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svetoff [14.1K]
3 years ago
10

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate

d by Lance Armstrong (m = 75.0 kg) is 6.50 W per kilogram of his body mass. (a) How much work does he do during a 189-km race in which his average speed is 12.5 m/s? (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule = 2.389 x 10-4 nutritional Calories.
Physics
1 answer:
disa [49]3 years ago
3 0

Answer:

E=1760.93 cal

Explanation:

a) First we find the total time

t=x/v

t=189000/12.5

t=15120 seconds

Now we multiply the time by his total wattage

E=t*W

E=(15120)(6.5)(75)=7371000 joules

As,

1 joule = 2.389 x 10-4

So, in term of nutritional Calories

E=7371000 * 2.389*10^{-4}

E=1760.93 Cal

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5 0
2 years ago
* 1a Average speed
ruslelena [56]

Explanation:

\implies   v_{av} =  \dfrac{total \: displacement}{total \: time}

\implies   v_{av} =  \dfrac{100}{10}

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4 0
3 years ago
At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur
kvasek [131]

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

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6 0
3 years ago
Your mother is sure that you were driving too fast because she knows
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5 0
3 years ago
A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).
mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa

(c) Same steps as in part (a)

P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa

ΔP=9.164-9.14

ΔP=0.0245kPa

8 0
3 years ago
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