In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
timates that it took 1.5 s for the ball to fall the last 32 m. T/I (a) Calculate the velocity of the ball when it is 32 m above the ground. [and: 14 m/s [downward]] (b) Calculate the total displacement of the ball. [and: 42 m [downward]]
1 answer:
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The correct answer is
<span>c) very small and very large
Let's see this with a few examples:
1) if we have a very small number, such as
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<span>we see that we can write it easily by using the scientific notation:
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<span>2) Similarly, if we have a very large number:
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<span>we see that we can write it easily by using again the scientific notation:
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<span>c) very small and very large
Let's see this with a few examples:
1) if we have a very small number, such as
</span>
<span>we see that we can write it easily by using the scientific notation:
</span>
<span>2) Similarly, if we have a very large number:
</span>
<span>we see that we can write it easily by using again the scientific notation:
</span>
</span>
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
