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VladimirAG [237]

3 years ago

8

Which is true regarding AC current? The electrons move back and forth. The +/- polarity is constant. It is common in portable de

vices. It can only produce low voltages.

2 answers:

Rus_ich [418]3 years ago

8 0

A. The electrons move back and forth.
Hope this helped. :L

andre [41]3 years ago

7 0

Answer: Option (a) is the correct answer.

Explanation:

AC current stands for alternating current. In an alternating current there is periodically reverse movement of current.

This means electrons move in one direction for a certain time and then they reverse their movement for the same time period and it continues in the back and forth position.

Thus, we can conclude that the statement electrons move back and forth is true regarding AC current.

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*<br> Check all of the things that can affect power<br><br> Time<br> Work<br> Distance<br> Force

gladu [14]

Answer:

fhddmvxmvydlyghclhchc

3 0

2 years ago

The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0

2 years ago

A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh

Debora [2.8K]

Answer:

A. 5/7h

B. Same height

C. See attachment below

Explanation:

Find the attachment below for better understanding.

7 0

2 years ago

Cual de las escalas de temperatura es la mas antigua

IrinaVladis [17]

Answer:

the translation I got for this question is

Which of the temperature scales is the oldest?

Explanation:

and i searched for it and got this=

Fahrenheit scale

6 0

2 years ago

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

a

The mass is $m_2 =21.75*10^{-27} \ kg$

b

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

So

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

So

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago