As the iron core isprogressively laid down, the mechanism
of iron oxidation changes from a protein dominated
process with H202 being the primary product of
O2 reduction to a mineral surface dominated process
where H20 is the primary product. These results emphasize
the importance of the apoferritin shell in facilitating
iron oxidation in the early stage of iron deposition
prior to significant development of the polynuclear
iron core.
The question is incomplete. The complete question is :
Two conducting spheres are mounted on insulating rods. They both carry some initial electric charge, and are far from any other charge. Their charges are measured. Then, the spheres are allowed to briefly touch, and the charge in one of them (sphere A) is measured again. These are the measured values:
a). Before contact:
Sphere A = 4.8 nC
Sphere B = 0 nC
What is the charge on sphere B after contact, in nC?
b). Before contact:
Sphere A = 2.9 nC
Sphere B = -4.4 nC
What is the charge on sphere B after contact, in nC?
Solution :
It is given that there are two spheres that are conducting and are mounted on an insulating rods which carry a initial charge and they are briefly touched and then one of the charge is measured.
Here the charge becomes divided when both the spheres are connected and then removed.
a). charge after they are charged
= 2.4 nC
b). The charge is
= -0.75 nC
Answer:
I believe the answer is D
Explanation: The doppler affect isan increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other.
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Water, land. breath using skin and lungs
