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slamgirl [31]
3 years ago
6

Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Assume the mower moves wit

h a 1 km/h speed, and has a 0.6 m width
Physics
1 answer:
bekas [8.4K]3 years ago
8 0

Answer:

t = 11.9 h = 0.49 day

Explanation:

First of all, we will find the area covered by the mower per unit time:

Rate\ of\ Area\ Covered\ by\ Mower = vw

where,

v = speed of mower = 1 km/h = 1000 m/h

w = width of mower = 0.6 m

Therefore,

Rate\ of\ Area\ Covered\ by\ Mower = (1000\ m/h)(0.6\ m)\\Rate\ of\ Area\ Covered\ by\ Mower = 600\ m^{2}/h

Now, the standard football ground has an area of:

A = 105 m x 68 m = 7140 m²

Therefore, The time required to mow the lawn will be:

Time\ to\ Mow\ the\ Lawn = t = \frac{A}{Rate\ of\ Area\ Covered\ by\ Mower}\\t = \frac{7140\ m^{2}}{600\ m^{2}/h}

<u>t = 11.9 h = 0.49 day</u>

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How does the kinetic energy from the forward motion of a car traveling at 16 m/s
elena55 [62]

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

  • For the first case, velocity v = 16m/s
  • For the second case, velocity = 8m/s
  • Let the mass of the car be m

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

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