William, a project manager, needs to prepare the budget for a new software development project. To do so, he takes inputs from o
1 answer:
You might be interested in
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Answer:
The answer is given in the explanation.
Explanation:
The circuit is as indicated in the attached figure.
From the analytical description the zener voltage is given as
Here
Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.
The equivalent model is shown in the attached figure.
From the above equation, Vzo is calculated as
Here Vz is given as 7.5 V
Iz is given as 10 mA
rz is given as 30 Ω
Thus the Vzo is given as
The value of I_L is given as 5 mA
Now the expression of current is as
Now the resistance is calculated as
So the value of resistance is 186.66 Ω.
Considering the supply voltage is increased by 10%
V is 10-10%*10=10+1=11 so the
Considering the supply voltage is decreased by 10%
V is 10-10%*10=10-1=9 so the
Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA
so
Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus
The load voltage is 7.485 V
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
