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serious [3.7K]
3 years ago
12

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
Physics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

Explanation:

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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Answer:

1.1 m/(s)^2

Explanation:

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t=10s

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=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

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2 years ago
At this rate, how long would it take for two continents 3500 kilometers apart to collide?
meriva

<span>Given:

3,500 kilometers

Find:</span>

 

Years for two continents to collide = ?

 

<span>Solution:

We know that </span>typical motions of one plate relative to another are 1 centimeter per year.

So first, we convert 3,500 km to cm.<span>
</span><span>

</span>

The solution would be like this for this specific problem:

 

1 km = 100,000 cm

3,500 km x 100,000 = 350,000,000 cm

Since we know that 1 cm = 1 year, then that means 350,000,000 cm is equivalent to 350,000,000 years.

 

Therefore, it would take 350 million years for two continents that are 3500 kilometers apart to collide.

<span>
To add, </span>a phenomenon of the plate tectonics of Earth that occurs at convergent boundaries is called the continental collision.

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The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (g=9.8 m/s^2 towards the ground). Therefore, we can use the following suvat equation:

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In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

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Solving for s, we find the maximum height reached by the ball:

s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m

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