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3 years ago

For which of the following structures would a pile foundation be appropriate? (Select all that apply.)

Engineering

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

Football stadium on rocky soil

Skyscraper on bedrock

Apartment building on sandy soil

Explanation:

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Hareem and Shahad are on a road trip to Florida. They pull over to get gas, and as you may know it is sold by the gallon in the

Alona [7]

Answer:

53.3

Explanation:

4×14=56

56÷1.05=53.3

4 0

3 years ago

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

4 years ago

An oscilloscope display grid or scale is called?

zaharov [31]

Answer:

An oscilloscope display grid or scale is called a graticule.

Explanation:

5 0

3 years ago

Read 2 more answers

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago

A motorist is driving his car at 60km/hr when he observes that a traffic light 250m ahead turns red. The traffic light is

Alecsey [184]

Explanation:

Okay soo-

Given-

u = 60 km/hr = 60×1000/3600=50/3 m/s

t = 20 s

s = 250 m

a = ?

v = ?

Solution -

Here, acceleration is uniform.

(a) According to 2nd kinematics equation,

s = ut + ½at^2

250 = 50/3 ×20 + 0.5×a×20×20

250-1000/3=200a

(750-1000)/3=200a

a = -250/(3×200)

a = -5/12

a = 0.4167 m/s^2

The required uniform acceleration of the car is 0.4167 m/s^2.

(b) According to 1st kinematics equation

v = u + at

v = 50/3 + (-5/12)×20

v = 50/3-25/3

v = 25/3

v = 8.33 m/s

The speed of the car as it passes the traffic light is 8.33 m/s.

Good luck!

5 0

3 years ago