Answer:
the velocity of the car is 0.875 m/s
Explanation:
therefore the V of car is 0.875 m
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2 answers:
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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
Given Information:
Potential difference = V = 100 V
Capacitance C₁ = 10 mF
Capacitance C₂ = 5 mF
Capacitance C₃ = 4 mF
Required Information:
a. Charge q₃
b. Potential difference V₃
c. Stored energy U₃
d. Charge q ₁
e. Potential difference V₁
f. Stored energy U₁
g. Charge q ₂
h. Potential difference V₂
i. Stored energy U₂
Answer:
a. Charge q₃ = 0.4 C
b. Potential difference V₃ = 100 V
c. Stored energy U₃ = 20 J
d. Charge q ₁ = 0.33 C
e. Potential difference V₁ = 33 V
f. Stored energy U₁ = 5.445 J
g. Charge q ₂ = 0.33 C
h. Potential difference V₂ = 66 V
i. Stored energy U₂ = 10.89 J
Explanation:
Please refer to the circuit attached in the diagram
a. Charge q₃
As we know charge in a capacitor is given by
q₃ = C₃V₃
q₃ = 4x10⁻³*100
q₃ = 0.4 C
b. Potential difference V₃
The potential difference V₃ is same as V
V₃ = 100 V
c. Stored energy U₃
Energy stored in a capacitor is given by
U₃ = ½C₃V₃²
U₃ = ½*4x10⁻³*100²
U₃ = 20 J
d. Charge q ₁
Since capacitor C₁ and C₂ are in series their equivalent capacitance is
Ceq = C₁*C₂/C₁ + C₂
Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³
Ceq = 3.33x10⁻³ F
q ₁ = Ceq*V
q ₁ = 3.33x10⁻³*100
q ₁ = 0.33 C
e. Potential difference V₁
V₁ = q ₁/C₁
V₁ = 0.33/10x10⁻³
V₁ = 33 V
f. Stored energy U₁
U₁ = ½C₁V₁²
U₁ = ½*10x10⁻³*(33)²
U₁ = 5.445 J
g. Charge q ₂
q₂ = Ceq*V
q₂ = 3.33x10⁻³*100
q₂ = 0.33 C
h. Potential difference V₂
V₂ = q ₂/C₂
V₂ = 0.33/5x10⁻³
V₂ = 66 V
i. Stored energy U₂
U₂ = ½C₂V₂²
U₂ = ½*5x10⁻³*(66)²
U₂ = 10.89 J
