There are multiple reasons for this. First of all, water is available in almost every place on the Earth. It doesn't pollute the air, doesn't cause health use and is easily handle.
Other factor is the fact that water has a really high specific heat. This means that water, and more specifically steam, can aborb and transport more energy. A lower heat capacity would imply the need to boil more of the liquid to obtain the same amount of energy. This combine with the fact that water expands at a large rate when boiling, combine with everything mentioned previously, and you get a liquid with all the characteristics that a efficient turbine requires to work.
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:
Fg=Fcp
m*g=m*(v²/R),
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.
Masses cancel out and we have:
G*(M/r²)=v²/R, R=r so:
G*(M/r)=v²
r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,
r=G*(M/ω²r²),
r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:
r³=G*(M/(4π²/T²)), and finally we take the third root to get r:
r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.
“The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.”
Answer:
3 mA.
Explanation:
The following data were obtained from the question:
Resistor (R) = 500 Ω
Potential difference (V) = 1.5 V
Current (I) =.?
Using the ohm's law equation, we can obtain the current as follow:
V = IR
1.5 = I x 500
Divide both side by 500
I = 1.5 / 500
I = 3×10¯³ A.
Therefore, the current in the circuit is 3×10¯³ A.
Finally, we shall convert 3×10¯³ A to milliampere (mA).
This can be obtained as follow:
Recall:
1 A = 1000 mA
Therefore,
3×10¯³ A = 3×10¯³ × 1000 = 3 mA
Therefore, 3×10¯³ A is equivalent to 3 mA.
Thus, the current in mA flowing through the circuit is 3 mA.
