Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
Final Length = 30 cm
Explanation:
The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:
F = kΔx
where,
F = Force applied
k = spring constant
Δx = change in length of spring
First, we find the spring constant of the spring. For this purpose, we have the following data:
F = 50 N
Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m
Therefore,
50 N = k(0.05 m)
k = 50 N/0.05 m
k = 1000 N/m
Now, we find the change in its length for F = 100 N:
100 N = (1000 N/m)Δx
Δx = (100 N)/(1000 N/m)
Δx = 0.1 m = 10 cm
but,
Δx = Final Length - Initial Length
10 cm = Final Length - 20 cm
Final Length = 10 cm + 20 cm
<u>Final Length = 30 cm</u>
Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms
Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:
I(t)=io*Exp(-t/τ)
and also we consider that io=V/R=(1.5/6.43*10^3)
=233.28 A
then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6
=22.31 ms
Finally the time to reduce the current to 2.57% of its initial value is obtained from:
I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then
ln(0.0257)*τ =-t
t=-ln(0.0257)*τ=81.68 ms
