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sukhopar [10]
3 years ago
15

100Gw x 35ms (note: a watt is a joule per second)

Physics
1 answer:
Leviafan [203]3 years ago
7 0

'G'  =  "giga..."  =  billion  =  10⁹

'ms'  =  "millisecond"  =  0.001 second

100 Gw = 10¹¹ watts

35 ms  =  0.035 second

100 Gw x 35 ms  =  10¹¹ watts x 0.035 second  =  3.5 x 10⁹ J

                                                                     =  3.5 G-joules
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How are the climates of coastal regions affected by the specific heat capacity of water?
Katen [24]

Answer:

The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.

This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows  from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.

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3 years ago
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

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b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

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