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soldier1979 [14.2K]
3 years ago
15

What fraction of the worlds surface is estimated to be arable land?

Engineering
1 answer:
zzz [600]3 years ago
6 0

I believe the amount of arable land worldwide is 1/3 as a fraction.

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When you arrive at an intersection with a stop sign in your direction, if there is no marked stop
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C: Stop before entering the pedestrian crosswalk.

Explanation:

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I’m doing a project on renewable energy. There are 6 energy sources. Solar, wind, geothermal, hydroelectric, tidal, and biomass.
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"Biofuels"

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A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
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Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

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                                                  = $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

            $=2 \pi (0.23)^2$

            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

     = 7043.42 N

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2 years ago
For a short time a rocket travels up and to the left at a constant speed of v = 650 m/s along the parabolic path y=600−35x2m, wh
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Answer:

Detailed working is shown

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The attached file shows a detailed step by step calculation..

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3 years ago
Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w
I am Lyosha [343]

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

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