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PilotLPTM [1.2K]
3 years ago
9

An "energy bar" contains 26 g of carbohydrates. For the steps and strategies involved in solving a similar problem, you may view

a Video Tutor Solution.
Part A If the energy bar was his only fuel, how far could a 68 kg person walk at 5.0 km/h?
Physics
1 answer:
masha68 [24]3 years ago
7 0

To solve this problem we will apply the definition of Power and Speed. In turn, we will consider that one gram of carbohydrate, according to numerous scientific studies, contributes around 17kJ of energy. Therefore, if this were true, the total energy of 26 grams would be

E = (26)(17000) = 4.42*10^5J

Power can be described as the amount of energy applied at a given time, that is,

P = \frac{E}{t} \rightarrow t = \frac{E}{P}

t = \frac{4.42*10^5}{380}

t = 1.16*10^3s

The speed is described as the distance traveled in a certain time, and its units in international system is m / s, converting and replacing we will have

v = 5km/h(\frac{1000m}{1km})(\frac{1h}{3600s})

v = 1.388m/s

Now,

v = \frac{d}{t} \rightarrow d = vt

The distance is,

d = vt

d = (1.388)(1.16*10^3)

d = 1610.08m

Therefore the distance walked is 1610.08m

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
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Answer:

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Explanation:

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Read 2 more answers
What is the speed of an 800 kg automobile if it has a kinetic energy of 9.00 x 10^J?
MA_775_DIABLO [31]

Ek = 1/2 mv^2

9 × 10^4 = 1/2 × 800 × v^2

9 × 10^4/400 = 400 v^2 / 400

9 × 10^4/400 = v^2

√225 = v

15 ms⁻¹ = v

That's the only way I know how to work it out

I think in this case velocity and speed would be considered the same because me

s = d/t and v=d/t

one is distance travelled and the other is displacement of a body

7 0
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