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3 years ago

Why is concrete on its own not a good material to use

Engineering

1 answer:

Inga [223]3 years ago

7 0

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

Concrete exhibits high compressive strength when used. However, it has very low compressive strength. This is the reason why concrete is normally combined with steel to make a composite building material called reinforced concrete. The steel reinforces concrete hence increasing the tensile strength in RC buildings. The end composite is durable and fireproof. Generally, the main reason why concrete is not use on its own is due to its poor tensile strength.

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Consider a Mach 4.5 airflow at a pressure of 1.25 atm. We want to slow this flow to a subsonic speed through a system of shock w

marissa [1.9K]

Answer:

a. 130.73 atm

b. 102.62 atm

c. 87.1 atm

Explanation:

See the attached pictures.

6 0

3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$

$a = -0.128\,\frac{ft}{s^{2}}$

7 0

3 years ago

If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s

Archy [21]

s 0Miles (short), 150 Miles(medium), and 300 Miles (long).

Explanation:

4 0

3 years ago

9. How is the Air Delivery temperature controlled during A/C operation?

Rzqust [24]

Answer:

i believe the answer is a but i could be wrong

Explanation:

i hope it helps

6 0

3 years ago

I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to tr

Alina [70]

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = $\sqrt{(100*2)/32.15}$ = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.

8 0

2 years ago