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lesya [120]
3 years ago
9

Why we cannt cook food on high attitude (like mountain where exist low temperatures )? ​

Physics
1 answer:
zloy xaker [14]3 years ago
4 0
Cooking takes longer because water and other liquids evaporate faster and boil at lower temperatures.At sea level water boils at 212degrees Fahrenheit but at an altitude of 7,500 feet, it boils at about 198 degrees. Foods that are prepared by boiling or simmering will cook at a lower temperature and it will take longer to cook them. Hope this helps!! ; )
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Answer:

Sir Richard Branson's personal dilemma is that he is concerned about the environment and climate change, but he has made his fortune with an airline industry that contributes to the greenhouse gases.

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If the released energy of one earthquake is 100 times that of another, how much greater is its magnitude on the richter scale?
Alekssandra [29.7K]
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¿Por qué en algunas partes el agua tiene mayor presión?
Phantasy [73]
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7 0
2 years ago
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

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2 years ago
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