Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T= 
now putting the values we get
= 
= 1.71×10^{-3} sec
b) note that the time 
it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Answer:
22,800 years
Explanation:
Half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.0625 = (½)^(t / 5700)
log 0.0625 = (t / 5700) log 0.5
4 = t / 5700
t = 22,800
It takes 22,800 years.
Answer:
Yes.
Explanation:
Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.
250/4= 62.5 mph
to find the mph of a car, you need to divide the number of miles traveled by the hours that it took to travel that many miles
