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lukranit [14]
3 years ago
7

An olympic hurdler accelerates at a rate of 15.5 m/s2 . what is the rate in miles/min2 ?

Physics
2 answers:
Alborosie3 years ago
6 0

Answer: 34.68 mile/min^2

Explanation:

1 mile = 1609 meter

1 min = 60 s then squared will be 3600s

Mile/min^2 = 2.24

Give 15.5m/s^2

We multiply

15.5 by 2.24

= 34.68 mile/min^2

Wittaler [7]3 years ago
5 0
You have to use conversion units for this. For every 1 mile, there is an equivalent amount of 1,609 meters. For every 1 minute, there is an equivalent amount of 60 seconds. Using the dimensional analysis approach, the solution is as follows:

Acceleration = 15.5 m/s²*(1 mile/1,609 m)*(60 s/1 min)² = 34.68 miles/min²
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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:

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Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

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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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