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nydimaria [60]
3 years ago
11

.It takes time 8 min 20 seconds for light to reach from sun to the earth surface.If speed of light is taken to be 3times10^(5)km

/h.Find the distance of the sun from the earth in SI system.​
Physics
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

hjjiisisijjwjj

Explanation:

jjsjqiqu##uwuuw

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Pls help. This is about heat. I got the answer, just need the working
Mars2501 [29]

Answer:

See below

Explanation:

You have to heat the calorimeter to 100 C  from 20 C

  this will take   .20 kg *    390 j /kg-C   *  80 C = <u>6240 j</u>

You have to heat the mass of water to boiling point (100 C ) from 20C

this will take    

    .50 kg * 4182 j/kg-C  * 80 = <u>167,280 j </u>

AND you have to add enough heat to boil off  .03 kg of water:

   .03 kg * (2260000 j/kg-C )  =<u> 67,800 j</u>

<u />

Power = joules / sec =  (6240 + 167280 + 67800) / 274.8 =<u> 878 watts </u>

<u />

<u>Your answer may differ just a bit for slightly different or rounded values of specific heat or heat of fusion for water .....</u>

7 0
2 years ago
The caste system in India is illegal. True or False
anzhelika [568]
<span>Discrimination is illegal, but caste system is legal.
So answer: False</span>
4 0
3 years ago
PLEASE SOMEONE ANSWER!!! thank you so so so much!!!!
Oduvanchick [21]

Answer:

As much I know the gravity on moon is 1.62m/s२.

7 0
3 years ago
One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i
Musya8 [376]

The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

First we calculate the fuel consumed by each car in a year

Fuel consumed=6990/21.4=326.63 gallon

Now we calculate the amount of fuel consumed by 40 million cars in a year

Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

4 0
3 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
3 years ago
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