All categories

3 years ago

"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".

You might be interested in

Given the picture below, what is the volume of the object?

Valentin [98]

The answer is D, 10 ml

4 0

3 years ago

Which of the following is most likely to happen when energy is transferred to

ankoles [38]

Explanation:

the object will begin to move

4 0

3 years ago

Read 2 more answers

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b

kolbaska11 [484]

To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV

= 1.240x10^-10 m

= 1.240x10^-1 nm

3 0

3 years ago

Read 2 more answers

PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE

marysya [2.9K]

Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2 where h is the vertical distance thru which the pendulum travels

5 0

2 years ago