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natita [175]
3 years ago
14

What are some applications of Kepler’s laws still in use today?

Physics
2 answers:
Levart [38]3 years ago
7 0
<span>to plot and time the positions of comets and asteroids as they orbit the The application to Kepler's law are:
1. Sunplotting a course out of the solar system to explore other planets similar to Earth
2. plot the orbit of moons or man-made space satellites
</span><span>
An example of this is Kepler’s harmonic law is the third law of the Planetary Motion. He discovered this after ten years that there is a relation between the time of a planet’s orbit and its distance from the sun. The harmonic law states that the squares of the orbital periods of the planets around the Sun are proportional to the cubes of the planet’s orbital period</span>
tia_tia [17]3 years ago
7 0

Answer:

A. To plot and time the positions of comets and asteroids  

Explanation:

Kepler's law helps to understand the orbits of many planets and celestial bodies including comets and asteroids.

It was found that like other bodies or planets, comets also revolve around the sun in an elliptical path.

Consider sun is situated at one foci, when comets revolve around the sun the distance is very large but as soon as it comes closer to sun, it starts to melt thus forming a comet tail.

Hence time and position of a comet is described by Kepler's law.

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A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of
OLga [1]
Potential
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5 0
3 years ago
Read 2 more answers
A cylindrical metal can is to have no lid. It is to have a volume of 125π in3. What height minimizes the amount of metal used?
Alex Ar [27]

Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

6 0
3 years ago
During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial
Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

5 0
3 years ago
Help!!! Asap!!!
KiRa [710]

Answer:

Probs paper cuz the rest are metals

4 0
1 year ago
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