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Sindrei [870]
3 years ago
7

Why are manhole covers round in most of the cases?

Engineering
1 answer:
DiKsa [7]3 years ago
6 0

Answer:

Explanation:

Because there are no angles for alignment, the round shape makes these heavy covers easier to put back on once they're taken off. Round covers are also easier to manufacture. But the main reason manhole covers are round is so they won't accidentally fall into the manhole itself.

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53. The plan of a building is in the form of a rectangle with
schepotkina [342]

Answer: 150m

Explanation:

The following can be depicted from the question:

Dimensions of outer walls = 9.7m × 14.7m.

Thickness of the wall = 0.30 m

Therefore, the plinth area of the building will be:

= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)

= 10 × 15

= 150m

7 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
4 years ago
A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
allochka39001 [22]

Answer: 37.4K

Explanation:

See attachment

7 0
3 years ago
An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,
garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0
3 years ago
Isn't this website cheating?
Orlov [11]

Answer: What website?

Explanation:

8 0
3 years ago
Read 2 more answers
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