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3 years ago

Moishe can ride a bike without giving it any conscious thought moishe ability to ride a bike is an example of—-memory

Physics

2 answers:

s344n2d4d5 [400]3 years ago

8 0

Answer:

Moishe can ride a bike without giving it any conscious thought moishe ability to ride a bike is an example of implicit memory.

Explanation:

The information that you have to consciously remember is explicit memory (for example, the answers to an exam) and the information that you remember unconsciously and effortlessly is known as implicit memory (for example, driving or riding a bicycle). In other words, implicit memory (procedural memory or memory without consciousness) is a type of long-term memory that does not require the intentional recovery of previously acquired experience. This type of memory is unconscious and involuntary. It is also known as non-declarative memory since it is not possible to be aware, it is an automatic memory.

<u><em>Moishe can ride a bike without giving it any conscious thought moishe ability to ride a bike is an example of implicit memory.</em></u>

Anit [1.1K]3 years ago

7 0

Answer:

Implicit

Explanation:

Moishe can ride a bike without giving it any conscious thought moishe ability to ride a bike is an example of implicit memory .

Implicit memory refers to a sort of performing activities with near perfection as a result of numerous previous experiences.

Implicit memory is a demonstration of the learning curve that that says the more you do a thing the easier it becomes to do because you begin to do those things by default or without having to make conscious efforts.

Hence, Moishe's bike riding with unconscious memory is an act of past experiences and of implicit memory.

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Answer:

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Explanation:

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After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

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2. Then, the velocity will start to increase at the rate of the acceleration.

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$v^2 = v_0^2 + 2a*d$

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$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

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