Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

14

An exit sign must be:Colored in a way that doesn’t attract attentionIlluminated by a reliable light sourceAt least 3 inches tall

Approved by the fire department

1 answer:

Ksenya-84 [330]3 years ago

5 0

Answer:

Red

Explanation:

You might be interested in

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

1. One of these is NOT a type of pneumatic tool. Which one?

Serggg [28]

Answer:Circular

Explanation:

It’s the only thing not list under pneumatic tools‍♂️

5 0

3 years ago

Write in an Excel cell the appropriate @RISK function for a binomial distribution that results from 50 trials with probability o

borishaifa [10]

Answer:

Binomial Function

I got my output using the formula = 1 - BINOMDIST (17, 50, 0.3, TRUE)

Then got the cumulative probability distribution at 17 to be 0.2178

Explanation:

To draw a normal distribution:

1. Got to '@risk' and click on 'defined distribution'

2. Select 'binomial' in function block

3. enter formula in cell formula and click okay

The use of @RISK to draw a binomial distribution of 50 trials and probability of success as 0.3 by entering formular =RISKBINOMIAL (50, 0.3).

6 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

3 years ago

"A fluid at a pressure of 7 atm with a specific volume of 0.11 m3/kg is constrained in a cylinder behind a piston. It is allowed

AlekseyPX

Answer:

Work done by the fluid in the piston=164.5kJ/kg

Specific gas constant= 0.263 kJ/kg K

Molecular weight of gas= 31.54 kmol

7 0

4 years ago