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BaLLatris [955]
4 years ago
15

Miguel is learning to play the piano. He has learned to play some chord sequences with ease. In this scenario, the chord sequenc

es learnt by Miguel will be stored in his
Physics
2 answers:
AlexFokin [52]4 years ago
3 0

memory, be it his memory or the keyboard memory

hram777 [196]4 years ago
3 0

Answer:

procedural memory

Explanation:

The procedural memory is where we keep, the more or less automatic learning, escaping consciousness. Essentially motor skills and processes, such as writing or cycling. No one thinks of the sequence of movements we have to perform so that we can ride a bicycle, nor can we describe in words how to do it. It's automatic, which is beyond our consciousness, so we don't need to think to do it.

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Leonard is stressed and feeling overwhelmed by the college selection process. What should he do to combat this problem?
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He should ask other people for advice
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The image below shows two opposite forces acting on a rope, what can we say is true about the affect of the forces on the rope?
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It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
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What is the speed of light (f = 5.09 E14 Hz) in glycerol?
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Answer:

The speed of light in glycerol is 2.04×108 m/s .

Explanation:

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How does the diaphragm remove air from the lungs?
LekaFEV [45]

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3 years ago
During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f
dezoksy [38]

Answer:

F= 2.3733 x10^{20} N

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

M_ {sun} = 1.99 * 10^{30} Kg

M_ {Earth} = 5.98 * 10 ^ {24} Kg

M_ {Moon} = 7.36 * 10 ^{22} Kg

Average distances

\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m

\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m

Gravitational constant

G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}

The formula of the Gravitational Force between the Moon and the Earth would be,

F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}

F = 1.9908 * 10 ^{20} N

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}

F_2 = 4.3641*10^-{ 20} N

This force is in the direction of the Sun

The net force must be

F_ {net} = F_2-F

F_ {net} = 2.3733*10^{20} N

This in the direction of the Sun.

8 0
3 years ago
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