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Kipish [7]
3 years ago
7

A 150-lb person burns about 2700 Calories to run a marathon. How much energy is burned in kJ?

Chemistry
2 answers:
iris [78.8K]3 years ago
7 0

Answer:

150-lb person burned 11.30 kilo Joules of energy to ran a marathon.

Explanation:

Number of calories burned by 150-lb person = 2700 calories

1 cal = 0.001 kcal

2700 cal=2700\times 0.001 kcal=2.700 kcal

1 kilo calorie = 4.184 kilo Joules

Energy associated with 2.700 kcal:

2.700\times 4.184 kJ=11.30 kJ

150-lb person burned 11.30 kilo Joules of energy to ran a marathon.

Alex17521 [72]3 years ago
5 0
The question above is simply asking us to convert the value from calories into kilojoules. Therefore, we need some factor to multiply to be able to convert it to such units. For calories to kJ, the factor would be 4.184.

2700 cal ( 4.184 kJ/cal) = 11296.8 kJ
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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse
amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0
3 years ago
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Sulfur S 33.537%

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What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?
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Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

Explanation:

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