Ask question

Ask question

All categories

4 years ago

14

What is the set of processes used to encode, store, and retrieve information over different periods of time?

1 answer:

kakasveta [241]4 years ago

8 0

The set of processes used to encode, store, and retrieve information over different periods of time is called memory. Memory<span> makes us.</span><span> If we couldn't recall the who, what, where, and when of our everyday lives, we would never be able to manage.</span>

You might be interested in

An electric heater has a resistance of 50 ohms. If the applied voltage in a circuit containing the heater is 110 V, how many amp

Oksanka [162]

Use Ohm's Law:

V = I × R
(110) = I × (50)
2.2 Amps = I

8 0

3 years ago

Read 2 more answers

The majority of earthquakes worldwide occur at all but one location. That is A)at tectonic plate boundaries.B)where sediments de

DENIUS [597]

Answer:b

Explanation:

6 0

3 years ago

Heat always flows from a what place to a what place

Vera_Pavlovna [14]

Answer:

Heat is always the transfer of energy from an object at a higher temperature to an object at a lower temperature.

Explanation:

6 0

3 years ago

Read 2 more answers

A particular star is white. Another star that is much hotter is expected to be ______.

jenyasd209 [6]

Blue

hope this helps :)

4 0

4 years ago

Read 2 more answers

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago