Physics, 03.04.2020 06:48, itujggkmjf8067

Two point charges having charge values of 9.2 E -6 C and 3.5 E -6 C, respectively, are separated by 3 E -2 m. What is the absolute value of the mutual force between them

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answered: Guest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

answered: Guest

If you're looking for an even spread of clay, silt and sand, add a considerable amount of sand and some clay until the 3 even out". soil texture triangle. silty clay loam soil is approximately 20 percent or less sand, 25 to 40 percent clay, and 40 to 70 percent silt.

answered: Guest

To compare speeds speeds of wagons we will use second newton's law. it states: f = m * ahere we can see that force depends on the acceleration. if we know acceleration for both wagons we can compare them to make conclusion of speeds of wagons.we have: m1 = 10lbm2 = 20lbf1 = 2*f2also: m2 = 2*m1for bill we have equation: f1 = m1 * a1for tom we have equation: f2 = m2 * a2after inserting identities we have: 2*f2 =m1 * a1f2 = 2*m1 * a2in order to make conclusions about speeds we will solve these equations for a and find their ratio: a1 = 2f2 / m1a2 = f2 / 2m1a1 / a2 = (2f2/m1) / (f2/2m1)a1 / a2 = (2f2*2m1) / (m1*f2)a1 / a2 = (4f2m1) / (m1f2)a1 / a2 = 4this ratio is bigger than 1 and this means that a1 is greater than a2. a1 is 4 times greater than a2. speed of bill's wagon is 4 times greater than speed of tom's wagon.

answered: Guest

25,000 = 2.5 x 10⁴250 = 2.5 x 10^20.025 = 2.5 x 10⁻²0.00025 = 2.5 x 10⁻⁴

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Two point charges having charge values of 9.2 E -6 C and 3.5 E -6 C, respectively, are separated by...