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Levart [38]
3 years ago
10

How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including t

he requirements needed?
Engineering
1 answer:
Stella [2.4K]3 years ago
3 0

Answer:

The application of regulations in locations are very important.

Explanation:

The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.

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please answer correctly if it is not correct, report it to them thank you for answering correctly.god bless you​
Sergio [31]
1 is c 2 d 3 a 4 b5is I do
8 0
3 years ago
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For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep
Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that                              

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now  the metal   removal  rate   given as

MRR= 12 f V d

d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0
3 years ago
Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,
77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

L_{BC}^2 =  L_{AB}^2 + L_{AC}^2

L_{BC}^2 = (3.5+2.5)^2+ 4^2

L_{BC}= \sqrt{(6)^2+ 4^2}

L_{BC}= \sqrt{36+ 16}

L_{BC}= \sqrt{52}

L_{BC}= 7.2111 \ m

The cross -sectional of the cable is calculated by the formula :

A = \dfrac{\pi}{4}d^2

where d = 4mm

A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length \delta ; we can calculate for the force F_{BC by using the formula:

\delta = \dfrac{F_{BC}L_{BC}}{AE}

F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}

where ;

E = modulus elasticity

L_{BC} = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for L_{BC} and 0.006 m for \delta ; we have:

F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}

F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN     ---- (1)

Similarly; we can determine the force F_{BC} using the allowable  maximum stress; we have the following relation,

\sigma = \dfrac{F_{BC}}{A}

{F_{BC}}= {A}*\sigma

where;

\sigma = maximum allowable stress

Replacing 190 × 10⁶ Pa for \sigma ; we have :

{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \  kN     ------ (2)

Comparing (1) and  (2)

The magnitude of the force F_{BC} = 2.09676 \ kN since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

\sum M_A = 0

3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0

3.5 P - 3.328 F_{BC} = 0

3.5 P = 3.328 F_{BC}

3.5 P = 3.328 *2.09676 \  kN

P =\dfrac{ 3.328 *2.09676 \  kN}{3.5 }

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0
3 years ago
Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b
coldgirl [10]

Answer:

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

7 0
4 years ago
Only answer this if your name is riley
Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

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3 years ago
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