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!
Answer:
B,D,E
Explanation:
I got you
B. Experiment with a wider range of materials.
D.Use a laboratory galvanometer to make precise measurements.
E. Test the strength of the electromagnet by varying the number of wire coils.
BMI is a measure of body fat determined by ones Height, Weight, and Gender.
Answer:
e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.
Explanation:
This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.
The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.
<u>Mathematical expression for the Newton's second law of motion is given as:</u>
............................................(1)
where:
dp = change in momentum
dt = time taken to change the momentum
We know, momentum:

Now, equation (1) becomes:

<em>∵mass is constant at speeds v << c (speed of light)</em>

and, 
where: a = acceleration

also

so, more the time, lesser the force.
<em>& </em><u><em>Impulse:</em></u>




∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.
So, the impulse in both the cases will be same.