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2 years ago

9

The law of suggests that the orbit of planets is not circular but.

2 answers:

adoni [48]2 years ago

7 0

Newton's law of universal gravitation ... massaged and manipulated
with a lot of calculus and geometry ... predicts that closed gravitational
orbits are ellipses.

The circle is such a precise special kind of ellipse that a circular orbit
never occurs. The chance of it is like the chance of flipping a coin and
having it land standing up on its edge ... possible, but very improbable.

anygoal [31]2 years ago

5 0

Answer:

Kepler's 2nd law describe the shape of orbit

Explanation:

Kepler's 2nd law describe the shape of orbit

A planet's orbit's around which it revolve around the Sun is not a circle. It's an ellipse — a ring that's "flattened." The Sun (which is also called the planet's center) holds one ellipse target. A concentration is among two inner points helping to decide an ellipse's form.

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How fast is a car going if it accelerates at 10m/s/s for 4 seconds

lara31 [8.8K]

Answer:

40 ms-1

Explanation:

v=u+at

v=0+10×4

v=40

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2 years ago

A portable CD player uses two 1.5 volt batteries. if the current in the CD player is 2amps ,what is resistance?

chubhunter [2.5K]

resistance=voltage/current

1.5/2=0.75ohms

7 0

2 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s

Mumz [18]

Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

$M = \dfrac{W}{g}$

$M = \dfrac{40}{9.8}$

M = 4.08 Kg

b) Mass on the surface of Lo

Mass of an object remain same.

Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

W' = m g'

W' =4.08 x 1.81

W' = 7.38 N

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2 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

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3 years ago

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svp [43]

D dissolving solid NaF in water

6 0

3 years ago

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