The units of G must be C. m³ / ( kg s² )
<h3>Further explanation</h3>
Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:
<em>F = Gravitational Force ( Newton )</em>
<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>
<em>m = Object's Mass ( kg )</em>
<em>R = Distance Between Objects ( m )</em>
Let us now tackle the problem !
To find unit of Gravitational Constant can be carried out in the following way:
![{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}](https://tex.z-dn.net/?f=%7B%5BN%5D%7D%3D%20G%5Cfrac%7B%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%7D%7B%7B%5Bm%5E2%5D%7D%7D)
![{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}](https://tex.z-dn.net/?f=%7B%5Bkg%20~%20m%20%2F%20s%5E2%5D%7D%3D%20G%20%5Cfrac%7B%7B%5Bkg%5E2%5D%7D%7D%7B%7B%5Bm%5E2%5D%7D%7D)
![G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bkg%20~%20m%20%2F%20s%5E2%5D%7D%7B%5Bm%5E2%5D%7D%7D%20%7B%7B%5Bkg%5E2%5D%7D%20%7D)
![G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bkg%20~%20m%5E3%20%2F%20s%5E2%5D%7D%7D%20%7B%7B%5Bkg%5E2%5D%7D%20%7D)
![G = \frac{{[m^3 / s^2]}} {{[kg]} }](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7B%7B%5Bm%5E3%20%2F%20s%5E2%5D%7D%7D%20%7B%7B%5Bkg%5D%7D%20%7D)
![\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}](https://tex.z-dn.net/?f=%5Cboxed%20%7BG%20%3D%20%5Cfrac%7B%7B%5Bm%5E3%5D%7D%7D%20%7B%7B%5Bkg%20~%20s%5E2%5D%7D%20%7D%7D)
The unit of G must be 
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Gravitational Fields
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
Answer:
D
Explanation:
I hope you get a good grade!
The answer to this question would be:3850ft
To answer this question, you need to convert the speed velocity from miles/hour into feet/second. The equation would be: 750 miles/hour x 5280 foot/mile x 1 hour/3600second = 1100 ft/s
Then multiply the time with the velocity= 3.5 second x 1100 ft/s= 3850ft
Answer:
They don’t ‘represent’ anything, they are properties of the wave.
Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.
If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.
For many waves, we don’t perceive them at all (e.g. radio waves).
For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.
Answer:
20 m/s/s
Explanation:
F=ma, 350=17.5 * a, a=20 m/s/s
