All categories

3 years ago

The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

Engineering

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

Answer is c 0.500

Explanation:

$SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}$

We know that $\rho_{water}=62.42lb/ft^{3}$

Applying values we get

$SpecificGravity=\frac{31.2}{62.4}=0.5$

You might be interested in

Need some help with these plz

pashok25 [27]

100: D, third law of motion

101: D, second law of motion

5 0

3 years ago

AveGali [126]

Answer:

Explanation:

4 0

3 years ago

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i

koban [17]

Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N

(b) 17.1 m

Explanation: The length of wall under the surface can be given by

$b=25m/sin(60)\\=28.867$

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

$F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N$

Noting from the Bernoulli equation that

$Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\$

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0

3 years ago

Before a rotameter can be used to measure an unknown flow rate, a calibration curve of flow rate versus rotameter reading must b

allochka39001 [22]

This statement is b which is true: hope this helped

6 0

3 years ago

The motor of an electric vehicle runs at an average of 50 hp for one hour and 25 minutes. Determine the total energy. Write the

Sunny_sXe [5.5K]

Answer:

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

Explanation:

Power is the rate of change of work in time, since given input is average power, the total energy ( $\Delta E$ ) of the motor of the electric vehicle, measured in joules, is determined by this formula: