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Vlada [557]
4 years ago
7

Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

length 0.500 mm . What is the electric potential energy UUU of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Physics
1 answer:
kirza4 [7]4 years ago
6 0

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

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Power = 150 * 10/120

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A 1250 kg car starts from rest and speeds up to 14.0 m/s in 7.25s. How much force acted on the car ?
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M = 1250 kg
V = 14.0 m/s
t = 7.25 s
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Where on this diagram does the ball have the highest point of gravitational potential energy?
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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

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