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3 years ago

Elaborate on the reason that "cola" type drinks are used to make effective marinades.

Physics

2 answers:

Nadya [2.5K]3 years ago

8 0

Answer: The phosphoric acid in cola helps to break down the meat.

Explanation: The acid inside of the Cola breaks down the meat, making it softer. On top of this, the sweetness adds flavor. The phosphates helps in maintaining the pH and buffer properties of the meat, causes emulsification of fats, thereby making it softer.

emmainna [20.7K]3 years ago

5 0

Answer: because of the phosphoric acid.

Explanation: There is an acid in the "cola" drinks (phosphoric acid), that helps to "break" the meat and tenderize it, making it a little bit softer.

Also, the fact that the "cola" drinks have a lot of sugars, so it adds a lot of flavor to the end product.

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Explanation:

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3 years ago

Read 2 more answers

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

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3 years ago