Answer:
A dependent variable is a variable that is tested in an experiment. An independent variable is that can be modified. Depending on what you are testing, the dependent variable will change accordingly to the dependent variable.
- I'm reading this back and it doesn't make much sense, if you want me to reword this I can
Detailed Explanation:
1) Rusting of Iron
4Fe + 3O2 + 2H2O -> 2Fe2O32H2O
Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4
Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4
2) Fermentation of sucrose…
C12H22O11 + H2O -> 4C2H5OH + 4CO2
Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12
Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12
Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.
All the Best!
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
Answer:
s = 30330.7 m = 30.33 km
Explanation:
First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:
v = v₀√[T/273 k]
where,
v = speed of sound at given temperature = ?
v₀ = speed of sound at 0°C = 331 m/s
T = Given Temperature = 10°C + 273 = 283 k
Therefore,
v = (331 m/s)√[283 k/273 k]
v = 337 m/s
Now, we use the following formula to calculate the distance traveled by sound:
s = vt
where,
s = distance traveled = ?
t = time taken = 90 s
Therefore,
s = (337 m/s)(90 s)
<u>s = 30330.7 m = 30.33 km</u>
