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New answer posted
a year agoContributor-Level 10
Electric field squared is proportional to the power P of the bulb (E² ∝ P).
(E' / E)² = (60 / 100) ⇒ E' = E * √* (3/5)*
New answer posted
a year agoContributor-Level 9
Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100
New answer posted
a year agoContributor-Level 9
T = 2π√ (l/g) ⇒ g = 4π²l/T²
Percentage error: Δg/g = Δl/l + 2 (ΔT/T) = (0.1/10.0) + 2 (0.005/0.5) = 0.03
Percentage error = (Δg/g) * 100 = 3%
ω = 2πf = 100π rad/s
i_rms = i? /√2
While current changes from its maximum to its rms value, its phase changes by π/4 rad.
t = (π/4)/ω = π/ (4 * 100π) = 2.5 * 10? ³ s = 2.5ms.
New answer posted
a year agoContributor-Level 10
Assuming the rope in the boy's hand is vertical and using a Free Body Diagram (FBD):
f? = T
R + T = 90 ⇒ R = 90 - T
For the piece of wood not to move, f? ≤ µR:
T ≤ 0.5 (90 - T) ⇒ T ≤ 30N
New answer posted
a year agoContributor-Level 10
Given the refractive index μ = λ? / λ = 3/2 and v = 10m, the object distance is u = - (3/2)v = -15m.
Using the lens maker's formula:
μ/v - 1/u = (μ - 1)/R
(3/2)/10 - 1/ (-15) = (3/2 - 1)/R
This gives the radius of curvature R:
R = -30/13 m
New answer posted
a year agoContributor-Level 9
Resonance frequency is independent of R.
Quality factor = ωL/R ⇒ Quality factor decreases with increase in R.
Bandwidth of resonance circuit = R/L ⇒ increases with increase in R.
New answer posted
a year agoContributor-Level 10
The x and y coordinates of the center of mass are given by:
x? = y? = 4a / 3π
New answer posted
a year agoContributor-Level 10
The minimum force F? is calculated as:
F? = (μmg) / √* (1 + μ²)* = ( (1/√3) * 1 * 10 ) / √* (1 + (1/√3)²) = 5N
New answer posted
a year agoContributor-Level 10
Let the charge on C? be q µC. For the capacitor network:
q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16
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