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disa [49]
2 years ago
11

What did johannes kepler contribute to the study of planets? he formulated the universal law of gravitation. he identified the f

irst moons of jupiter. he collected detailed data that led to the proposal of the heliocentric model. he provided mathematical support for the heliocentric model.
Physics
1 answer:
Deffense [45]2 years ago
7 0

Johannes Kepler contributed to the study of planets : He provided mathematical support for the heliocentric model.

<h3>What is Kepler's first law of planetary motion?</h3>

Kepler's first law of planetary motion states that the planets revolve around the Sun in orbits elliptical in shape with the Sun at its one of the focus.

As the planets rotate around the Sun, the relative distance between the Sun and planets increases and decreases.

The heliocentric model in which the Earth and planets revolve around the Sun at the center of the universe. Johannes gave the mathematical calculations of this model.

Thus, last option is correct.

Learn more about Kepler's first law of planetary motion.

brainly.com/question/4978861

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Ocean breezes are due to which method of heat transfer?
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Ocean breezes are due to the convection method of heat transfer
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3 years ago
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Two different liquids are poured into a jar until it is half full. The jar is then sealed shut and shaken. The liquids undergo a
Lilit [14]

Answer:

A closed system.

Explanation:

The three major types of system are: open, closed and isolated. Open system interacts with its surroundings with respect to its particles and energy. A closed system interacts with its surroundings with respect to energy but not its particles. While an isolated system does not interact with its surroundings in any way.

Therefore, after the jar is sealed, it is an example of a closed system. This is because the emitted gas could not escape into the surroundings, but thermal energy was emitted into its surroundings after the chemical reaction has taken place.

7 0
3 years ago
What is conclusion about potential difference (voltmeter readings) in a parallel electric circuit ?
Ainat [17]

Answer: the conclusion is that

Explanation:

5 0
3 years ago
An object with a mass 4.0 kg has a momentum of 64 kgm/s . How fast is the object traveling ?
Anastasy [175]

Answer:

The object will travel at the speed of 16 m/s.

Explanation:

Given

  • Mass m = 4.0 kg
  • Momentum p = 64 kgm/s

To determine

How fast is the object traveling?

<u>Important Tip:</u>

The product of the mass and velocity of an object —  momentum.

Using the formula

p = mv

where

  • m = mass
  • v = velcity
  • p = momentum

Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula

p = mv

64\:=\:4\times v

switch the equation

\:4\times \:v\:=64

divide both sides by 4

\frac{4v}{4}=\frac{64}{4}

simplify

v=16 m/s

Therefore, the object will travel at the speed of 16 m/s.

3 0
3 years ago
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
4 years ago
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