<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
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Refer to the diagram shown below.
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
Answer:
C. It helps scientists detect dark matter.
Explanation:
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