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svetlana [45]
3 years ago
11

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

charges of magnitude 2*Q and 2*Q when held at a distance r/2.?
Physics
1 answer:
Pepsi [2]3 years ago
4 0

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

F=\frac{kQ^2}{r^2}

Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}

F'=16F

F'=197.504 N

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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
4 years ago
Read 2 more answers
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