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mars1129 [50]
3 years ago
13

A 0.5 m^3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water at 25 °C. Find the weight of the l

iquid.
Engineering
1 answer:
svp [43]3 years ago
7 0

Answer:

total weight of liquid = 4788.25 N or 488.09 kg

Explanation:

given data

total volume = 0.5 m³

volume of ethanol = 10 %  of volume = 0.10 × 0.5 = 0.05 m³

volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³

to find out

weight of the liquid

solution

we know that density of water at 25  is 997 kg/m³

and density of ethanol is 789 kg/m³

so weight of water is = density × volume × g

put here value and we take g = 9.81

weight of water is = 997 × 0.45 × 9.81

weight of water = 4401.25 N     ......................1

weight of ethanol is = density × volume × g

put here value and we take g = 9.81

weight of ethanol is = 789 × 0.05 × 9.81

weight of ethanol = 387.00 N       ...............2

so total weight of liquid = sum of equation 1 add 2

total weight of liquid =  4401.25 + 387

total weight of liquid = 4788.25 N or 488.09 kg

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A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
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Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

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Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

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3 years ago
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Answer:

(b) False

Explanation:

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7 0
3 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

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Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

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Answer:

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