Answer:
t = 1.06 sec
Explanation:
Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:
R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.
The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:
C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d
The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:
V = V₀*e(-t/RC)
The product RC (which is called the time constant of the circuit) can be calculated as follows:
R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d
Simplifying common terms, we finally have:
R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec
If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:
V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1
⇒ t = RC = 1.06 sec.
Answer:
Basically there are two principal differences between the convection and conduction heat transfer
Explanation:
The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.
We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ
