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3 years ago

Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

Physics

1 answer:

spin [16.1K]3 years ago

3 0

Explanation:

The water cycle basically involves five steps:

evaporation and transpiration ⇄
condensation, ⇄
precipitation, ⇄
runoff, ⇄
infiltration ⇄

So when a thunderstorm occurs it helps in completing the precipitation process by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water runoffs to the surface of the ground, on plants, into rocks, rivers, and lakes.

Next, the Infiltration process enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the evaporation and transpiration process begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back into the atmosphere.

It then condenses and falls back as precipitation again.

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How long does water evaporation take? What factors influence it?

svetoff [14.1K]

There is no certain time on how long it takes. Because the factors will always be different and the factors heavily affect the evaporation time. Some factors include: humidity, heat, how the sun is visible (whether clouds are covering it or not)

7 0

3 years ago

Read 2 more answers

Does the ke of a car change more when it accelerates from 11 km/h to 21 km/h or when it accelerates from 21 km/h to 31 km/h?

svp [43]

Titty milk I think because it taste amazing so you can go 21km/h

7 0

3 years ago

A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the

nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c) 1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

Ke = 1/2mv²

= (1.67×10⁻²⁷×)(1.4106)²/2

= 1.661×10⁻²⁷KJ

5 0

3 years ago

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8

hammer [34]

Missing figure and missing details can be found here:
http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....

Solution:
(a) The work done by the spring is given by
$W= \frac{1}{2} k (\Delta x)^2 
$
where k is the elastic constant of the spring and $\Delta x$ is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
$W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J$

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
$W_W = -F_{//} (x_2 -x_1)$
where the negative sign is given by the fact that $F_{//}$ points in the opposite direction of the displacement of the cart, and where
$F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N$
therefore, the work done by the weight is
$W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J$

8 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$