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kenny6666 [7]
3 years ago
14

calculate how much black eyes seeds are necessary to plant a 6- hectare( 14.425 acres) field. given that the weight of 1000 blac

k eye peas is 230 g ( take 1 lb=453.4 grams), consider that the planting area is 100 m wide​
Engineering
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer: 1.38g

Explanation:

Width of planting area = 100m

Field size = 6-hectares(14.425 acres)

Weight of 1000 black eye seed = 230g

1 lb = 453.4g

1 black eye seed = 230g/1000 = 0.23g = 0.00023kg

1 hectare = 10,000sq metre

6 hectare = 60,000sq metre

(Weight/Area) kg/m2

0.00023kg / 10,000 = 2.3×10^-8kg/m^2

And field size = 6hectares = 60,000m^2

(2.3×10^-8) × 60,000 = 0.00138kg of black eye seed

0.00138kg × 1000 = 1.38g

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STP stands for standard temperature pressure and NTP stands for normal temperature pressure

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Economics is the study of how individuals and societies make choices under the condition of
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1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

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Saturated liquid water flows through 2 cm ID stainless steel tubes at 200 g/s. The water is at 80oC and the inside surface of th
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A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
2 years ago
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