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MatroZZZ [7]
3 years ago
12

One baked potato provides an average of 30 mg of vitamin C. If 70 potatoes weigh 20 lb., how many milligrams of vitamin C are pr

ovided per pound of potatoes?
Physics
1 answer:
JulsSmile [24]3 years ago
5 0

Answer:

105 mg

Explanation:

Given that:

1 baked potato provides 30 mg of vitamin C.

So,

70 baked potatoes provide 30\times 70 mg of vitamin C

Also,

70 potatoes = 20 lb

So,

20 lb potatoes provide 30\times 70 mg of vitamin C

Thus,

1 lb potatoes provide \frac {30\times 70}{20} mg of vitamin C

<u>Thus, 105 mg of Vitamin C are provided per pound of the potatoes.</u>

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Answer:

thermal mass is the ability of a material to absorb, store and release heat... Materials such as concrete, bricks, and tiles absorb and store heat, And then they have Thermal mass.

Explanation:

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Each large pizza can feed three teenage boys. How many pizzas would be needed to feed the
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19 pizzas

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Which one of the following accurately describes the force of gravity? A. The gravity acceleration has no effect on a body moving
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All objects that have mass are subject to the earth's gravitational acceleration. The product of mass and the earth's gravitational acceleration defines the weight of an object as its gravitational force.
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A convection cell is BEST described as ___.
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Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri
Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

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3 years ago
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