Answer:
Your question is lacking some information attached is the missing part and the solution
A) AB = AD = BD = 0, BC = LC
AC =
B) AB = AD = BC = BD = 0
AC =
Explanation:
A) Forces in all members due to the load L in position A
assuming that BD goes slack from an inspection of Joint B
AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C
B) steps to arrive to the answer is attached below
AB = AD = BC = BD = 0
AC =
Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m
Answer:
a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr
Explanation:
Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2
Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr
1W = 3.41 BTU/hr
Given parameters:
thickness, t = 7.5mm = 7.5/1000 = 0.0075m
Temperatures 150 C = 150 + 273 = 423 K
50 C = 50 + 273 = 323 K
Temperature difference, T = 423 - 323 = 100 K
We are assuming steady heat flow;
a.) Heat flux, Q" = kT/t
K= thermal conductivity of the material
The thermal conductivity of brass, k = 109.0 W/m.K
Heat flux,
b.) Area of sheet, A = 0.5m2
Heat loss, Q = kAT/t
Heat loss,
Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr
c.) Material is now given as soda lime glass.
Thermal conductivity of soda lime glass, k is approximately 1W/m.K
Heat loss,
Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr
d.) Thickness, t is given as 15mm = 15/1000 = 0.015m
Heat loss,
Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr