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Damm [24]
3 years ago
8

If a long distance runner with a weight of 596.82 newtons does 35,674.7 joules of work during a portion of a race, what distance

will she cover during that portion?
Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

\boxed {\boxed {\sf About \ 59.77 \ meters}}

Explanation:

Work is equal to the product of force and distance.

W=F*d

We are solving for the distance, d. We know the force is 596.82 Newtons and the work is 35,674.7 Joules.

  • If we convert the units for Joules, the problem will be simpler later.
  • 1 Joule is equal to 1 Newton meter.
  • The work done by the runner, 35,674.7 J, equals 35,674.7 N*m

F= 596.82 \ N\\W= 35,674.7  \ N*m

Substitute the values into the formula.

35,674.7 \ N*m= 596.82 \ N * d

We want to solve for the distance, so we must isolate d on one side of the equation.

d is being multiplied by 596.82 Newtons. The inverse of multiplication is division. Divide both sides of the equation by 596.82 N

\frac{35,674.7 \ N*m}{596.82 \ N} =\frac{596.82 \ N*d}{596.82 \ N}

\frac{35,674.7 \ N*m}{596.82 \ N}=d

The Newtons will cancel, hence the prior unit conversion.

\frac{35,674.7 \ m}{596.82} =d

59.7746389 \ =d

Let's round to the nearest hundredth, so the answer is concise. The 4 in the thousandth place tells us to leave the 7 in the hundredth place.

59.77 \ m  \approx  d

The distance covered during the portion of the race is about <u>59.77 meters.</u>

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goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

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