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ICE Princess25 [194]
2 years ago
12

How many kcalories are provided by a food that contains 25 g carbohydrate, 6 g protein, and 5 g fat?

Physics
1 answer:
MrRissso [65]2 years ago
7 0

<u>169 Kcalories</u> are provided by a portion of food that has 25 grams of carbs, 6 grams of protein, and 5 grams of fat.

Kcalories mean kilo-calories. Basically, kilo-calorie or kcal refers to 1,000 calories. To get the Kcalories of food, you have to add the kcal of carbohydrates, protein, and fat.

Get the product by multiplying the number of grams of carbohydrate, protein, and fat by 4,4, and 9, respectively. So if you want to get the energy or Kcal available from a meal, you must then combine the outcomes.

Simply put it, take note of the following conversions:

  • 1 gram of carbohydrate is 4kcal
  • 1 gram of protein is also 4kcal
  • Though, 1 gram of fat is 9kcal

So here's how to compute the Kcalories of food that contains 25g carbs, 6g protein, and 5g fat.

1.    25g x 4kcal/g = 100kcal

2.    6g x 4kcal/g = 24kcal

3.    5g x 9kcal/g = 45kcal

4.    100kcal + 24kcal + 45kcal = 169kcal!


Therefore, the food contains 169 kilo-calories!

You might be interested in nutrient density of an orange juice per kcalorie. Look here: brainly.com/question/26495283

#SPJ4

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Learning Goal:
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Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

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A. Substituting for C in U,

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B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

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