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3 years ago

7

The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i

nclined plane?

1 answer:

Nadya [2.5K]3 years ago

7 0

Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

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A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

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Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

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Given

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Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

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Answer:

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Let the distance moved by Joe is "d"

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$v_{avg} = \frac{2d}{0.0404d}$

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