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
Answer:
Q=339.5W
T2=805.3K
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.
3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature
Answer:
Do you mean 4m^3 and 3.0 tones?
Explanation:
solution:
Mass = m = 3.0 tones
- 1 ton = 1,000 kg
= 3.0 × 1,000
= 3,000 kg
volume = v = 4m^3
Required:
Mass density of oil = p = ?
We know that;
The answer is:
750kg / m^3
Answer:
Max shear = 8.15 x 10^7 N/m2
Explanation:
In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;
Allowable Shear Stress = Torque x Radius / pi/2 x radius^4
Putting the values we have;
T = 2000 N/m
Radius = Diameter/2 = 0.05 / 2 = 0.025 m
Putting values in formula;
Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4
Max shear = 8.15 x 10^7 N/m2