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Tom [10]
3 years ago
5

How did kepler descrube the planets orbits​

Physics
1 answer:
Natalka [10]3 years ago
4 0

Answer:

The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit. Kepler's Second Law: the imaginary line joining a planet and the sons sweeps equal areas of space during equal time intervals as the planet orbits.

(Brainliest, please?)

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Which approach would be the most interested in studying Phineas Gage
viva [34]
Biological because its studies the function of the brain’s lobes
7 0
3 years ago
Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t
Law Incorporation [45]

Answer:

The block has an acceleration of 3 m/s^{2}

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

\sum F_{r} = ma   (1)

Where \sum F_{r} represents the net force, m is the mass and a is the acceleration.

F_{x} + F{y} = ma  (2)

The forces present in x are F = 10 N and f = 4 N (the friction force):

F_{x} = 10 N - 4 N

Notice that f subtracts to F since it is at the opposite direction.

F_{x} = 6 N

The forces present in y balance each other:

F_{y} = 0

Therefore:

6 + 0 = ma  

6 N = (2kg)a  (3)

But 1 N = 1 Kg.m/s^{2} and writing (3) in terms of a it is get:

a = \frac{6 Kg.m/s^{2}}{2 Kg}  

a = 3 m/s^{2}

So the block has an acceleration of a = 3 m/s^{2}.

4 0
3 years ago
1000 kg has 50, 000 joules of kinetic energy. what is it's speed?
kicyunya [14]
The answer is 36 kilometers per hour, or 10 meters per second.
7 0
3 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
4 years ago
What is the density of a piece of quartz with a mass of 30g and a volume of 6cm^3
FromTheMoon [43]

Answer:

\boxed {\boxed {\sf d= 5 \ g/cm^3}}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the quartz is 30 grams and the volume is 6 cubic centimeters.

m=30 \ g \\v= 6 \ cm^3

Substitute the values into the formula.

d=\frac{30 \ g}{6 \ cm^3}

Divide.

d= 5 \ g/cm^3

The density of this piece of quartz is 5 grams per cubic centimeter.

7 0
3 years ago
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