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
Answer:
Not knowing the units the tolerance is 0.02. I would presume mm but hopefully your question has more detail.
Explanation:
The tolerance is the portion after the main dimension (+/- 0.02). In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative). If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
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