Answer:
Just draw a line from point D join to point E
The triangle formed DME will be congruent to AMC
Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars
Explanation :
A)
Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}
= 3780kJ
And 1 hour = 3600s
Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W
B)
At 15km/hour a 15km run takes 1 hour.
1 hour is 3600s and the runner burns 1050 joule per second.
Energy used in 1 hour = 3600 x 1050 J/s
= 3780000 J or 3.78MJ
C)
1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km
15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ
Finally,
1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ
This means that the runner needs 5320/1008 = 5.3 bars
Answer:
Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop
Explanation:
Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.
example of feedforward is
Shower
which consist of following control points
Hear toilet flush (measurement)
Customize water to compensate
feedback refers to that point when water turns hot before the configuration changes
Answer:
(a) Current density at P is 
.
(b) Total current I is 3.257 A
Explanation:
Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;
where 
and 
unit vectors.
(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore
(b) Total current flowing outward can be calculated by using the relation,
where integral is calculated through the circular band given in the question. We can write the integral as below,
due to unit vector multiplication. Then,
where 
. Therefore
Answer:
4.83m/
Explanation:
For a particle moving in a circular path the resultant acceleration at any point is the vector sum of radial and the tangential acceleration
Radial acceleration is given by 
r
Applying values we get 
X0.3m
Thus 
At time = 2seconds 
The tangential acceleration is given by 
Thus the resultant acceleration is given by
