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2 years ago

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A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push caus

es the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters?

1 answer:

Marina CMI [18]2 years ago

7 0

Answer:

$v_{f} =4.9\frac{m}{s}$ : velocity of the crate after the person has pushed the crate a distance of 6 meters

Explanation:

Crate kinetics

Crate moves with uniformly accelerated movement

v f²=v₀²+2a*d (formula 1)

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Known data

v₀=0 The speed of the crate is equal to zero because part of the rest.

a= 2m/s²

d= 6m

Distance calculating

We replace data in the Formula (1)

v f²=0+2*2*d

v f²=2*2*6

v f²=24

$v_{f} =\sqrt{24}$

$v_{f} =4.9\frac{m}{s}$

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Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

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Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

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Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

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luda_lava [24]

Answer:

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Explanation:

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2 years ago

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sergeinik [125]

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2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

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Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

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Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

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The other expressions are incorrect, let’s prove it:

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$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

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2 years ago