Answer:
VB − VA = g tAB & (VA + VB)/2 = h / tAB
Explanation:
s = h = Displacement
tAB = t = Time taken
VA = u = Initial velocity
VB = v = Final velocity
a = g = Acceleration due to gravity = 9.8 m/s²




Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used
Answer:
17.6 m/s²
Explanation:
Given:
= 90 m/s (final velocity)
= 2 m/s (initial velocity)
Δt = 5s (change in time)
The formula for acceleration is:
= Δv / Δt
We can find Δv by doing
Δv =
- 
Replace the values
Δv = 90m/s - 2m/s
Δv= 88m/s
Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.
= Δv / Δt
= 
acceleration = 17.6 
E S *
The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).
The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years
Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.
To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.
As a result, you get a mere 268335.7 centimeters.
To put that into perspective, that's only about 1.7 miles
A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
The average kinetic energy<span> of a </span>gas<span> particle is </span>directly proportional<span> to the </span>temperature<span>.</span>