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3 years ago

What part do tiny, solid particles in the atmosphere play in cloud formation?

1 answer:

5 0

Well,

The tiny, solid particles you find in the atmosphere serve as cloud condensation nuclei.

You see, in order for water to condense, the water must have a direct object, grammatically speaking.

Your answer: Tiny, solid particles in the atmosphere play a role in cloud formation by providing the object upon which water may condense to form clouds.

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Calculate the kinetic energy of a 64.g bullet moving at a speed of 411.ms . Round your answer to 2 significant digits.

Mandarinka [93]

Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the bullet.

v = speed of the bullet.

Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

Then, the kinetic energy will be the following:

KE = 1/2 · 0.064 kg · (411 m/s)²

KE = 5.4 × 10³ J

8 0

3 years ago

Trình bày về Thí nghiệm tán xạ Rutherford, phát hiện proton.

Setler [38]

Answer:

Any other language I don't know this language

6 0

3 years ago

If you adjust the brightness of a light until it matches the intensity of sourness of the taste of a lemon, you are engaging in

Butoxors [25]

Unrestricted Submarine Warfare

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4 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

4 0

2 years ago