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natali 33 [55]
3 years ago
6

Explain Kepler's laws

Physics
2 answers:
cupoosta [38]3 years ago
7 0
Kepler's three laws of planetary motion can be stated as follows: (1) All planets move about the Sun in elliptical orbits, having the Sun as one of the foci. (2) A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.
riadik2000 [5.3K]3 years ago
5 0

Kepler's three laws of planetary motion are:

The 1st law is that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. The 2nd law is the speed of the planet increases as it nears the sun and decreases as it recedes from the sun. The 3rd law is the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

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How t calculate gravitational force
lions [1.4K]

Answer:

F=G(m1m2)/Rsquare if radius is given

F=G(m1m2)/dsquare if distance is given

where,

f =gravitational force

G =gravitational constant

m1=mass of one object

m2=mass of another object

d=distance between two object from their center r=radius of earth/planet

6 0
3 years ago
A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h
babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

8 0
3 years ago
A 56.0 kg box hangs from a rope. what is the tension in the rope if: the box has vy = 5.30 m/s and is speeding up at 5.10 m/s2 ?
Minchanka [31]

T = tension force in the rope in upward direction

m = mass of the box attached at end of rope = 56 kg

W = weight of the box in downward direction due to gravity

a = acceleration of the box in upward direction = 5.10 m/s²

weight of the box is given as

W = mg

inserting the values

W = (56) (9.8)

W = 548.8 N


force equation for the motion of the box is given as

T - W = ma

inserting the values

T - 548.8 = (56) (5.10)

T = 834.4 N

5 0
3 years ago
Which equation would you use to solve the following problem? An object is moved 29 m with a force of 289 N. What is the work don
Ksivusya [100]
Work, scientifically speaking, is done when a force is applied to an object which consequently moves the object at a certain direction. Work in formula, is force multiplied with distance.
 W = F x d
4 0
3 years ago
If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi
saul85 [17]
The gravitational potential energy:
E p = m x g x h
where m is the mass and h is the height;
m = 50 kg,  g = 9.81 m/s² ,  h = 40 m
E p = 50 kg  x 9.81 m/s²  x  40 m
Answer:
E p = 19,620 J = 19.62 kJ
8 0
3 years ago
Read 2 more answers
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