A beaker is filled to the 500 ml mark with alcohol. What increase in volume (in ml) the beaker contain when the temperature changes from 5°C to 30°C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) C°, Palcohol = 1.12 x 104/ °C

Answer:

You cant update it if there isin´t any update available lol

Explanation:

Answer:

4

or

22

Explanation:

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°

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]

Differentiating the above equation:

[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]

[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]

        [tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]

Answer:

silos to store grains

Explanation:

I got it from PLATO... hope this helps...

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Answer:

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Explanation:

Answer:

made essentially from a nonmetallic mineral (such as clay) by firing at a high temperature.

Explanation:

Answer:

for 13

Explanation:

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