Displacement/distance metres
Time seconds
Force Newtons
Energy Joules
Voltage Volts
Current intensity Amperes
Resistance Ohms
Light intensity Candella
Pressure Pascals
Charge Coulombs
Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
<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>
A :-) it was given the name Newton (N). from this, the derived unit of energy (or work) is defined ,as the work produced when the unit of force causes a displacement equal to the unit of length of its point of application along its direction . It was given the name Joule (J).
All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>
