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3 years ago

Why does it rain a lot?

Physics

1 answer:

Vaselesa [24]3 years ago

3 0

Answer:

Clouds are made up of tiny water droplets. ... As more and more droplets join together they become too heavy and fall from the cloud as rain. Warm air can hold more moisture than cool air. When the warmer air is cooled and the moisture condenses, it often rains more heavily.

Explanation:

hope it helps

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Which of the following are examples of projectile motion?

liubo4ka [24]

Answer: A. a basketball being shot toward the basket

Explanation: The definition of projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. So, the basketball is the object being thrown and the person throwing the ball is aiming it to go into the basket making that the path of trajectory. Hope that makes sense and helps!

5 0

2 years ago

In order for a ball to move upward, the initial velocity of the ball must be greater than _____.

ch4aika [34]

Answer:

The answer is zero please Give me Brainly

Explanation:

5 0

2 years ago

Please help me out as soon as you can. Really need help.

olga nikolaevna [1]

The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

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4 0

3 years ago

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

3 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago