Answer: 
= 28800 Pa or 28.8 kPa
Explanation: To determine the pressure of a liquid in a rotating tank,it is used:
p = 
- γfluid . z + c
where:
is the liquid's density
w is the angular velocity
r is the radius
γfluid.z is the pressure variation due to centrifugal force.
For this question, the difference between a point on the circumference and a point on the axis will be:
= 
- γfluid.
- (
- γfluid.
)
= 
- γfluid( -)
Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:
=
= 28800
The difference in pressure between two points, one on the circumference and the other on the axis is = 28800 Pa or 28.8 kPa
Answer:
Both Technicians A & B are correct
Explanation:
In diagnosing wheel systems, both steps said by technicians A & B are correct because it's the normal procedure to diagnose wheel speed sensor circuit faults.
Answer:
hello your question lacks the required image attached to this answer is the image required
answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)
