Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the
path of the blade
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:
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
Answer:
The options that apply are:
B, C and D.
Explanation:
There have been a number of accidents all over the world resulting from Acts of God, professional negligence amongst other things.
These may not be avoided completely but the actions above speak to how they can be mitigated or reduced.
Cheers!
