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:
it would be 39.2 m/s
Explanation:
After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.
Answer:
The observer sees the space-probe 9.055m long.
Explanation:
Let
be the length of the space-probe when measured at rest, and
be its length as observed by an observer moving at velocity
, then

Now, we know that
and
, and putting these into
we get:


Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.
Answer:
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Explanation:
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To solve this problem we will use a free body diagram that allows us to determine the Normal Force.
In general, the normal force would be equivalent to

Since the skier is standing on two skis, his weight will be divide by two

Pressure is given as the force applied in a given area, that is

Replacing F with N'


Our values are given as,




Replacing we have that


Therefore the pressure exerted by each ski on the snow is 776.01Pa