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Alex Ar [27]
4 years ago
6

Acid rain falling on a rock outcrop over a period of many years can cause the rock on the service to dissolve. this is an exampl

e of____.
A.erosion
B.the rock cycle
C.chemical weathering
D.mechanical weathering
Physics
1 answer:
Bogdan [553]4 years ago
6 0
C. Chemical weathering
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Is brainly even trying anymore?
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Yes because it helps with my work
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3 years ago
If the temperature at the surface of Earth (at sea level) is 100°F, what is the temperature at 2000 feet if the average lapse ra
oksano4ka [1.4K]

Answer:

107 °F

Explanation:

Given that

The temperature at sea level = 100°F

height ,h= 2000 feet

The average lapse rate = 3.5°F/1000 feet

Given that rise in temperature 3.5°F per 1000 feet.

1000 feet ⇒ 3.5°F

Given that 2000 feet

2000 feet ⇒ 3.5°F x 2 +100°F

2000 feet ⇒ 107 °F

Therefore the temperature will be 107 °F  .

6 0
3 years ago
A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th
Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

C = \epsilon_0 (\frac{A}{d})

Here,

\epsilon_0 =  Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})

C = 1.5901*10^{-10}F

PART A) Energy stored in the capacitor is

E = \frac{1}{2} CV^2

E = \frac{1}{2} (1.5901*10^{-10})(575)^2

E = 2.628*10^{-5}J

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

Q = CV

Q = (1.59*10^{-10})(575)

Q = 9.1425*10^{-8}C

Now if d = 10.04*10^{-3}m we have that the capacitance is

C = \epsilon_0 (\frac{A}{d})

C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})

C = 3.9754*10^{-11}F

Then the energy stored is

E = \frac{1}{2} \frac{Q^2}{C}

E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})

E = 1.051*10^{-4} J

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

W = 1.051*10^{-4} J -2.628*10^{-5}J

W = 7.882*10^{-5} J

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