All categories

2 years ago

Explain how the atmosphere and hydrosphere interact to create weather

Physics

1 answer:

ELEN [110]2 years ago

3 0

Answer:

The interaction between the atmosphere and hydrosphere usually consist of the water cycle.

Explanation:

The atmosphere and the hydrosphere interact to create weather activity by the change of states of water, which is usually known as the Water Cycle.

The sun that heats the water from the hydrosphere produces the water evaporation, forming water vapor into the atmosphere. This water vapor is then cooled and condensed into the atmosphere to form liquid water, that is released to the hydrosphere as precipitation in form of rain or snow, to star over the cycle.

I hope it helps you!

You might be interested in

What is the difference between speed and velocity?

olga55 [171]

The answer is c.velocity is speed and direction

3 0

2 years ago

Please help. It’s probably easy

iragen [17]

I believe it’s 60km/h

I divided the total distance (120 km) by the time it took to get there (2h) to get this.

7 0

2 years ago

The brick wall (of thermal conductivity 1.12 W/m · ◦ C) of a building has dimensions of 2.8 m by 5 m and is 8 cm thick. How much

hichkok12 [17]

Answer:

Explanation:

If H be the heat flowing in time t through an area of A having thickness d

H = k A x ( θ₂ - θ₁ ) t / d , k is thermal conductivity , ( θ₂ - θ₁ ) is temperature difference of walls

putting the given values

= (1.12 x 2.8x 5 x 9 x 16.7 x 60 x 60) / .08

= 1.06 x 10⁸ J .

5 0

2 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago

working alongside the pharmacist , one of the duties a medication reconciliation technician would perform is to a ) deal with va

Vanyuwa [196]

Answer:

okkiikkkkkkkkoiiiiiiii8iiii

5 0

2 years ago