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AveGali [126]
3 years ago
6

Echoic sensory memory is used when a lightning bolt flashes across the sky.

Physics
2 answers:
Ahat [919]3 years ago
6 0
If the statement above answers whether the statement is true or false, then the answer is false. It is not true that the echoic sensory memory is used when a lightning bolt flashes across the sky because it does not have that kind of capacity. The echoic sensory memory is the one responsible for the auditory information.
Alex73 [517]3 years ago
3 0

It's false on E2020.

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What subatomic particle has a nuetral charge?​
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An atom is a particle with a neutral charge
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When rust forms on a piece of iron, what evidence do you have that a chemical reaction has taken place?
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The rust is the product and evidence
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An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seis
vivado [14]

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be d, the time taken by P waves be t_P and the time taken by the S waves be t_S.

\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}

\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}

For the P waves,

t_P=\dfrac{d}{8000}

d=8000t_P

For the S waves,

t_S=\dfrac{d}{4500}

d=4500t_S

Equating the d,

8000t_P=4500t_S

Divide both sides of the equation by 500 to reduce the terms.

16t_P=9t_S

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

t_S-t_P=120

t_S=120+t_P

Substitute this in the equation of the distance.

16t_P=9(t_P+120)

16t_P=9t_P+1080

7t_P=1080

t_P=\dfrac{1080}{7}

Substitute this in the equation for d involving t_P.

d=8000t_P

d=8000\times\dfrac{1080}{7}

d=1234285.7 \text{ m }= 1234.3 \text{ km}

4 0
3 years ago
How does the gravitational force between two objects change if the distance between the objects doubles?
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if the distance between the objects is doubled the force is reduced by a factor of 4

<h3>Whats is gravitational force?</h3>

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G = gravitational constant

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m2 = mass of object 2

r = distance between the objects

From the formular, the gravitational force and the distance is an inverse relationship so increasing the distance by a factor results to reduction of the force by the square of the factor. hence doubling the distance which is distance mutiplied by 2 will lead to reduction of the force by 2^2 = 4

Therefore: The force decreases by a factor of 4.

hope it helps

4 0
2 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}

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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,

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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

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

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

E = 2.628*10^{-5}J

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Q = CV

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Q = 9.1425*10^{-8}C

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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

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