All categories

3 years ago

What are the 3 main types of global winds in the northern hemisphere?

Physics

1 answer:

sveta [45]3 years ago

8 0

Answer:

The trade winds, the prevailing westerlies, the polar easterlies.

Explanation:

There are three prevailing wind belts associated with these cells: the trade winds, the prevailing westerlies, and the polar easterlies (Fig. 3.10). Fig. 3.10 only shows the circulation cells and winds in the Northern Hemisphere

You might be interested in

What are the issues that hinders efforts to achieve sustainability?

Nadya [2.5K]

Answer:

who will solve environmental problems, who is responsible for environmental problems, and who pays the cost of implementing solutions

Explanation:

4 0

3 years ago

A sample of an ideal gas initially occupies a volume of 6 L. The pressure of the sample is then doubled while it is cooled to on

Troyanec [42]

Answer:

V₂= 1 L

Explanation:

Given that

Volume occupies V₁= 6 L

Initial pressure = P₁

Initial temperature = T₁

The final pressure =P₂ = 2 P₁

Final volume =V₂

Final temperature = T₁/3

As we know that equation for ideal gas

P V = m R T

P=pressure, V=volume, T=temperature

m=mass ,R=gas constant

Now from mass conservation

$m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1\times 6}{RT_1}=3\times \dfrac{2P_1V_2}{RT_1}$

6 = 3 x 2 V₂

V₂= 1 L

So the final volume will be 1 L

4 0

3 years ago

How is it that even light precipitation can still cause the collection of large amounts of water.

Firdavs [7]

Answer:

the heat of the light.

Explanation:

no matter the light, there's always heat being produced from it. and heat makes liuqid rise

4 0

3 years ago

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$