All categories

3 years ago

What type of plans have to do with earth, soil, excavation, and location of a house on a lot?

Engineering

1 answer:

lbvjy [14]3 years ago

5 0

plants will give eggs and milk Explanation:

You might be interested in

Are there engineering students here?

leva [86]

Uh, I’d assume so because Brainly has a whole section of questions for them.

7 0

3 years ago

Read 2 more answers

A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

AleksandrR [38]

Answer:

The work furnished by the compressor is $69.77kJ/s$

The minimum work required for the state to change is $55.26kW$

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0

3 years ago

Durante el segundo trimestre de 2001, Tiger Woods fue el golfista que más dinero ganó en el PGATour. Sus ganancias sumaron un to

ehidna [41]

Answer: a. 0.4667

b. 0.4667 and C 0.0667

Explanation:

Given Data:

N = population size (10)

n = random selection (2)

r = number of observations = 7

Therefore

f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )

When y = 1

f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 2

f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 0

f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )

= 1 / 15

= 0.0667

8 0

3 years ago

Water flows at low speed through a circular tube with inside diameter of 2 in. A smoothly contoured body of 1.5 in. diameter is

Art [367]

Answer:

Pressure = 11.38 psi

Force = 13.981 Ibf

Explanation:

Step by step solution is in the attached document.

5 0

3 years ago

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of

Alexxandr [17]

Answer:

$P_m_i_n = 442KPA$

Explanation:

We are given:

m = 1.06Kg

$T_H = 1.2T_L$

T = 22kj

Therefore we need to find coefficient performance or the cycle

$COP_R = \frac {1}{(T_R/T_l) -1}$

$= \frac {1 }{1.2-1}$

= 5

For the amount of heat absorbed:

$Q_l = COP_R Wm$

= 5 × 22 = 110KJ

For the amount of heat rejected:

$Q_H = Q_L + W_m$

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= $= \frac{132}{1.06}$

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

$T_L = \frac{T_H}{1.2}$ ;

$T_L = \frac{342.5}{1.2}$

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at $T_L$ = 12.4°c we have 442KPa

6 0

3 years ago