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Aleksandr [31]
3 years ago
12

18. Which would be the most reliable source of information to use for a history report? (2 points)

Physics
2 answers:
miv72 [106K]3 years ago
5 0
An encyclopedia because it isn’t biased
Paladinen [302]3 years ago
3 0

Answer:

encyclopedia most reliable I think

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Its Exothermic.......
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What is the line running through earth from pole to pole
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The Longitutes are <em>imagenary lines that run from the North pole to the South pole.</em>
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What is convection ?
bagirrra123 [75]
Convection is the rise of hot air and the drop of cold air due to a large amount of molecules being moved usually in gas form.\

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3 years ago
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6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating
Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P  =  55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

8 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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