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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.

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A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0

3 years ago

What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af

sladkih [1.3K]

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$ .

4 0

2 years ago

What energy output objects work with the turbine?

Sonbull [250]

Answer:

Energy output from Solar panel is Electric Energy so any object which require electric energy an input will run by turbine. for example Electric Bulb,Electric Water Heaters.

hope it helps you!!

6 0

3 years ago

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr

mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out