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Degger [83]
2 years ago
11

One acre is equal to 43560ft².How much is it in m²?​

Physics
2 answers:
Goryan [66]2 years ago
4 0

1 ft²= 0.09290304m²

so:

43560ft²= 0.09290304m²×43560

= 4,046.8564224 m²

pychu [463]2 years ago
4 0

Answer: 4046.9m² (approx.)

Explanation:

An acre is 43560 ft², but the US has two feet, the International foot, and the (old) Survey foot. Since 1983, the answer varies by state according to which foot the State has adopted for surveying. (Previously, it was always the Survey foot).

International: 43560 ft² x (0.3048 m/ft)² = 4046.856 422 m²

Survey: 43560 ft² x (1200/3937 m)² = 4046.872 610 m²

Both round to 4046.9 m² if that’s good enough, but there is a tiny difference.

To know more about conversions, refer to:

Reference link - brainly.com/question/1462592

#SPJ2

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yarga [219]

Answer:

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Explanation:

When the water is heated, thermal expansion happens so the particles move farther apart and then the hot water moves up because it is less dense then the colder water. ... Convection is happening because when air is heated by the fire, the air becomes less dense and moves up because denser things sink.

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3 years ago
The maximum allowed leakage of microwave radiation from a microwave oven is 5.0 mw/cm2. if microwave radiation outside an oven h
irina [24]

Answer:

194 V/m

Explanation:

In order to find electric field, we can use the formula of power density

i.e Pd = E^2 / Z

where:

Pd = power density in W/m^2

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Z = impedance of free space = 120 * π

E = sqrt(Pd * Z) -----> re-arranging it for E

before solving, convert Pd unit into W/m^2  

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Solving for E:

E= sqrt(50 * 120 * π)

E = 137.3 V/m  

the above value is RMS value

In order to find the peak amplitude of the oscillating field will therefore be 137.3 * sqrt(2) = 194 V/m

8 0
4 years ago
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An electron in the beam of a TV picture tube is accelerated through a potential difference of 2.00 kv.?It then passes into a mag
Stells [14]

Answer:

The magnitude of the field is 8.384×10^-4 T.

Explanation:

Now, i start solving this question:

First, convert the potential difference(V) 2 kv to 2000 v.

As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so

r = 0.36/2 = 0.18 m.

As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.

We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.

qV = 1/2mv^2

(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2

v = 2.65×10^7 m/s.

These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:

qvB = mv^2/r

qB = mv/r

B = mv/qr

B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)

Hence, B = 8.384*10^-4 T.

5 0
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Answer:

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