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Sergio039 [100]

3 years ago

8

The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg,

what is the runner's average kinetic energy during the run

1 answer:

Semenov [28]3 years ago

5 0

Answer:

the runner's average kinetic energy during the run is 476.96 J.

Explanation:

Given;

mass of the runner, m = 85 kg

distance covered by the runner, d = 42.2 km = 42,200 m

time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)

= 12,600 s

The speed of the runner, v = d/t

v = 42,200 / 12,600

v = 3.35 m/s

The runner's average kinetic energy during the run is calculated as;

K.E = ¹/₂mv²

K.E = ¹/₂ × 85 × (3.35)²

K.E = 476.96 J

Therefore, the runner's average kinetic energy during the run is 476.96 J.

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a_sh-v [17]

Answer:

$x\approx 0.948$

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

$U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}$

Where $x$ is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

$F = -\frac{dU}{dx}$

The function is derived in terms of distance:

$F (x) = \frac{84}{5\cdot x^{9}} -\frac{104}{5\cdot x^{5}}$

Then, it is needed to find at least of x so that F(x) equals to 0.

$\frac{84}{5\cdot x^{9}}-\frac{104}{5\cdot x^{5}}=0$

$\frac{84}{x^{4}}-104 = 0$

$84-104\cdot x^{4} = 0$

$x=\sqrt[4]{\frac{84}{104} }$

$x\approx 0.948$

7 0

3 years ago

How fast would a 20 kg dog have to run, to reach a momentum of 120 kg x m/s? Include unit in your answer. Worth 40 points!

Ivahew [28]

Answer:

6 m/s

Explanation:

momentum= mass times velocity

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8 0

3 years ago

Read 2 more answers

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

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Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

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Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

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3 years ago

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Answer:

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4 years ago

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