Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q =
<em>d</em>T
=
(T₂ - T₁)
From the above equations, the underlying assumption is that
remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let
be heat constant of calorimeter
Q₂ =
ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m
ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m
' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
<h2>
Answer:</h2>
7532V
<h2>
Explanation:</h2>
For a given transformer, the ratio of the number of turns in its primary coil (
) to the number of turns in its secondary coil (
) is equal to the ratio of the input voltage (
) to the output voltage (
) of the transformer. i.e
=
----------------(i)
<em>From the question;</em>
= number of turns in the primary coil = 8 turns
= number of turns in the secondary coil = 515 turns
= input voltage = 117V
<em>Substitute these values into equation (i) as follows;</em>
= 
<em>Solve for </em>
<em>;</em>
= 117 x 515 / 8
= 7532V
Therefore, the output voltage (in V) of the transformer is 7532
Answer:
A). Dry unit weight = 1657.08Kg/m3
B). Porosity = 0.37
C). Void ratio = 0.593
D). 0.712
Explanation:
Total unit weight, Y = 120pcf =1922.2 Kg/m3
Specific gravity of solids, Gs = 2.64
Water content, w = 16%
A). Dry unit weight
Yd = Y/(1+w)
= 1922.2/(1+0.16) = 1657.08Kg/m3
B). Porosity
However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3
Void ratio = 2.64×1000/1657.08 = 0.593
And porosity = e/(1+e) =0.593/(1+0.593) = 0.37
C). void ratio, e = 0.593
D). Degree of saturation, S = m×Gs/e where m =water content
S = 0.16×2.64/0.593 = 0.712