Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
Answer:
W = - 184.8 kW
Explanation:
Given data:





We know that work is done as
![W = - [ Q + \dor m[h_2 - h_1]]](https://tex.z-dn.net/?f=W%20%3D%20-%20%5B%20Q%20%2B%20%5Cdor%20m%5Bh_2%20-%20h_1%5D%5D)
for
density of air is 1.22 kg/m^3 and 
for 


W = -[14 + 1.708[400-300]]
W = - 184.8 kW
Answer:
B. 180 million joules
Explanation:
Apply the formula for heat transfer given as;
Q=m*c*Δt where
Q = electrical energy consumed by the heater in joules
m= mass of air in the chamber in kg
c= specific heat of air in joules per kg degrees Celsius
Δt= change in temperatures in degrees Celsius
Given in the question;
m= 1200 kg
c= 1000 J/°C /kg
Δt = 180°-30°= 150° C
Substitute values in the equation to get Q as;
Q=m*c*Δt
Q= 1200 * 1000* 150
Q= 180000000 joules
Q = 180 million joules
<u>The correct answer option is B : 180 million joules.</u>
True, In the STEM field engineers use and think in terms of sciences like physics and math.