Answer:
A longitudinal section
Explanation:
A longitudinal section is a section drawn along the length of an object, as opposed to a cross section, which is drawn across the width or diameter of an object.
The question is incomplete. The complete question is :
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is 
in.
A) Determine the maximum shear stress developed in the shaft.
= ?
B) Also, what is the "wind up," or angle of twist in the shaft at full power?
= ?
Solution :
Given :
Angular speed, ω = 1700 rpm
Power 
= 1424500 ft. lb/s
Torque, 
= 8001.27 lb.ft
A). Therefore, maximum shear stress is given by :
Applying the torsion formula
= 2.93 ksi
B). Angle of twist :
= 0.08002 rad
= 4.58°
Answer:
1609.344 metres
I think, correct me if I'm wrong
Answer:
Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally
updated information on model changes that do not affect the technician. Who is correct? the answer is c
Answer:
The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵ °C⁻¹.
Explanation:
We know that Linear thermal expansion is represented by the following equation
Δ L = L x ∝ x Δ T ---- (1)
where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and Δ T is the change in temperature.
Given that:
L = 0.6 m
T₁ = 15° C
T₂ = 37° C
Δ L = 0.28 mm
∝ = ?
Solution:
We know that Δ T = T₂ ₋ T₁
Putting the values of T₁ and T₂ in above equation, we get
Δ T = 37 - 15
Δ T = 22 °C
Also Δ L = 0.28 mm
Converting the mm to m
Δ L = 0.00028 m
Putting the values of Δ T, Δ L, L in equation 1, we get
0.00028 = 0.6 x ∝ x 22
Rearranging the equation, we get
∝ = 0.00028 / (0.6 x 16)
∝ = 0.00028 / 13.2
∝ = 2.12 x 10⁻⁵ °C⁻¹
