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Stella [2.4K]
1 year ago
5

PLEASE HELP

Physics
1 answer:
Paul [167]1 year ago
7 0
B would be your right answer. Because Jupiter’s gravity pull is much stronger than earths
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I dont think so. It also probably matters what type of str it is
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A man can walk three blocks east and four blocks north in 20 min. He can drive 12 times that distance in the same amount of time
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Answer:  25.2 km

<u>Explanation:</u>

3 blocks east and four blocks east = 7 blocks

1 block = 0.3 km  --> 7 blocks = 2.1 km

12 times 2.1 km = 25.2 km

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Which of the following are examples of projectile motion?
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A or C    I would pick C

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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

6 0
3 years ago
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