Answer:
The specific gravity of the unkown liquid is 15.
Explanation:
Gauge pressure, at the bottom of the tank in this case, can be calculated from
where 
and 
are the height of the column of oil and the unkown liquid, respectively. Writing for 
, we have
Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.
I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W
<span>Well, It is the aphelion point, When the Earth is farthest away from the Sun, when the Northern Hemisphere is warm. the Earth is closest to the Sun, or at the perihelion, 2 weeks after the June Solstice, when the Northern Hemisphere is enjoying warm summer months. Well this kind of weather is very nice.</span>
