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user100 [1]
3 years ago
6

Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given

as 1:2:3 for cement, sand, and gravel with w/c of 0.50. The specific gravities of sand and gravel are 2.60 and 2.70 respectively. Entrained air content is 7.5%. How many pounds of cement, water, sand, and gravel are needed for the driveway?
Engineering
1 answer:
Akimi4 [234]3 years ago
4 0

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A
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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car  to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both  cars which is equal from instant point of abreast.

So; we can say :

D_{pursuit} =D_{police}

By using the second equation of motion to find the distance S;

S= ut + \dfrac{1}{2}at^2

D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)

D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)

D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)

D_{police} = ut _P + \dfrac{1}{2}at_p^2

where ;

u  = 0

D_{police} =  \dfrac{1}{2}at_p^2

D_{police} =  \dfrac{1}{2}*(1.96)*(t+12)^2

D_{police} = 0.98*(t+12)^2

D_{police} = 0.98*(t^2 + 144 + 24t)

D_{police} = 0.98t^2 + 141.12 + 23.52t

Recall that:

D_{pursuit} =D_{police}

(187.8+15.65 \ t-1.525 t^2)=  0.98t^2 + 141.12 + 23.52t

(187.8 - 141.12)  + (15.65 \ t  -  23.52t)  -( 1.525 t^2    - 0.98t^2)  =   0

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR  t = 3.02

Since we can only take consideration of the value with a  positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time  required for the police car  to over take the automobile = 12 s + 3.02 s

Total time  required for the police car  to over take the automobile = 15.02 sec

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