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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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What is the velocity?
Blababa [14]

Answer:

B. 0.98 m/s

Explanation:

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I hope this made sense and hoped it helped. Good luck with your test luv :)

8 0
3 years ago
Please help me do this problem
Digiron [165]

Answer:

i) acceleration from B to D is 0, because the velocity is constant (stays the same)

ii) whatever units of distathat might be, we can calculate the number:

for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

4 0
3 years ago
A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

(c) energy is being delivered by the battery?

The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

4 0
3 years ago
Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation  that is

for red light

h ( c / λr ) = Ф +  kinetic energy

for green light

h ( c / λg ) = Ф +  1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr  - 1/ λg)

put all value

Ф =( 6.63 ×10^{-34} (3 ×10^{8} )  / 0.50) × ( 1.50/ λr  - 1/ λg)

Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

Ф = 1.5305  ×10^{-19} J  × ( 1ev / 1.6 ×10^{-19} J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0
3 years ago
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A car is driving along a road at a speed of 50 km/hr. How far will the car travel in 5 hours?​
BARSIC [14]

Answer:

I think it's 250

Explanation:

If the car is traveling 50 km/hr that means every hour, the car drives 50 km. So if you want to know how far it will go in 5 hours you do 50x5.

4 0
2 years ago
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