Answer:
F' = F
Explanation:
The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

where,
F = Force of attraction
G = Universal gravitational costant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects
Now, if the masses and the distance between them is doubled:

<u>F' = F</u>
First off, you need to know the weight of the projectile, lift and drag coefficients something like a high Reynolds number is preferred, then use the gravitational constant of 9.8 meters per second squared those would be a good start to get closer to your goal
<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

Which is the final velocity of the two railroad cars
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:



Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:



Replacing the values we have:


Therefore the refractive index for the liquid is 1.7323
Answer: You will only see the color that cellophane lets through
Explanation:
Let's begin by the fact the whole electromagnetic spectrum is known as "white light", which is composed by a range of colors (wavelengths).
Now, if we have a source with white light (the Sun, for example) and we interpose a cellophane of any color (let's choose red), this cellophane will act as a filter and will only let pass the color of the cellophane.
This is because the filter will absorb the other colors of the spectrum.