Answer:
Under normal conditions, a magnetic material like iron doesn't behave like a magnet because the domains don't have a preferred direction of alignment. On the other hand, the domains of a magnet (or a magnetized iron) are all aligned in s specific direction.
<span>3933 watts
At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3
The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams.
Rounding to 4 significant figures gives me 4705 grams of water.
The heat of vaporization for water is 2257 J/g. So the total energy applied is
2257 J/g * 4705 g = 10619185 J
Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds.
Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So
10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3)
= 3933 watts</span>
It would have to be cardio i think not totally sure
The heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises and the cooler air from the sea moves in to replace the risen air. The correct answer is option A
There will be heat transfer from a region of higher temperature to the region of lower temperature. But in the case of land and sea breeze, the transfer of heat are the result of convectional current in nature. Because the land is a better absorber of heat and also has a lower specific heat capacity compare to sea, during the day, the heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises.
The cooler air from the sea moves in to replace the risen air.
Why do ocean winds or sea breezes blow toward shore during the day ? It is because air over the beach heats up, rises and is replaced by ocean air.
Therefore, option A is correct
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Answer:
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
Where 
and 
are the orbital periods of Mercury and Earth respectively. We have 
and 
. Replacing this and solving for
