Answer: At 520 feet between the piers, the center arch of Eads Bridge was the longest rigid span ever built at the time of its construction (only a few suspension bridges had longer spans).
Explanation:
Answer:
12.332 KW
The positive sign indicates work done by the system ( Turbine )
Explanation:
Stagnation pressure( P1 ) = 900 kPa
Stagnation temperature ( T1 ) = 658K
Expanded stagnation pressure ( P2 ) = 100 kPa
Expansion process is Isentropic, also assume steady state condition
mass flow rate ( m ) = 0.04 kg/s
<u>Calculate the Turbine power </u>
Assuming a steady state condition
( p1 / p2 )^(r-1/r) = ( T1 / T2 )
= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )
= ( 9 )^0.285 = 658 / T2
∴ T2 = 351.22 K
Finally Turbine Power / power developed can be calculated as
Wt = mCp ( T1 - T2 )
= 0.04 * 1.005 ( 658 - 351.22 )
= 12.332 KW
The positive sign indicates work done by the system ( Turbine )
Answer:
(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
Explanation:
Solution
(a) Let Mn denotes the number of maintenance visits after the nth journey
Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3
We therefore, notice that M follows an arithmetic sequence
So,
Mn = M₁ + (n-1) (M₂ -M₁)
= 1 + (n- 1) 1 = n
or Mn =n
(b) For what value of n we will get fro Mn > 10
Thus,
n > 10 (exceed 10) or n =11
(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused
So,
n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field.
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.
