"The gravity of the Sun causes the planets to move in a circular path."

1 answer:

3 0

You're thinking of Kepler's first law of planetary motion, but that's not
what it says.

First of all, Kepler didn't say anything about gravity. He only described
the orbits. And he said that the orbit of each planet is an ellipse, with the
sun at one focus.

If by some chance you're thinking of Newton ... he showed that if his formula for
gravity is correct, then the orbit of a planet must be an ellipse. But Newton's law
of gravity is not one of his so-called "three laws of motion".

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A bike rider was moving to the right at a constant speed. Suddenly the wind then starts blowing against her with 3 N of force to

stiks02 [169]

Answer:

The correct option is D

Explanation:

This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). This wind/force would cause her speed to reduce or slow down (as posited by the law).

8 0

3 years ago

If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?

jeka94

Answer:

$v=60m/s$

Explanation:

Use the Kinematic Equation:

$v=v_o+at$

Plug in what is given and solve

$v=0+(12)(5)$

$v=60m/s$

3 0

2 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$