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3 years ago

The three main stages to the memory process are encoding, storage, and retrieval.

Physics

2 answers:

never [62]3 years ago

8 0

Answer:

Explanation:

oksian1 [2.3K]3 years ago

7 0

Answer:

a- true

Explanation:

i did the test and its correct

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Can someone help me please? I’ve been trying to solve these questions all day.

qwelly [4]

#16

If we put a resistor in circuit it will slow the speed of current

Let's check ohms law

$\\ \rm\Rrightarrow \dfrac{V}{I}=R$

So if resistance is more current is less

#17

Again use ohms law

$\\ \rm\Rrightarrow V=IR$

$\\ \rm\Rrightarrow V\propto I$

Voltage must be increased

4 0

2 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

I WILL GIVE BRAINLIEST!! Record the volume of the liquid in the graduated cylinder. The volume of the ------------liquid is mill

MariettaO [177]

Answer:

53 milliliters

Explanation:

5 0

3 years ago

Communication definition

PtichkaEL [24]

Answer:

the imparting or exchanging of information or news.

Explanation:

4 0

3 years ago

Building codes usually limit the current carried by a No. 14 copper wire to 15 A. Many household circuits are wired with this si

Sedbober [7]

Answer:

The drift velocity of the electrons in this case is c) 5.52 x 10-4 m/s

Explanation:

First of all, we need to find the volume occupied by 63.3g of copper, so $V=\frac{m}{\rho}= \frac{0.0633Kg}{8,900kg/m^{3}} =7.11 \times 10^{-6} m^{3}$ , then if we assume each atom of copper contributes with one free electron to the material body $n=\frac{N_{A}}{V}=\frac{6.02 \times 10^{23} electrons}{7.11 \times 10^{-6} m^{3}} =8.46 \times 10^{28} electrons/m^{3}$ .

Finally, we apply $v_{d}=\frac{I}{nqA}$ , where A is area so, $A=\pi (\frac{d}{2})^{2}= \pi (\frac{0.0016m}{2})^{2}=2 \times 10^{-6} m^{2}$ , thus $v_{d}=\frac{I}{nqA}=\frac{15C/s}{(8.46 \times 10^{28} m^{-3})(1.60 \times 10^{-19} C)(2 \times 10^{-6} m^{2})}=5.50 \times 10^{-4} m/s$ .

As we can see c. 5.52 x 10-4 m/s is the nearest one.

3 0

3 years ago