Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
Please see the attached file for the complete answer.
Explanation:
Answer:
Explanation:
R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k
Vt = 428V
The series and parallel circuit combination is as follow:
(R6║R7 + R5) + R4 + R3║R2 + R1
(6*2/6 + 2) + 5 = 13/ k
(13/2*6/13/6 + 6) = 78/31k
78/3 + 3 = 171/3 = 57k
57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1)k = 115/58k = 2k
It = Vt/2k = 428/2000 = 0.2A
∴ I1 = 0.2A
I1 = I2 + I3
Using current divider rules to obtain I2 and I3
∴ I2 = I1 X (I2/I2 + I3) = 0.2 X ( 1/4) = 0.05A
and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A
I3 = I4 + I5, using current divider
I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08
I5 = 0.15 - 0.08 = 0.07A
I5 = I6 + I7, using current divider
I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A
I7 = 0.07 - 0.05 = 0.02A
Most obviously and effectively by stretching it. Heating it up, lifting it up or throwing it across the room will all store a small amount of energy in it for variously short spans of time
Answer:
The sign board must be placed 573 ft ahead of the exit.
Explanation:
Distance needed for reducing the speed from 55 mph to 25 mph is given as
Here
-
is the velocity at the end which is 25 mph 
is the velocity at the start which is 55 mph- a is the rate of deceleration which is -5 ft/s^2
- G is the Road grade which is 1% or 0.01
- g is the gravitational acceleration whose value is 32.2 ft/s^2
Now the perception time is 2.5 second, 20/20 vision person can read 6 inch letters from 60 x 6 ft.
For 20/40 vision person can read 6 inch letters from 30 x 6 ft=180 ft.
SSD is 
So the minimum distance is given as
So the sign board must be placed 573 ft ahead of the exit.
