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Vinvika [58]
3 years ago
10

if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud

Physics
1 answer:
sveta [45]3 years ago
3 0

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

          (1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass . 

On Earth, that mass weighs  4,900,000 newtons .

                                            (about  1,102,300 pounds)

                                            (about  551 tons)

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In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.
lawyer [7]

The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.

The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz

Learn more: brainly.com/question/14588679

6 0
2 years ago
Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1
max2010maxim [7]

Answer:

2.5 m/s

Explanation:

Mechanical energy is the sum of the potential and kinetic energy.

E = PE + KE

E = mgh + ½mv²

172.1 J = (7.26 kg) (9.8 m/s²) (2.1 m) + ½ (7.26 kg) v²

v = 2.5 m/s

7 0
3 years ago
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True or false: Scientific models are always built to scale, and are fully functional representations of the product they are tes
vodka [1.7K]

Answer:

False

Explanation:

Scientific models are not always built to scale and may not be fully accurate

3 0
2 years ago
4.
tatiyna
The answer is A. A folkway is a closely held belief by a specific group of people.
6 0
3 years ago
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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
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