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Y_Kistochka [10]
2 years ago
15

What are the philological elements of interior design most like?

Engineering
1 answer:
sp2606 [1]2 years ago
7 0

Answer:

Ea public address glven via the intercom system of a large buildingxplanation:

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‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

3 0
2 years ago
The diagram illustrates a method of producing plastics called​
hodyreva [135]

Answer:

polymerisation,

Explanation:

6 0
2 years ago
2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate
Hitman42 [59]

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

Explanation:

The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate

Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

7 0
3 years ago
Read 2 more answers
In a manufacturing facility, 2-in-diameter brass balls (k = 64.1 Btu/h·ft·°F, rho = 532 lbm/ft^3, and cp = 0.092 Btu/lbm·°F) ini
bekas [8.4K]

Answer:

Explanation:

First we compute the characteristic length and the Biot number to see if the lumped parameter

analysis is applicable.

Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an

analysis, the time to reach a certain temperature is given by the following

From the data in the problem we can compute the parameter, b, and then compute the time for

the ratio (T – T)/(Ti

– T)

4 0
3 years ago
Read 2 more answers
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera
saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \overline T = 548 \ K

\rho = 7900 \ kg/m^3

K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K

\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\  B_i = \dfrac{h(\rho/4)}{K} \\ \\  =0.657

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\

0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\  0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o   \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81

t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\  t_f= 2162.5 \\ \\ t_f = 36 mins

However, on a single rod, the energy extracted is:

\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) )  \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\  \theta = 1.54 \times 10^7 \ J

Hence, for centerline temperature at 50 °C;

The surface temperature is:

T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}

5 0
2 years ago
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