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denpristay [2]
3 years ago
10

Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?

Physics
1 answer:
Elanso [62]3 years ago
7 0

Answer: The displacement is 1 block.

Explanation:

Let's define:

The right is the positive side.

The left is the negative side.

Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

A - M.

In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

Then her new position is:

-3 + 5 = 2

Now she walks 3 blocks to the left, then her new position is:

2 - 3 = -1

The displacement will be equal to the difference between the final position (-1) and the initial position (-2)

Then the displacement is:

D = -1 - (-2) = -1 +2 = 1

The displacement is 1 block.

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4 years ago
A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc
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(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

  • <em>first charge, q₁ = 8.75 μC </em>
  • <em>second charge, q₂ = -6.5 μC  </em>
  • <em>electric field, E = 1.85 x 10⁸ N/C</em>
  • <em>distance between the two charges, r = 2.5 cm</em>

<em />

(a)

The attractive force between the charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N

The force on the negative charge due to the electric field is calculated as follows;

F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N

The tension on the wire is the resultant of the two forces and it is calculated as follows;

T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N

The force on the first negative charge due to the electric field is calculated as follows;

F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N

The force on the second negative charge due to the electric field is calculated as follows;

F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N

The tension on the wire is the resultant of the three forces and it is calculated as follows;

T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N

Learn more here:brainly.com/question/19565286

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