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WINSTONCH [101]
3 years ago
7

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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A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a
lilavasa [31]

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2   B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula      Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a  and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a  and a = 1.215 m/s2.

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