Answer:
Explanation:
At resonance ω₀L = 1 / ω₀C , L is inductance and C is capacitance .
C = 1 / ω₀²L , ω₀ = 5.1 x 10⁵ . ( given )
voltage over resistance = R I , R is resistance and I is current
voltage over inductance = Iω₀L
R I / Iω₀L = 60 / 40
R / ω₀L = 3 / 2
L = 2 R / 3 ω₀
= 2 x 121 / 3 x 5.1 x 10⁵
= 15.81 x 10⁻⁵
C = 1 / ω₀²L
= 1 / (5.1 x 10⁵)² x 15.81 x 10⁻⁵
= .002432 x 10⁻⁵
= 24.32 x 10⁻⁹ F
Let the angular frequency required be ω
Tan 45 = (ωL - 1 / ωC) / R
ωL - 1 / ωC = R
ω²LC - 1 = R ωC
ω²LC = 1 + R ωC
ω² x 15.81 x 10⁻⁵ x 24.32 x 10⁻⁹ = 1 + 121 x ω x 24.32 x 10⁻⁹
ω² x 384.5 x 10⁻¹⁴ = 1 + 2942.72 x10⁻⁹ω
ω² - 7.65 x 10⁶ ω - 1 = 0
ω = 7.65 x 10⁶
frequency = 7.65 x 10⁶ / 2π
= 1.22 x 10⁶ Hz
These waves most likely belong to the part<span> of the electromagnetic spectrum that contains radio waves, since radio waves have the lowest frequency of any of the other waves.</span>
No, it is not possible for thermal energy to be equal in both bowls.
Answer:
The uncertainty in momentum changes by a factor of 1/2.
Explanation:
By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.
So, the uncertainty in momentum changes by a factor of 1/2.
Answer: The answer is False
Explanation: This is for the one's in apex <>