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3 years ago

Explain an experiment of the phenomenon of rainfall

Physics

1 answer:

maks197457 [2]3 years ago

7 0

Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.

Explanation:

Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.

The water cycle makes rainfall possible:

First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.

Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.

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The weight of an astronaut plus his space suit on the Moon is only 250 N. (a) How much does the suited astronaut weigh on Earth?

Evgesh-ka [11]

Answer:

a)1500N

b)153.06kg

Explanation:

F = ma

g(moon) = is the acceleration due to gravity on the moon

g(earth) is the acceleration due to gravity on the earth

g(moon) = 1/6g(earth)

g(earth) =6g(moon)

F(gearth) = mg(earth)

= m 6g(moon)

= 6 × 250

= 1500N

b) F(gearth) = mg(earth)

m = F /g

= 1500/9.8

= 153.06kg

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3 years ago

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stellarik [79]

Answer: This is not easy lol

Explanation:

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3 years ago

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A rocket has total mass Mi=360 kg , including Mf=330kg of fuel and oxidizer. In interstellar space, it starts from rest at the p

steposvetlana [31]

let the parameters are:

M=instantaneous mass of the rocket

v=velocity of the rocket

t=time

initial mass=Mi=360 kg

fuel mass=Mf=330 kg

relative speed=Ve=1500 m/s

rate of mass decay=k=2.5 kg/s

writing newton’s second law of motion:

M*du/dt=-Ve*dM/dt

==>du=-Ve*dM/M

Integration of both sides,

u=-Ve*ln(M)

using the limits with at t=0, u=0 and velocity being v(t),

at t=0, mass=Mi, at any time t, mass=Mi-k*t

v(t)-0=-Ve*ln((Mi-k*t)/Mi)

v(t)=-Ve*ln(1-(k*t/Mi))

<h3>What is needed for rocket propulsion?</h3>

Rocket propulsion is the strategy utilized to make the basic thrust to lift a rocket into the air. The force that the rocket employments to lift off from the soil is known as rocket propulsion. The third law of movement by Newton serves as the foundation for rocket propulsion. Here, the fuel is shot out from the exit mightily in arrange to cause an equal and converse reaction. The some sorts of rocket propulsion are fuelled by liquid, strong, cold gas, and particles. The rocket's mass, fuel burn rate, and weaken speed all impact speeding up.

To learn more about Rocket propulsion, visit;

brainly.com/question/15363207

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1 year ago

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2× $10^{-6}$ J

(b) KE = 2× $10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

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3 years ago

How would playing a game of soccer, baseball, or basketball be different if inertia didn’t exist?

Elina [12.6K]

This would be a funny game.

Inertia, according to first Law of Newton, means that when you throw a ball it will continue moving unless a force acts over it.

Without inertia, the ball instead of continue moving, would stop as soon as it leaves your hand or foot. You had to stay pasted to the ball unitl it reaches the desired destiny for it continue moving.

5 0

4 years ago

Explain an experiment of the phenomenon of rainfall​

Explain an experiment of the phenomenon of rainfall