Burn in hell i watched your stupid video and i still could not get the answer

Explain the 11 sections that a typical bill of quantity is divided into

tresset_1 [31]

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

5 0

3 years ago

During a cold winter day, wind at 50 km/h is blowing parallel to a 4-m-high and 10-m-long wall of a house. If the air outside is

fenix001 [56]

Answer:

to determine the rate of heat loss from that wall by convection = 12780 watts

Explanation:

8 0

3 years ago

The reason for the nature of the signal's value is: Not Linear Humans are Continuous The stock market closes once per day, and s

Serga [27]

Answer:

Explanation:

* Daily close of the stock market

The nature of the signal in value is <u>Discrete</u>, as the stock market closes once per day,and so is discrete time.
And the nature of the signal in time is continuous as the company are continuous.

8 0

3 years ago

Refrigerant R-12 is used in a Carnot refrigerator

Sphinxa [80]

Answer:

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

6 0

3 years ago

I need this asap thank you :) plzzzzz When the spring on a mousetrap car is fully unwound, the force acting on the car is _____.

Usimov [2.4K]

Answer: (only friction) the friction lets it keep its speed and not slow down and it creats volocity between the serface of where the mousecar is running and the wheels on the ground

sorry if im wrong i tried my best

6 0

3 years ago