Answer:
1.4 mL
Explanation:
Given that:
volume of alcohol [tex]V_0[/tex] = 500
Temperature change ΔT = (30° - 5°)C
=25° C
[tex]\beta_{ alcohol }= 1.12 \times 10^{-4} / ^0C[/tex]
The increase in volume ΔV = [tex]\alpha \times V_o \times \Delta T[/tex]
[tex]= 500 \times 1.12 \times 10^{-4} / ^0 C\times 25^0 C[/tex]
= 1.4 mL
definition of ceramics
Answer:
Ceramics are generally made by taking mixtures of clay, earthen elements, powders, and water and shaping them into desired forms. Once the ceramic has been shaped, it is fired in a high temperature oven known as a kiln. Often, ceramics are covered in decorative, waterproof, paint-like substances known as glazes.
Explanation:
Answer:
made essentially from a nonmetallic mineral (such as clay) by firing at a high temperature.
Explanation:
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 2490 hp and causes the shaft to
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 [tex]$\frac{3}{8}$[/tex] in.
A) Determine the maximum shear stress developed in the shaft.
[tex]$\tau_{max}$[/tex] = ?
B) Also, what is the "wind up," or angle of twist in the shaft at full power?
[tex]$ \phi $[/tex] = ?
Solution :
Given :
Angular speed, ω = 1700 rpm
[tex]$ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$[/tex]
[tex]$= 56.67 \pi \text{ rad/s}$[/tex]
Power [tex]$= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$[/tex]
= 1424500 ft. lb/s
Torque, [tex]$T = \frac{P}{\omega}$[/tex]
[tex]$=\frac{1424500}{56.67 \pi}$[/tex]
= 8001.27 lb.ft
A). Therefore, maximum shear stress is given by :
Applying the torsion formula
[tex]$\tau_{max} = \frac{T_c}{J}$[/tex]
[tex]$=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$[/tex]
= 2.93 ksi
B). Angle of twist :
[tex]$\phi = \frac{TL}{JG}$[/tex]
[tex]$=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$[/tex]
= 0.08002 rad
= 4.58°
