Ben and Dan both weigh 600 N. They are doing pull-ups
1 answer:
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Answer:
e.)At twice the distance, the strength of the field is E/4.
Explanation:
The strength of the electric field at a certain distance from a point charge is given by:
where
k is the Coulomb's constant
Q is the charge
r is the distance from the point charge
In this problem, the distance from the point charge is doubled:
r' = 2r
So the new electric field strength is
so, at twice the distance the strength of the field is E/4.
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!