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Grace [21]
3 years ago
12

Give an example of a situation when it would be appropriate to round a number and a situation in which it would not be appropria

te to round a number. Explain the difference in the situations.
Physics
1 answer:
Ivan3 years ago
6 0

it would be appropriate to round a number when you are talking about people or animals. Ex. it fits 10 1/2 people. it will most likely fit 9-11 people.

You it is not appropriate to round with money.

Ex. If you owe someone $9.43 you don't just give them $9


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A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small × ’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion.

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New research is using functional magnetic resonance imaging (MRI), a scan of the brain that shows specific areas that are activa
kolezko [41]

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-Frontal Lobe

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4 0
3 years ago
A soccer ball is kicked from the ground with an initial speed of 19.3 m/s at an upward angle of 45°. a player 54.7 m away in the
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Hello there

the answer is

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3 0
3 years ago
Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.
Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s  c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = \frac{1}{2}*a*t^{2} = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

a = \frac{2*3.00m}{1s2} = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + \frac{1}{2}*a*t^{2} = 6.00m/s*1s +\frac{1}{2}*6.00 m/s2*1s^{2}  = 9.00 m

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0
3 years ago
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