Answer:

Explanation:
Recall the formula for acceleration:
, where
is final velocity,
is initial velocity, and
is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:

Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!
It's true answer that interference is the evidence of the particle nature of light..
p=F/A
or,P=d×V×G/A (m=d×V)
or,p= d× A×h×g/A (A and A are cut)
or,P=d×H×G
To me, that sounds like the "Law of Conservation of Energy".
The equation
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:


Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

Therefore, the equation
represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!