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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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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
Tanzania [10]

Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0
3 years ago
A car starting from rest has a constant acceleration of 4 meters per second per second. How fast will it be going in 5 seconds?
irinina [24]

Answer:

18 m

Explanation:

Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s

solving for the final velocity, v

v = a * t

v = 4 m/s^2 * 3 s

v = 12 m / s

Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

4 0
2 years ago
Where is the safest spot in the house during a tornado, hail storm, and earthquake? Include separate answers.
USPshnik [31]

Answer:

Explanation:

in a room with no windows that way it doesnt shatter. Usually a closet... or if you have a basement.sorry but that applies to all them

4 0
3 years ago
Read 2 more answers
A special triangular frame is being made for a piece of artwork. The the base of the triangular frame must be 90 cm. If the area
Mashutka [201]

Answer:

Explanation:

base of triangular frame, b = 90 cm

Area, A = 765 cm²

Let the height is h.

Area of a triangular frame = 1/2 x base x height

765 = 0.5 x 90 x h

h = 17 cm

Thus, the height of triangular frame is 17 cm.

5 0
3 years ago
You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr
Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

v_r = 27 - 10 = 17 m/s

relative deceleration of car

a_r = -7 m/s^2

now the distance before they stop with respect to each other is given by

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2 *(-7)*d

d = 20.6 m

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

v_f^2 - v_i^2 = 2 a s

0^2  - 27^2 = 2 * (-7)* s

s = 52.1 m

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

v_f - v_i = at

0 - 27 = (-7) * t

t = 3.86 s

now the distance covered by truck in same time

d = 3.86 * 10 = 38.6 m

now after the car will stop its distance from the truck is

D = 25 + 38.6 - 52.1 = 11.5 m

<em>so the distance between them is 11.5 m</em>

6 0
3 years ago
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