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adell [148]
3 years ago
13

What is the difference between condensation and precipitation

Physics
2 answers:
saw5 [17]3 years ago
5 0

Answer: The difference between condensation and precipitation is the is the force of gravity that acts on precipitation and draws down the heavy liquid water down to the earth in form of rain.

Explanation: Condensation is the physical change of gaseous phase of water that is water vapour into liquid phase as liquid water an example is fog. Condensation basically depends on temperature and pressure.

Precipitation is the condensation of gaseous phase of water i.e. atmospheric water vapour which falls under gravity from clouds back to the earth in form of drizzle and/or rain. Precipitation depends on the temperature and the concentration of the solution.

Alexxandr [17]3 years ago
3 0

Explanation:

The water cycle is based on three parts;

1. Evaporation

2. Condensation

3. Participation

Condensation:

It is the process in which water vapor changes into liquid water or in other words, it is the transition from the gaseous state to liquid state.

Precipitation:

It is the process in which any liquid or frozen water such as snow that forms in the atmosphere and falls back to the Earth

Condensation depends on temperature and pressure whereas precipitation depends on the temperature and the concentration of the solution.

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Explanation:

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People living at Earth’s equator are traveling at a constant speed of about 1,670 km/h as Earth spins on its axis. What are thes
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The people living around the equator experience acceleration. Acceleration is the change in speed or/and direction. When on a rotating body, the speed does not change it is only the direction that changes as it rotates.When on a rotating body you do not have to change speed but only direction. In a normal situation when you are traveling on a circle you will never have a straight line.
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3 years ago
Car A (mass 1100 kg) is stopped at a traffic light when it is rear­ended by car B(mass 1400 kg). Both cars then slide with locke
viva [34]

Answer:

Part a)

v_a = 3.94 m/s

Part b)

v_b = 3.35 m/s

Part C)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

a = -\frac{F_f}{m}

a = -\frac{\mu mg}{m}

a = - \mu g

now we will have

v_f^2 - v_i^2 = 2ad

0 - v_i^2 = 2(-\mu g)(6.1)

v_a = \sqrt{(2(0.13)(9.81)(6.1)}

v_a = 3.94 m/s

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

v_b = \sqrt{2(0.13)(9.81)(4.4)}

v_b = 3.35 m/s

Part C)

By momentum conservation we will have

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1400 v_b = 1100(3.94) + 1400(3.35)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

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3 years ago
¡¡¡AYUDA CON ESTOS EJERCICIOS DE FÍSICA!!!
asambeis [7]

Answer:

(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C

Explanation:

(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J

The electric potential is given by

W = q V

40\times10^{-6}=5 \times10^{-6}\times V\\\\V = 8 V

(b)

charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m

Let the potential is V.

V =\frac{k q}{r}\\\\V =\frac{9\times 10^{9}\times 8\times 10^{-6}}{0.5}\\\\V =144000 V

(c)

Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V

Let the charge is q.

W= q V

8\times10^{-5}= q\times 4000\\\\q =2\times 10^{-8} C

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3 years ago
If an object is moving up, is it considered to have a positive or negative velocity?
grin007 [14]

Answer:

A positive velocity

Explanation:

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