and B are the correct answer for this

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3 years ago

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Please write something What to do in case of a fire. ~write a few sentences~

Zarrin [17]

Follow stop drop and roll if the fire is in the room
Other wise exit immediately and find the nearest fire point. Then when you are out phone the fire services.
I hope this helps :)

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2 years ago

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A person runs with a constant velocity of 1.2 m/s for 30 seconds. how far do they travel?

Nataliya [291]

They will travel 36 m

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3 years ago

A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac

zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

( because at moon g will become g/6)

As we know that

$F=\dfrac{K}{x^2}$

Here x= 1100 miles

F 2 tons

$2=\dfrac{K}{1100^2}$

$K=2.4\times 10^6$

We know that

Work = F. dx

$W=\int_{x_1}^{x_2}F.dx$

$W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx$

$W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}$

$W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]$

W=76.55 miles.metric tons

6 0

3 years ago

What is the equivalent resistance of the

BigorU [14]

Answer:

Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

Since $R_{1}$ and $R_{2}$ are connected in parallel, the voltage across the two resistors would both be $V$ . Thus, the current going through the two resistors would be $(V / R_{1})$ and $(V / R_{2})$ , respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .

7 0

2 years ago