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A farmer has 12 hectares of land on which he grows corn, wheat, and soybeans. It costs $4500 per hectare to grow corn, $6000 to

maw [93]

The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans

<h3>How to solve algebra word problem?</h3>

He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;

C + W + S = 12 ----(1)

It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;

45C + 60W + 50S = 63750 -----(2)

He will grow twice as many hectares of wheat as corn. Thus;

W = 2C ------(3)

Put 2C for W in eq 1 and eq 2 to get;

C + 2C + S = 1200

3C + S = 1200 -----(4)

45C + 60(2C) + 50S = 63750

45C + 120C + 50S = 63750

165C + 50S = 63750 ------(5)

Solving eq 4 and 5 simultaneosly gives;

C = 250 and W = 500

Thus; S = 1200 - 3(250)

S = 450

Read more about algebra word problems at; brainly.com/question/13818690

5 0

1 year ago

What happens when force is placed on a square/rectangle?

mart [117]

im not sure i need to see a photo and also is this science

7 0

3 years ago

Read 2 more answers

The best saw for cutting miter joints is the

ZanzabumX [31]

Answer:

The best saw for cutting miter joints is the backsaw.

Add-on:

i hope this helped at all.

6 0

2 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, $A=2\pi rL\;\cdots(ii)$

Variation of temperature w.r.t the radius of the annalus is

$\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}$

$\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)$

Putting the values from the equations (ii) and (iii) in the equation (i), we have

$q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}$

$\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}$

$\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)}$ [as $q=2.4\pi L$ , and $T_2-T_1=10 ^{\circ}C$ ]

$\Rightarrow k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0

3 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

3 0

2 years ago