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3 years ago

Which of the following causes clouds to form?

2 answers:

5 0

The answer is B because this is part of precipitation and in the process Water vapor condensing as it rises into cooler air and the clouds gets heavy from the condensation and forms clouds then the clouds form rain and the rain gets heavy and falls down.

alisha [4.7K]3 years ago

4 0

The correct answer is B. Water vapor condensing as it rises into cooler air
Hope this helps have a nice day

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Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light

Nimfa-mama [501]

Answer:

$f = 5.95 \times 10^{14} Hz$

Explanation:

As we know that the frequency of the wave is given as

$f = \frac{c}{\lambda}$

here we know that

$c = 3 \times 10^8 m/s$

also we know that

$\lambda = 5.04 \times 10^{-7} m$

now we have

$f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}$

$f = 5.95 \times 10^{14} Hz$

6 0

3 years ago

Question 6 of 10

sergejj [24]

The answer is C.89,866. I multiplied 262x343.

3 0

3 years ago

The boy on the tower throws a ball 20m downrange. What is his pitching speed?

san4es73 [151]

Answer:

The pitching speed of the ball is 19.7 m/s

Explanation:

Here, we can use the third equation of motion, $v^{2} = u^{2} - 2as$

whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled

Here, the initial velocity of the the ball is given as zero and the acceleration due to gravity is 9.8 , the distance 's' is given as 20 m

Using the equation, $v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s$

Hence, the pitching speed of the ball is 19.7 m/s

5 0

3 years ago

A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

7 0

3 years ago

When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool

zubka84 [21]

This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster

5 0

4 years ago

Read 2 more answers