Answer:
3.62 V
Explanation:
L = 80 cm = 0.8 m
f = 15 rps
B = 60 m T = 0.060 T
ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s
v = r ω
here, r be the radius of circular path. Here r = length of rod = L
v = 0.80 x 94.2 = 75.36 m/s
The motional emf is given by
e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V
Four electrons are placed at the corner of a square
So we will first find the electrostatic potential at the center of the square
So here it is given as

here
r = distance of corner of the square from it center



now the net potential is given as


now potential energy of alpha particle at this position

Now at the mid point of one of the side
Electrostatic potential is given as

here we know that



now potential is given as


now final potential energy is given as

Now work done in this process is given as



E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V
You should charge it to 1000 volts to store 1.0 J of energy.
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
Answer:
Explanation:
To find the half-lifes of the isotope we need to use the following equation:
(1)
<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>
<em>: is the half-lifes.</em>
By solving equation (1) for t we have:
<u>Having that:</u>
Nt = 450
N₀ = 3150 + 450 = 3600,
The half-lifes of the isotope is:

Therefore, 3 half-lives of the isotope passed since the rock was formed.
I hope it helps you!