Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s
Answer:
The peak current carried by the axon is 5.85 x 10⁻⁸ A
Explanation:
Given;
distance of the field from the axon, r = 1.3 mm
peak magnetic field strength, B = 9 x 10⁻¹² T
To determine the peak current carried by the axon, apply the following equation;
where;
B is the peak magnetic field
r is the distance of the magnetic field from axon
μ is permeability of free space = 4π x 10⁻⁷
I is the peak current
Re-arrange the equation and solve for "I"
Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A
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:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
The formula for kinetic energy is
. Thus, the equation for velocity is
.
