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alexandr402 [8]
3 years ago
15

What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011

N –1. 1 × 1011 N –1. 6 × 1011 N 1. 8 × 1011 N.
Physics
1 answer:
nlexa [21]3 years ago
4 0

Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

\rm F= K\frac{q_2q_3}{r^2}

\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}

\rm F=  - 1. 1 \times 10^{11 }N

Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

To learn more about the electric force refer to the link;

brainly.com/question/1076352

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Jasmine is diving off a 3-meter springboard. her height in meters above the water when she is x meters horizontally from the end
olganol [36]

Answer:

5.25 m

Explanation:

Given;

The height equation h;

h=-x^2+3x+3

Where;

h = the height above water

x = horizontal distance from the end of the board

The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

2x = 3

x = 3/2

Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

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8 0
3 years ago
8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, T_P = 30 N

The tension in rope Q, T_Q = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, \Sigma F_y = T_P_y + T_Q_y + W = 0

∴ W = -(T_P_y + T_Q_y)

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, \Sigma F_x = T_P_x + T_Q_x  = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0
3 years ago
A race car reaches the finish line and the driver slows down to come to a stop. If it takes the car 10 seconds to stop at a cons
boyakko [2]

Answer:

53.64 m/s

Explanation:

Applying,

a = (v-u)/t............. Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make u the subject of the equation

u = v-at............. Equation 2

From the question,

Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)

Substitute these values into equation 2

u = 0-(-5.364×10)

u = 0+53.64

u = 53.64 m/s

6 0
3 years ago
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