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4 years ago

For precipitation to form, cloud droplets must grow in volume by roughly __________ times.

2 answers:

8 0

<span>For precipitation to form, cloud droplets must grow in volume by roughly one million times.

Hope I helped :)</span>

ziro4ka [17]4 years ago

7 0

The correct answer is by roughly one million times.

Any product of the condensation of atmospheric water vapor, which falls under gravity is known as precipitation. The main forms of precipitation are rain, drizzle, snow, sleet, and others.

All the precipitation give rise to clouds and the clouds are produced when water vapor in the atmosphere condenses and cools. With the condensation of water vapor, the formation of water droplets takes place, and if the formation of clouds takes place within or is moved into the part of the atmosphere, which is below freezing then the droplets turn into ice crystals.

In order to form precipitation, the cloud droplets must grow in volume by roughly one million times.

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When reading the printout from a laser printer, you are actually looking at an array of tiny dots.

solniwko [45]

Answer:

The value is $y = 3.097 * 10^{-5} \ m$

Explanation:

From the question we are told that

The diameter of the pupil is $d_p = 4.2 \ mm = 4.2 *10^{-3} \ m$

The distance of the page from the eye $d = 29 \ cm = 0.29 \ m$

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9} \ m$

The refractive index is $n_r = 1.36$

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

$y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d$

$y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29$

$y = 3.097 * 10^{-5} \ m$

7 0

3 years ago

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

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3 years ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$