All categories

3 years ago

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

2 answers:

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.

You might be interested in

Physicist ____ was found to have completely made up some of the data on which his astounding discoveries were made.

GaryK [48]

Answer:

Jan Hendrik Schön I believe

6 0

3 years ago

Which statement is true about a planet’s orbital motion?

lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

$V = \sqrt{\frac{GM}{R + h} }$

Where

V - velocity of the orbit at a height h from the surface

R - Radius of the second object

G - Gravitational constant

h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

$1/2 mV^{2} = GMm/R$

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0

3 years ago

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

good morning you are you still

8 0

3 years ago

A small plane flies at a speed of 102 km/h in still air. Suppose the wind blows out from the west (with the air moving east) at

Semmy [17]

Answer:

2.68 hours

Explanation:

A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.

B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.

The plan resultant speed can be calculated by using pythagorean theorem.

Resultant Speed = Sqrt( 102^2 + 46^2 )

Resultant Speed = Sqrt( 12520)

Resultant speed = 111.89 km/h

From the definition of speed,

Speed = distance/time

Where distance = 300 km

Substitute the resultant speed and the distance into the formula.

111.89 = 300/time

Time = 300/111.89

Time = 2.68 hours

Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point

7 0

4 years ago