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3 years ago

8

Where is information first stored in the human brain?

1 answer:

sergejj [24]3 years ago

3 0

When you first learn some information, the first place it goes in your brain is to your short-term memory. So, you read something, or hear something, and it stays in your short-term memory for a little while. If you don't hear it again, or that information doesn't get repeated, it is likely to disappear from your memory - you will forget it. However, the more you repeat that information, the more likely it is that it will be remembered, that is, it will enter your long-term memory and stay there for a long time.
Hope this helps :)

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How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

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4 years ago

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Harrizon [31]

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3 years ago

Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

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3 years ago

A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football

KengaRu [80]

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

$p_{i} = p_{f}$

$m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}$ (1)

Where:

$m_{a}$ : is the mass of the first football player = 110 kg

$m_{b}$ : is the mass of the second football player = 90 kg

$v_{a_{i}}$ : is the initial velocity of the first football player = 5.0 m/s

$v_{b_{i}}$ : is the initial velocity of the second football player = 0 (he is at rest)

$v_{a_{f}}$ : is the final velocity of the first football player = 0 (he stops after the impact)

$v_{b_{f}}$ : is the final velocity of the second football player =?

By solving equation (1) for $v_{b_{f}}$ we have:

$110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}$

$v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s$

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!

8 0

3 years ago