Answer:
b
Explanation:
only if there signal is turned on
Answer:
Explanation:
Using the kinematics equation
to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:

Then:

The distance traveled by car B in the given time (t) is expressed as:

For car A, the needed time (t) to come to rest is:

Also, the distance traveled by car A in the given time (t) is expressed as:

Relating both velocities:





t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.



d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Answer:
True
Explanation:
Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true
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:
Gravitational force (pulled downward by the Earth)
Normal force (pushed upward by the ground)
Applied force (pushed by the person)
Friction force (pulled opposite the direction of motion by the roughness of the ground)