Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
Answer:
A) and B) are correct.
Explanation:
Let's take a look at the attached picture. Now
The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE
The equivalent capacitance is going to be: 
And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE
And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE
Answer:
Protons and electrons are charged particles. Neutrons have no charge.
I believe that would be D.
Cardiovascular disease generally refers to conditions that involve narrowed or blocked blood vessels that can lead to a heart attack, chest pain, or stroke. Which it has nothing to do with being infectious.
Answer:
θ = 1.591 10⁻² rad
Explanation:
For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source
Let's write the diffraction equation for a slit
a sin θ = m λ
The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ
θ = λ / a
In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.
θ = 1.22 λ / D
Where D is the diameter of the pupil
Let's apply this equation to our case
θ = 1.22 600 10⁻⁹ / 0.460 10⁻²
θ = 1.591 10⁻² rad
This is the angle separation to solve the two light sources