Answer:
the correct answue are B, A, C, C, B
Explanation:
1) The electric field is requested, let's approximate the membrane by a parallel plate with surface charge density
E =
E =
E = 5.65 10⁵ N / C
the correct answer is B
2) A calcium ion has two positive charges, so the force applied by each side of the membrane (plate)
F = q E
F = 2 1.6 10⁻¹⁹ 5.65 10⁵
F = 1.8 10⁻¹³ N
the total force is the sum of the force of each membrane and the two forces go to the same side
F = total = 2 F
F_total = 3.6 10⁻¹³ N
the correct answer is A
3) the field and the electric potential are related
ΔV = - E s
ΔV = - 5.65 10⁵ 10 10⁻⁹
ΔV = - 5.65 10⁻³ V
the correct answer is C
4) In the exercise they indicate that the outer wall has a positive charge, therefore, as they indicate that we approximate the system to a capacitor, the inner wall must be negatively charged.
The electric field goes from the positive to the negative charge, which is why it goes from the outer wall to the inner wall
the correct answer is C
5) For this part we use conservation of energy
starting point. On the inside wall, brown
Em₀ = U = qV
final point. On the outside
Em_f = K
energy is conserved
Em₀ = Em_f
q V = K
K = 3 10⁻¹⁵ 5.65 10⁻³
K = 1.7 10⁻¹⁷ J
the correct answer is B
Answer:
I'm pretty sure it shows a direct relationship
Explanation:
If 56.5kJ are needed to raise the temp by 90°C and if the heater is 60% efficient that means that:
60% X y = 56.5kJ
where y is the electrical energy in kJ that the heater will use.
y = 94.2kJ
Answer:

Explanation:
Since, as we know, the potential difference 'ΔV' is the difference of between the Potential energy per unit charge U/qo at one point 'B' to Potential energy per unit charge at other point 'A'. It so happens when a test charge 'qo' moves from point A to B, the potential difference becomes the change of potential energy of the system, i.e.