Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
Answer: (B) 100
Explanation:
Given that;
Pstatic = 20 psig , hz = 160ft, hf = 20ft
Now total head will be;
T.h = hz + hf
T.h= 160 + 20
T.h = 180ft
Minimum pressure = Psatic + egh
we know that specific weight of water is 62.4 (lb/ft3)
so
P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr
P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)
P.min = 20 + 78
P.min = 98 lbf/in²
Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100
Answer:
b. Technician B only
Explanation:
A watchdog timer is a circuit that automatically monitors the MCU (Microcontroller Unit) for any anomaly, detects it and helps the MCU to recover from the malfunction it has detected.
If the input signal turn-on time is too fast for the input circuit, that is a malfunction and this activates the watchdog timer circuit to correct this malfunction immediately. So Technician B only is correct as the watchdog timer is activated immediately once there is a malfunction.
Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80 
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
