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s344n2d4d5 [400]
3 years ago
9

The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU

Physics
2 answers:
Vsevolod [243]3 years ago
6 0

Answer:

29 AU

Explanation: trust me

valina [46]3 years ago
4 0

Answer:

The answer is 29 AU

Hoped I helped

mark me as brainliest

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What Is An Insulator Give An Example
kati45 [8]

plastics, Styrofoam hOPE THIS HELPS

8 0
2 years ago
Read 2 more answers
1. When an object is at rest, not moving, and is crashed into by another
wlad13 [49]

Answer:

both experience forces or at least a force

Explanation:

it would go in the direction the other object

(second object, the one that crashed) was going

si if going right then right if left then left

plus or minus

4 0
3 years ago
What is the 5 quantitative research problems?<br>​
iogann1982 [59]

Answer:

There are four main types of Quantitative research: Descriptive, Correlational, Causal-Comparative/Quasi-Experimental, and Experimental Research. attempts to establish cause- effect relationships among the variables. These types of design are very similar to true experiments, but with some key differences.

Explanation:

Quantitative research is defined as a systematic investigation of phenomena by gathering quantifiable data and performing statistical, mathematical, or computational techniques.

<h3>I hope it will help you</h3>

<h3><em>please</em><em> make</em><em> me</em><em> brainlest</em></h3>

<h2>THANK U</h2>

7 0
2 years ago
How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for
kumpel [21]

As we know that electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

F_g = 1.98 \times 10^{20} N

here we can assume that both will have equal charge of magnitude "q"

now we have

1.98 \times 10^{20} = \frac{kq^2}{r^2}

1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}

1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2

now we have

q = 5.7 \times 10^{13} C

6 0
3 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0
3 years ago
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