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xeze [42]
3 years ago
7

Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that

the recent statics show a that after the first journey to the outer space, the rocket was sent to service once, and after that the number of maintenance visits increase by 1 after every journey, for instance, after the second journey, the rocket had to go to service twice before being re-used, three-times after the 3rd use and so on.
A. Generate a rule for the number of visits to the maintenance department with respect to space journeys, is this rule arithmetic or geometric.
B. How many journeys will it take to for the number of service visits after a certain journey exceed 10?
C. The rocket is not to be re-used again if the total number of maintenance visits exceed 50, then how many journeys will a single rocket serve before being out of service?
Engineering
1 answer:
Ainat [17]3 years ago
7 0

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

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(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

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(a) 8 MPa

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(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

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(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740  × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

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(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

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