Answer:
P= 5.5 bar
Explanation:
Given that
L= 4000 m
d= 0.2 m
Friction factor(F) = 0.01
speed V= 2 m/s
Head = 5 m
Head loss due to friction
So the total head(H) = 5 + 40.77 + 10.3 =56.07
Where 10.3 m is the atmospheric head.
We know that
P=ρ g H
So total Pressure
P= 1000 x 9.81 x 56.07 Pa
P= 5.5 bar
Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
Answer:
Hearing protection would be your answer!
Explanation:
This includes earplugs,muffs etc.
Hope it helps!
Answer:
The final velocity of the rocket is 450 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 0
constant upward acceleration of the rocket, a = 18 m/s²
time of motion of the rocket, t = 25 s
The final velocity of the rocket is calculated with the following kinematic equation;
v = u + at
where;
v is the final velocity of the rocket after 25 s
Substitute the given values in the equation above;
v = 0 + 18 x 25
v = 450 m/s
Therefore, the final velocity of the rocket is 450 m/s.
Answer:
The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v
Explanation:
Solution
Given that:
V+ = 20v
Re = 2kΩ
Rc = 1kΩ
Now we will amke use of the method KVL in the loop.
= - Ve + IE . Re + VEB + VB = 0
Thus
IE = V+ -VEB -VB/Re
Which gives us the following:
IE = 20-0.7 - 10/2k
= 9.3/2k
so, IE = 4.65 mA
IB = IE/β +1 = 4.65 m /101
Thus,
IB = 0.046039 mA
IB = 46.039μA
IC =βIB
Now,
IC = 100 * 0.046039
IC is 4.6039 mA
Now,
VB = 10v
VE = VB + VEB
= 10 +0.7 = 10.7 v
So,
Vc =Ic . Rc = 4.6039 * 1k
=4.6039 v
Finally, this is the table summary from calculations carried out.
Summary Table
Parameters IE IC IB VE VB Vc
Unit mA mA μA V V V
Value 4.65 4.6039 46.039 10.7 10 4.6039
