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lbvjy [14]
3 years ago
14

Which contributions did Johannes Kepler make? Check all that apply. He revived Aristotle's model of the solar system. He solved

Ptolemy's model by proving elliptical orbits. He proved Galileo's calculations were incorrect. He determined that planets move faster when closer to the Sun. He discovered laws of planetary motion.
Physics
2 answers:
Arada [10]3 years ago
3 0

The correct answers are: Options 2,4 and, 5

2)He solved Ptolemy’s model by proving elliptical orbits.

4)He determined that planets move faster when closer to the Sun.

5)He discovered laws of planetary motion.

Ilya [14]3 years ago
3 0

Answer:

He solved Ptolemy's model by proving elliptical orbits.

He determined that planets move faster when closer to the Sun.

He discovered laws of planetary motion.

Explanation:

The laws of planetary motion were published by Johannes Kepler. The first law tells us that all planets move around the Sun describing elliptical orbits, with the Sun at one of the two foci. The second law tells us that planets travels faster when closer to the Sun,  then slower when farther from the Sun. The third law tells us that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

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A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the
Elan Coil [88]

Answer:

The horse is going at 12.72 m/s speed.

Explanation:

The initial speed of the horse (u) = 3 m/s

The acceleration of the horse (a)= 5 m/s^{2}

The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m

Applying the second equation of motion to find out the time,

s=ut+\frac{1}{2}at^{2}

15.3=3t+2.5t^{2}

2.5t^{2}+3t-15.3=0

Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.

Now applying first equation of motion, to find out the final velocity,

v=u+at

v=3+1.945*5

v=3+9.72

v=12.72 m/s

The horse travels at a speed of 12.72 m/s after covering the given distance.

7 0
4 years ago
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic
miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let X_A be its distance. In t hours he has walked

X_A=5t\ \text{miles}

Office B is walking at 6 mph directly west. In t hours his distance is

X_B=6t\ \text{miles}

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

D=\sqrt{X_A^2+X_B^2}

D=\sqrt{(5t)^2+(6t)^2}

D=\sqrt{61}t

This distance is known to be 2 miles, so

\sqrt{61}t=2

t =\frac{2}{\sqrt{61}}=0.256\ hours

t is approximately 15 minutes

3 0
3 years ago
Show the equation of simple pendulum to be dimensionally consistent
nataly862011 [7]
T is in seconds (s) 

<span>2pi is dimensionless </span>

<span>L is in meters (m) </span>

<span>g is in meters per second squared (m/s^2) </span>

<span>so you can write the equation for the period of the simple pendulum in its units... </span>

<span>s=sqrt(m/(m/s^2)) </span>

<span>simplify</span>

<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>

<span>s=sqrt(s^2) </span>

<span>s=s </span>

<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
6 0
3 years ago
The brakes application to a car produce an acceleration of 6ms2 in the opposite direction to the motion .If the car takes 2 seco
Anna [14]

Answer:12 meter

Explanation:

acceleration(a)=6m/s^2

Time(t)=2 seconds

Distance =(a x t^2)/2

Distance =(6 x 2^2)/2

Distance=(6 x 2 x 2)/2

Distance=24/2

Distance =12

Distance is 12 meters

7 0
3 years ago
If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti
Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

\frac{T^2}{a^3}=const.

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

Learn more about Kepler's third law:

brainly.com/question/11168300

#LearnwithBrainly

5 0
3 years ago
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