Answer:
Yes.
Explanation:
Displacement is a vector quantity, meaning it has to be a straight graph jointing from starting point to final point and has a direction as well as magnitude and it can also be negative as well as positive and zero.
Displacement is also changes in position (points) of distance. A distance is scalar quantity meaning it only has magnitude and does not have direction. Since distance is scalar quantity, it cannot be negative. Also graph of distance can be a curve graph or any graph.
So when does displacement equal to distance? If an object is moving in straight path or line in one/fixed direction, both displacement and distance are same.
Because if a distance is a straight path and the displacement is simply a straight line (or ray) jointing from starting point to end point, both distance and displacement are both straight path or rays. Since distance and displacement have same graph, we can conclude that both are same in values.
If we go by calculus, given x = 2t as example of straight graph where x stands for position and t stands for time. We can find the displacement from b to a by using the following formulas:
<u>Displacement</u>
<u /><u />
<u>Distance</u>
<u /><u />
v stands for velocity which we can find from:
<u>Velocity</u>
<u /><u />
If you have learnt calculus, first, differentiate x = 2t with respect to t.
Then substitute v = 2 in s and l to find find if both displacement and distance are equal in straight path.
<u>Displacement</u>
<u /><u />
<u>Distance</u>
<u /><u />
Since both displacement and distance are equal when integrating on straight graph, we can conclude that an object moving in straight fixed point has same distance and displacement.
Answer:
x rays is the correct answer
When a wave and/or the observer are moving, the frequency changes according to the Doppler Effect.
When the source is at rest and the observer moves away from it, the wave looks slower and its velocity will be:
v' = vsound - vobserver
In this case,
f₁ = v'/λ = (vsound - vobserver) / λ = f₀ · (<span>vsound - vobserver) / vsound
= f</span>₀ · (1 - vobserver/vsound)
On the contrary, when the source is moving away from an observer at rest, the wave appears with a different wavelengh:
λ' = (vsound + vsource) / f₀
In this case,
f₂ = v/<span>λ' = vsound / ((vsound + vsource) / f₀)
= </span>f₀ ·
Now, let's put together these two cases:
f' = v' / <span>λ' =
= </span>
<span>
= f</span>₀ ·
= f<span>₀ · </span>
A. Pitch
okay bye have a nice day
From: Charli
1, 5 and 6 are the rocks your looking for, they make up the other ones on your list