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3 years ago

The ease with which a memory can be retrieved is influenced by all of the following EXCEPT

Physics

2 answers:

34kurt3 years ago

4 0

Hi! I just took this lesson, the answer is: "The person's belief in the information"

I hope this helps!

sladkih [1.3K]3 years ago

4 0

1.) memory

2.) Accuracy

3.) encoding, storage, retrieval

4.) stores

5.) encoding

6.) retrieval

7.) the Maintenance store (MS)

8.) encoding

9.) the person’s belief in the information

10.) Because you cannot remember something that you have not learned.

Just took the quiz 100%

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Buoyant force is the net upward force that affects on the object in a fluid

Katarina [22]

Answer:True

Explanation:

Buoyant force is the net upward force, that affect on the object in a fluid

4 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

brilliants [131]

Answer:

P = VI = (IR)I = I2R

Explanation:

What the equation means is that if you double the current you end up with 4 times the power loss. It's like the area of carpet you need for a room - if you make the room twice as long and twice as wide you need 4x as much carpet. The physical explanation is that the voltage difference along a wire depends on the current - more current flowing with a resistance means more voltage (pressure of electricity if you like) is built up.

This extra voltage means more power. So if you double the current your would double the power, but you also double the voltage which doubles the power again = 4x as much power. P = VI = (IR)I = I2R

I hope this helps you out, if I'm wrong, just tell me.

8 0

3 years ago

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Check and double-click on the red polygon for the Henry Mountains laccolith complex in the folder labeled Problem 1. This comple

MArishka [77]

Answer: ghg

Explanation:

4 0

3 years ago

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Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

Which statement best describes metallic bonding

horrorfan [7]

Answer:

It's a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal bars. It can also be described as the sharing of free electrons among a structure of positively charged ions

4 0

3 years ago