The electrostatic force between two charges q1 and q2 is given by

where

is the Coulomb's constant and r is the distance between the two charges.
If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
Answer:
a
The orbital speed is 
b
The escape velocity of the rocket is 
Explanation:
Generally angular velocity is mathematically represented as
Where T is the period which is given as 1.6 days = 
Substituting the value


At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as

Where G is the universal gravitational constant with a value 
M is the mass of the earth with a constant value of 
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
![r = \sqrt[3]{\frac{GM}{w^2} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7Bw%5E2%7D%20%7D)
![= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B6.67%2A10%5E%7B-11%7D%20%2A%205.98%2A10%5E%7B24%7D%7D%7B%284.45%2A10%5E%7B-5%7D%29%5E2%7D%20%7D)

The orbital speed is represented mathematically as

Substituting value

The escape velocity is mathematically represented as

Substituting values


Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface, 
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :




So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
We know that
• The mass of the elevator is 5000 kg.
Let's draw a free-body diagram.
As you can observe, there are just two forces involved, the weight of the elevator and the tension force. Let's use Newton's Second Law.

But, W = mg = 5000kg*9.8m/s^2 = 49,000 N, and m = 5000 kg, a = 0 (because the speed is constant).

<h2>Therefore, the tension in the cable is 49,000 N.</h2>
Answer:
29.0 g
Explanation:
The mass of the piece of gold is given by:
m = dV
where
m is the mass
d is the density
V is the volume of the piece of gold
The density of gold is
d = 19.3 g/cm^3
while the volume of the sample is equal to the volume of displaced water, so
V = 64.5 mL - 63.0 mL = 1.5 mL
And since
1 mL = 1 cm^3
the volume is
V = 1.5 cm^3
So the mass of the piece of gold is:
m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g