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3 years ago

Energy

Physics

2 answers:

Ahat [919]3 years ago

7 0

I think the answer may be D- mechanical

Setler79 [48]3 years ago

3 0

The answer is OD) Mechanical

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An ideal air mass source region must meet two criteria. First, it must be an extensive and physically uniform area. The second c

Usimov [2.4K]

Answer:

The correct option is : B. stagnation

Explanation:

Air mass is a volume of air, which is described on the basis of water vapor content and temperature. It has the ability to adapt to the characteristics of surface below it and can travel thousands of miles from its source region.

An ideal source region of an air mass must fulfill the following two criteria:

1. It should be a large and uniform area.

2. the area should have stagnant circulation of atmospheric air .

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3 years ago

A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. i

ollegr [7]

Magnetic field outside it due to long wire is given by

$B = \frac{u_o i}{2 \pi r}$

Magnetic field due to long wire inside wire at any point

$B = \frac{u_o i r}{2 \pi R^2}$

Now the ratio of two magnetic field is given by

$\frac{B_{in}}{B_{out}} = \frac{r_1/R^2}{1/r_2}$

$\frac{0.285}{0.200} = \frac{4*10}{R^2}$

$1.425 = \frac{40}{R^2}$

R = 5.3 mm

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3 years ago

57. Estimate Potential Energy A boulder with a

Oksanka [162]

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

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3 years ago

An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .