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3 years ago

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Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force betw

een them?
A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive

If one could answer this ASAP, it will be greatly appreciated

1 answer:

mr Goodwill [35]3 years ago

5 0

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.

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Explanation:

Here is the complete question:

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s. During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s. Finally, during the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

Explanation:

(a) To determine how far the bicyclist has traveled during the entire trip, we will calculate the distance she covered in each part of the trip, and then sum up the distances to determine the total distance covered.

First, The distance covered in the first part of the trip

During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s

That is,

Average speed, $v$ = 7.57 m/s

and time, $t$ = 26.8 minutes

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∴ $t$ = 26.8 minutes = (26.8 × 60) secs = 1608 secs

$Average speed = \frac{Distance }{ Time}$

Then, $Distance = Average speed$ × $Time$

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For the distance covered in the second part of the trip

During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s

That is, Average speed, $v$ = 3.17 m/s

and time, $t$ = 42.4 minutes

Convert the time to seconds

∴ $t$ = 42.4 minutes = (42.4 × 60) secs = 2544 secs

From,

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For the distance covered in the third part of the trip

During the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s

That is, Average speed, $v$ = 15.0 m/s

and time, $t$ = 9.69 minutes

Convert the time to seconds

∴ $t$ = 9.69 minutes = (9.69 × 60) secs = 581.4 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 15.0$ × $581.4$

$Distance = 8721$ m

This is the distance covered in the third part of the trip.

Now for the distance covered during the entire trip,

Total distance = distance covered in the first part of the trip + distance covered in the second part of the trip + distance covered in the third part of the trip

Hence,

Total distance = 12172 m + 8064.48 m + 8721 m

Total distance = 28958.04 m

Hence, she has traveled a total distance of 28958.04 m during the entire trip.

(b) For her average velocity for the trip

Average velocity is given by

$Average velocity = \frac{Total distance traveled}{Total time}$

Total distance traveled = 28958.04 m

Total time = 1608 secs + 2544 secs + 581.4 secs

Total time = 4733.4 secs

Hence,

$Average velocity = \frac{28958.04}{4733.4}$

Average velocity = 6.1178 m/s

Average velocity ≅ 6.12 m/s

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