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lilavasa [31]
3 years ago
14

Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

can throw a small rock. one albert is 88 meters. how many square alberts is equal to one acre? (1 acre = 43,560 ft2 = 4050 m2)
Physics
1 answer:
True [87]3 years ago
7 0

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Therefore 1 acre is equivalent to about 0.52 square alberts.</span>

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The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not
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6 0
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Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
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Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

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       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

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It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

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Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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