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elena55 [62]
3 years ago
6

Halley’s Comet appears around the sun about once every 76 years. It passes through the part of its orbit nearest the sun in just

a few months but then moves out to great distances and takes more than half a century for the rest of its orbit. Explain these time differences in terms of Kepler’s laws.
Physics
1 answer:
Scilla [17]3 years ago
4 0

Answer:

There are actually three, Kepler’s laws that is, of planetary motion: 1) every planet’s orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit. As it’s the third which is most often used, Kepler’s law usually means Kepler’s third law (of planetary motion).

Explanation:

Kepler's third law would tell us that Halley's comet has an average distance much greater than that of the Earth. However, there is a time in Halley's comet's orbit that brings it closer to the Sun than the Earth. Kepler's third law is a mathematical relation between a planet's period and its average distance.

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Mars was formed by the __________ of smaller objects.
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How many electrons can a nucleas hold
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A nuclear can hold 2 electrons
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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
Why is hydroelectric power considered to be an example of multiple transfers of energy?
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Hydroelectric power is considered to be an example of multiple transfers of energy because potential energy when the water is in the reservoir turns into kinetic energy making the water move to the dam and then the dam moves into the turbine turning it to electric energy.  
7 0
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Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector
Slav-nsk [51]

Answer:

The resultant vector is 1 m/s

Explanation:

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3 years ago
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