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:
The angular velocity is 7.56 rad/s
the maximum water height is 2 ft
Explanation:
The z-position as a function of r is equal to
(eq. 1)
where
h0 = initial height = 1 ft
w = angular velocity
R = radius of the cylinder = 1.5 ft
zs(r) = 0 when the free surface is lowest at the centre
Replacing and clearing w

if you consider the equation 1 for the free surface at the edge is equal to

An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with a blue stripe or dot.
Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft