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Tamiku [17]
2 years ago
6

An extremely long thin wire carries a uniform linear charge density of 358 nC/m. Find the potential difference between points 5.

0 m and 6.0 m from the wire, provided they are not near either end of the wire. (
Physics
1 answer:
Romashka-Z-Leto [24]2 years ago
7 0

Answer:

1.2kV

Explanation:

We are given that

Linear charge density,\lambda=358nC/m=358\times 10^{-9} C/m

1 nC=10^{-9} C

r_1=5 m

r_2=6 m

We have to find the potential difference between poinst 5.0 m and 6.0 m from the wire.

We know that potential difference of wire

V=2k\lambda ln\frac{r_2}{r_1}

Where

k=9\times 10^9

Substitute the values

V=2\times 9\times 10^9\times 358\times 10^{-9}ln\frac{6}{5}

V=1174.8 V

V=1.17 kV\approx 1.2 kV

1kV=1000 V

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ryzh [129]
<h2>Answer:</h2>

<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>

<h3>Explanation:</h3>

Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.

In the given question the circular displacement is 25 rounds around a central point.

The angular displacement is measured in degrees and 1 round is equal to 360 degrees.

25 Rounds = 25 × 360 = 9000 degrees.

Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.

In SI,angular velocity is represented in radians per second.

So, 1 radian = 57.29 degrees

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2 years ago
Which term do modern psychologists prefer to use in place of short-term memory?
Vsevolod [243]

Correct answer: D). Working Memory

The short term memory is the part of the memory system in which the information is stored for 30 seconds. However, it can be increased by rehearsal. Short term memory is also called active memory or the working memory. The working memory is the part of the cognitive system that is responsible for storing information temporarily for processing. It is important for reasoning.

Hence, the correct answer would be option D.




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2 years ago
A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
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Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
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Answer:

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effort distance;2.6m

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effort:138.46N

<u>R</u><u>e</u><u>q</u><u>u</u><u>i</u><u>r</u><u>e</u><u>d</u><u> </u><u>e</u><u>f</u><u>f</u><u>o</u><u>r</u><u>t</u><u> </u><u>i</u><u>s</u><u> </u><u>1</u><u>3</u><u>8</u><u>.</u><u>5</u><u>N</u><u>.</u>

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