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Vanyuwa [196]
3 years ago
10

An engineering drawing shows the: (A) dimensions, tolerances, cost, and sales or use volume of a component.(B) dimensions, toler

ances, materials, and finishes of a component.(C) materials, finishes, machining operations, and dimensions of a component.(D) cost, materials, tolerances, and lead-time for a component.(E) cost, dimensions, and machining operations for a component.
Engineering
1 answer:
Leto [7]3 years ago
3 0

Answer:

(B) dimensions, tolerances, materials, and finishes of a component.

Explanation:

An engineering drawing :

  An  engineering drawing is a technical drawing which draws the actual component .

An engineering drawing shows

1. Materials

2.Dimensions

3.Tolerance

4.Finishes of a component

Engineering drawing does not shows any information about the cost of component.

So the option B is correct.

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A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It
topjm [15]

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

7 0
3 years ago
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

8 0
4 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
AVH
BigorU [14]

Answer: preventive maintenance

Explanation:

3 0
3 years ago
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