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posledela
2 years ago
12

(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

charge on each? (b) What is the charge on each if the force is attractive?
Physics
2 answers:
Mrrafil [7]2 years ago
8 0

Answer:

3x10-7C

Explanation: Use Coulombs Law. Find q2 and derive the equation. Make sure to convert micro to decimals for accurate results

Anastaziya [24]2 years ago
7 0
3 is jxhuneuxndnzixbf
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How much is the moment due to force P about point B. P(unit=N) vector is equal to (150i+260j ) and vector BA (unit=meter) is equ
artcher [175]

Answer:

The moment (torque) is given by the following equation:

\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)

Explanation:

The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.

4 0
1 year ago
In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

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A light ray strikes a plane mirror at an angle of 23° to the normal. What is the angle between the reflected ray and the mirror?
tiny-mole [99]

Answer:67^{\circ}

Explanation:

Given

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Thus incident angle i=23^{\circ}

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4 0
2 years ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
svlad2 [7]

Answer:

Electric field, E = 45.19 N/C

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It is given that,

Surface charge density of first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density of second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a point between the two surfaces is given by :

E=\dfrac{\sigma}{2\epsilon_o}

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So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

6 0
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