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dangina [55]
4 years ago
11

Possible data where using a bar graph would be better than using a line graph.

Physics
1 answer:
docker41 [41]4 years ago
3 0

Answer:

Make a graph that represents exotic pet ownership in the United States. There are 8,000,000 fish, 1,500,000 rabbits, 1,300,000 turtles, 1,000,000 poultry and 900,000 hamsters.

Explanation:

The biggest difference is that bar graphs are more versatile while line graphs are better for showing trends over time or another measure with a logical progression of values (such as distance from a given point). Bar graphs can also show frequency distributions (how often you observe different outcomes) much more effectively than line graphs.

Line charts should be used only for time series (chronological) or when there is some other sequence to the dimensions on the x-axis, e.g. dates, months, sequence of stages of a project, sequence of meters along on a gas pipeline, and they should be used to detect trends and patterns, not to give people exact quantitative readings.

Bar charts should be used for comparing specific x-axis values, though they can certainly be used for time series, like line charts. They can also be used to display parts of a whole in favor of pie charts, in which case, the space between the bars should be reduced.

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A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
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The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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