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
Explanation:
The tangential speed of Andrea is given by :

Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,

New angular speed is twice that of the Chuck's speed.
The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
- <em>Current flowing in the wire, I = 4.00 mA</em>
- <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
- <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
- <em>Length of wire, L = 2.00 m</em>
- <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>
<em />
The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
Learn more here: brainly.com/question/22406248
Answer:
= 2.33
Explanation:
.According to snell's law:
n1sin i = n2sin r ,
where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,
i is angle of incidence,
r is angle of refraction.
Given that,
n1 = 1,
i = 51 degrees,
r = 19.5 degrees. ,
n2= ?
So,
1*sin 51 = n2 sin 19.5
=> n2 = sin51 / sin19.5
= 2.33