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VMariaS [17]
2 years ago
13

The location movement and intensity of precipitation can be found by using?

Physics
1 answer:
vaieri [72.5K]2 years ago
4 0

<span>Answer: Doppler radar
Doppler radar or weather radar is used to locate precipitation,</span> forecast future intensity, calculate or detecting its motion and estimates precipitation whether it is a rain, snow, hail or storm. They are also helpful in determining the structure of storms and its potential to cause severe weather.

Moreover, if a portion of the atmosphere becomes saturated with water vapor, it will condense and precipitate allowing Doppler radar to analyze and observe precipitation.

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A farmer mated two ducks. They produced three offspring. Two of the offspring were then selected to mate with each other because
cestrela7 [59]

it's called inbreeding


7 0
3 years ago
A 4-kg object falls vertically a distance of 5 m. its potential energy has changed by approximately how much?
Brums [2.3K]
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen

Neglect air resistance.

The PE (potential energy) is 
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J

The PE is converted into KE (kinetic energy) after the fall. 
Therefore the PE decreased by 196 J ≈ 200 J

Answer: d. It has decreased by 200 J
7 0
3 years ago
what is the power of an electrical device which operates with a current of 12.4 A and a potential difference of 12 V​
jek_recluse [69]

148.8 Watts

Explanation:

P = VI

= (12 V)(12.4 A)

= 148.8 Watts

3 0
2 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

        F = g m₁m₂ / r²

        Σ F = 0

       F1- F2 = 0

       F1 = F2

We set the reference system in the body of greatest mass (M) the planet

       F1 = g m₁ M / x²

       F2 = G m1 m / (D-x)²

      G m₁ M / x² = G m₁ m / (D-x)²

      M (D-x)² = m x²

      MD² -2MD x + M x² = m x²

     x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

     x = [2MD  ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

     x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

     x = D {M  ±  Ra (2M²-mM)} / (M-m)

    x = D (M ± M √ (2-m/M)) / (M-m)

    x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

 

    x = D (M / (M-m)) (1 + √2)

     x = D (M/M-m) 2.41

b) X = 2/3 D

     x = D (M/M-m) 2.41

     2/3 D = D (M/(M-m)) 2.41

     2/3 (M-m) = M 2.41

     2/3 M - 2/3 m = 2.41 M

     1.743 M = 0.667 m

     M/m = 0.667/1.743

     M/m =  0.38

3 0
2 years ago
If you have 5.112 grams of ammonia (NH3), then how many molecules of ammonia do you have?
Dominik [7]
I hope it helps! good luck in chem!

7 0
2 years ago
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