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11Alexandr11 [23.1K]
2 years ago
14

Explain why the baking instructions on a box of cake mix are different for high and low elevations. Would you expect to have a l

onger or shorter cooking time at a high elevation?
Physics
1 answer:
Rina8888 [55]2 years ago
6 0

Answer:

All this is due to <u>atmospheric pressure</u> (which decreases with height) and affects the boiling point of water. That is why for recipes that carry some liquid (cake or bread, for example) it is necessary to modify both the ingredients and the baking time.

To better understand this, let's begin by explaining that the boiling point of water at an atmospheric pressure at sea level is 100\°C and as the height increases and the atmospheric pressure decreases, this boiling point  will diminish, as well.

This means the water will not boil at the same temperature at different pressures, because at lower pressure the boiling temperature will be lower.

In other words, as at this point the water boils at a lower temperature, it <u>heats less</u>, so that the <u>food takes longer to cook </u>(because it is necessary to increase the cooking time in order to have a well done food).

This is because as the height increases, the atmosphere becomes drier, the air has less oxygen and moisture evaporates quickly, therefore, it is sometimes necessary to add more water to compensate this evaporation during cooking.

Hence, at higher altitudes, the cooking time will be longer.

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9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

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To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
2 years ago
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