+2 electron charges = 2x1.6x10^-19Coulombs
Answer:
22,800 years
Explanation:
Half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.0625 = (½)^(t / 5700)
log 0.0625 = (t / 5700) log 0.5
4 = t / 5700
t = 22,800
It takes 22,800 years.
Answer:
Explanation:
a. The source of centripetal force on the car is (3) the static friction force.
b. ac = v²/R = (20²)/50 = 8 m/s²
c. Fc = m(ac) = 1500(8) = 12 kN
d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
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
