I thought you were going to ask for the resistance of the unknown
series resistor. Since you only want the equivalent resistance of the
circuit, you don't even need to know the resistance of the lamp.
I = E / R
Current through the circuit = (voltage of the battery) / (circuit resistance).
0.5 = (12) / R
Multiply each side by 'R' : (0.5) R = 12
Multiply each side by 2 : <em>R = 24 ohms</em>
(Since the resistance of the lamp is 10 ohms, the
unknown series resistor is the other 14 ohms.)
Answer:
The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.
The depth of the water has to be equal to 7.07 m in order to have a stationary volume.
Explanation:
In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:
0.002*h^2 = 0.1
h^2 = 0.1/0.002
h^2 = 50
h = sqrt(50) = 7.07 m
Answer:
Aluminium, Nickel, Cobalt and Iron plus varying levels of Copper, Titanium and Niobium.
MAGNETIC: Alni, Alcomax, Hycomax, Columax, and Ticonal.
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:


We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:

And the velocity achieved in part 1:

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

Then, to get
, we do:



And we substitute the values:
