Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound =
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds =
kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer: 101 means "introductory something". The allusion is to a college course with the course code 101, which in the American system and probably others indicates an introductory course, often with no prerequisites.
Explanation:
The name Topher means Christ Bearer and is of American origin. Topher is a name that's been used primarily by parents who are considering baby names for boys. Short form of Christopher.
I am not sure but if this is a trick question then the answer lies with his name, so this is the best I got. Hope it helps!
Answer:
Explanation:
The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.
Round Robin:
we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A
C* = 8
D*=17
B*=23
E*=28
AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES
B) PRIORITY SCHEDULING:
1-6 7-14 15-24 25-26 27-30
B E A C D
AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.
C)FCFS
1-10 11-16 17-18 19-22 23-30
A B C D E
AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES
D)SJF
1-2 3-6 7-12 13-20 21-30
C D B E A
AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.