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:
One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m
These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.
Applying Gauss's theorem
Flux through spherical surface = charge inside sphere / ε₀
= 0 / ε₀
= 0 Ans .
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
Answer:
Explanation:
Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:
where:
variation of the magnetic flux
is the variation of time
#The magnetic flux through the coil is expressed as:
Where:
N- number of circular loops
A-is the Area of each loop(
)
B-is the magnetic strength of the field.
- is the angle between the direction of the magnetic field and the normal to the area of the coil.
=0.0250T/s is given as rate at which the magnetic field increases.
#Substitute in the emf equation:
Hence, the induced emf is
Answer: The reference frame of a passenger in a seat near the center of the train
Explanation:
the speed of light is the same for the passenger and the bicyclist
then the avents are simultaneous fo the passenger not for the bicyclist
the delay between the two events for the bicyclist is
Δt=Δd/vs
where
Δd= lenght of train
vs=speed of sound
the reference frame of a passenger in a seat near the center of the train
Solution:
The space and time transformations are:
x' = γ(x - vt)
t' = γ(t - vx/c²).
In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)
We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)
Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =
30/(2c - 0.5c) = 6.7 x 10^(-8)s
