Answer: motion parallax
Explanation:
Motion parallax refers to a form of depth perception whereby objects that are closer to an individual appears to move at a faster speed than the objects that are far.
Therefore, Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the mountains in the distance is an example of motion parallax.
Answer:
The speed of the ball was, v = 3 m/s
Explanation:
Given data,
The time period of the ball, t = 8 s
The distance the ball rolled, d = 24 m
The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,
v = d / t m/s
Substituting the given values in the above equation,
v = 24 / 8
= 3 m/s
Hence, the speed of the ball was, v = 3 m/s
Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀

∴ Capacitance is 0.572×10⁻¹⁰ Farad
I think metal, steel and copper.
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''


Vy in this problem will follow this equation =

where g is the gravity acceleration

This is equation (1)
For Y(t) :

We suppose yi = 0

This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s

The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m