Answer:
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Explanation:
No, because the distance-time would show a constant velocity but the velocity-time graph shows an increasing velocity.
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Answer:
effort force: The force used to move an object over a distance. resistance force: The force which an effort force must overcome in order to do work on an object via a simple machine. ideal mechanical advantage: The factor by which a mechanism multiplies the force put into it.
Explanation:
Answer:
Explanation:
Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².
Now, the pilot is traveling in a circle of radius
r = 3340 m
And the speed is
v = 495 m/s
Then, acceleration?
The acceleration of a circular motion can be determine using centripetal acceleration
a = v² / r
a = 495² / 3340
a = 73.36 m/s².
Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully
