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2 years ago

The metric ruler is typically divided into

Engineering

2 answers:

Anika [276]2 years ago

7 0

Answer is b because it is the nearest to

Bas_tet [7]2 years ago

4 0

Either B or C depends

Explanation:

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How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor

zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0

3 years ago

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof th

ryzh [129]

Answer:

$5.184

Explanation:

The cost can be calculated using the formula: $Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\$

Before using this, we require the following conversions:

$\frac {320}{1000} = 0.32$

<em>30 Days → Hours:</em>

$30 \times 24 = 720$

Using the above stated formula:

$M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184$

4 0

2 years ago

QUESTION 3

lianna [129]

D D D D D D D D D D D D D D D DdDdddddf

6 0

3 years ago

A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

Hitman42 [59]

Answer:

radius = 9.1 × $10^{-3}$ m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support = 45 mm

solution

we apply here minimum radius formula that is

radius = $\sqrt[3]{\frac{FL}{\sigma \pi}}$ .................1

here F is applied load and is length

put here value and we get

radius = $\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}$

solve it we get

radius = 9.1 × $10^{-3}$ m

8 0

2 years ago

For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin

Nitella [24]

The correct answer is B

4 0

3 years ago