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3 years ago

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A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

all day. Given that there are 86,400 seconds / day, how much energy, in Joules, would you burn in climbing stairs all day?

1 answer:

defon3 years ago

8 0

Answer:

2940.1 joules would you burn in climbing stairs all day.

Explanation:

Work = W = F $\times$ d

going up stairs would be against force of gravity

W = mgh

where h is the height

the question is not complete because we need speed or distance

h = v $\times$ t

so assuming 1 step per second

h = 86,400 steps $\times$ 7inchs/step $\times$ 0.0254 m/inch

h = 15362 m

so from this

W = 800 N $\times$ 15362

= 12289600 J

that means YOU need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal

Kcal = $\frac{W}{(J/Kcal)}$

= $\frac{12289600}{4180}$

= 2940.1 Kcal

if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.

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Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

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Intensity of light is given by

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3 years ago

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Explanation:

What characteristics must the movement of a person have so that the value of the displacement is equal to the distance traveled?

Displacement is equal to the shortest path covered by an object. It is given by the difference of final position and the initial position.

Distance is equal to the total path covered by an object during the journey.

When an object moves in a straight line path, in this case, the displacement is equal to the distance traveled.

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3 years ago

Rex decides to launch a priceless vase into the air at a 90-degree angle. The initial velocity vase is +7.0 m/s. Dylan claims th

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An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.

Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range

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2 years ago

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Explanation:

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In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

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$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

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Substituting (6) in (3):

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$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

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