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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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Answer:

P=45.2W

Explanation:

From the question we are told that:

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Answer:

0.00479 volts

Explanation:

From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change

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r = 1m, t = r÷V = 0.036

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How many changes can this undergo?Does it mean from this current state or how it can change.Please explain it to me.If you were
Fofino [41]

Let's assume this is a drawing of particles of a gas substance. This assumption is made upon the fact that these particles are not close and are represented in motion characteristic for gases. Gases can become solid by skipping the liquid phase. This process is called deposition. Also, a gas can become a liquid through the process of condensation as a result of energy loss at molecular level. Likewise, this is enabled thanks to heat loss or applied pressure.

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

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D = Screen distance = 3 m

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d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

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dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

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