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Genrish500 [490]
3 years ago
14

6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4

to 3.6. If a 150-lb bag of sand is used, how much cement and gravel must be used?
Engineering
1 answer:
S_A_V [24]3 years ago
7 0

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb

Mass of cement used is 62.5 lb

\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb

Mass of gravel used is 225 lb

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thrust washer

can be used to eliminate rubbing friction of wheel touching frame

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3 years ago
The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery
Lynna [10]

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

3 0
3 years ago
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe T_p = -15° C = (-15 +273)K =258 K

Temperature of water T_w = 3° C = (3 + 273)K = 276 K

Temperature of ice T_i = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient Lh_l =2000 W/m².K

The energy balance can be expressed as:

q_{conduction} =q_{convention}

where;

q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}       -------------   equation (1)

q_{convention} = \pi DLh_l(T_w-T_i)  ------------ equation(2)

Equating both equation (1) and (2); we have;

\frac{2\pi LK(T_i-T_p)}{In(R/r)} = \pi DLh_l(T_w-T_i)

Replacing the given data; we have:

\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)} = \pi (0.1)*2000(276-273)

\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96

In(\frac{0.05}{r})*1884.96 = 182.84

In(\frac{0.05}{r}) = \frac{182.84}{1884.96}

In(\frac{0.05}{r}) =0.0970

\frac {0.05}{r} =e^{0.0970}

\frac {0.05}{r} =1.102

r=\frac{0.05}{1.102}

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

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