Answer: a) 135642 b) 146253
Explanation:
A)
1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.
3- reserving all resources in advance helps would happen most likely if the algorithm has been used.
5- Resource ordering comes first before detection of any deadlock
6- Thread action would be rolled back much easily of Resource ordering precedes.
4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread
2- only option practicable after thread has been killed.
Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.
Answer:
A benchmark
Explanation:
Most times a benchmark serves as the better measure when assessing a computer's performance, this is because CPU speeds can only evaluate an aspect of a computer's performance whereas a benchmark offers the advantage of measuring all the aspects of a computer's performance for a specific type of computing problem.
Answer:
Could ask a family member to help
Explanation:
Answer:
In Btu:
Q=0.001390 Btu.
In Joule:
Q=1.467 J
Part B:
Temperature at midpoint=274.866 C
Explanation:
Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= ![\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B3600%7D%20%28Btu%2Fs%29%2F%28ft.F%29%3D8.33%2A10%5E%7B-3%7D%20%20%28Btu%2Fs%29%2F%28ft.F%29)
Thermal Conductivity is SI units:
![k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K](https://tex.z-dn.net/?f=k%3D30%28Btu%2Fhr%29%2F%28ft.F%29%20%2A%20%5Cfrac%7B1055.06%7D%7B3600%2A0.3048%2A0.556%7D%20%5C%5Ck%3D51.88%20W%2Fm.K)
Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft
Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft
T_1=500 C=932 F
T_2=50 C= 122 F
Part A:
In Joules (J)
![A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2Ar%5E2%5C%5CA%3D%5Cpi%20%2A%280.002%29%5E2%5C%5CA%3D0.00001256%20m%5E2)
Heat Q is:
![Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7Bk%2AA%2A%28T_1-T_2%29%7D%7BL%7D%20%5C%5CQ%3D%5Cfrac%7B51.88%2A0.000012566%2A%28500-50%7D%7B0.2%7D%5C%5C%20Q%3D1.467%20J)
In Btu:
![A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2Ar%5E2%5C%5CA%3D%5Cpi%20%2A%280.00656%29%5E2%5C%5CA%3D0.00013519%20m%5E2)
Heat Q is:
![Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7Bk%2AA%2A%28T_1-T_2%29%7D%7BL%7D%20%5C%5CQ%3D%5Cfrac%7B8.33%2A10%5E%7B-3%7D%2A0.00013519%2A%28932-122%7D%7B0.656%7D%5C%5C%20Q%3D0.001390%20Btu)
PArt B:
At midpoint Length=L/2=0.1 m
![Q=\frac{k*A*(T_1-T_2)}{L}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7Bk%2AA%2A%28T_1-T_2%29%7D%7BL%7D)
On rearranging:
![T_2=T_1-\frac{Q*L}{KA}](https://tex.z-dn.net/?f=T_2%3DT_1-%5Cfrac%7BQ%2AL%7D%7BKA%7D)
![T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C](https://tex.z-dn.net/?f=T_2%3D500-%5Cfrac%7B1.467%2A0.1%7D%7B51.88%2A0.00001256%7D%20%5C%5CT_2%3D274.866%5C%20C)
Answer:
Following is attached the solution or the question given.
I hope it will help you a lot!
Explanation: