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Ivanshal [37]
1 year ago
8

The positions of four objects as a function of time are shown

Physics
1 answer:
KengaRu [80]1 year ago
6 0

There is no movement in line C and the greatest velocity occurs at line D. The answers are:

1. 0.5 m/s

2. 0.25 m/s

3. 14m and -2m

4. -1 m/s

<h3>What is Position - time Graph ?</h3>

Position time graph is the graph of distance or displacement against time. The slope of the graph is velocity.

The given positions of four objects as a function of time are shown

on the graph to the right.

1.) The velocity of object A will be the slope m of the line A.

Slope m = Δx / Δt

m = (4 - 0) / (8 - 0)

m = 4 / 8

m = 0.5 m/s

Velocity at A = 0.5 m/s

2.) The average velocity of object B will be the slope m of the line B.

Slope m = Δx / Δt

m = (6 - 4) / (8 - 0)

m = 2 / 8

m = 0.25 m/s

The average velocity of object B is 0.25s

3.) The object moved a total distance during the first eight seconds will be 4m for A, 2m for B, and 8m for D

Total distance = 4 + 2 + 8 = 14m

It’s net displacement during the same time will be 2. That is,

Displacement = 8 - 6 = -2m

4.) The greatest speed occurred at line D. The velocity of the object moving at the greatest speed will be the slope of the line D

V = -Δx / Δt

V = -8/8

V = -1 m/s

Therefore, there is no movement in line C and the greatest velocity occurs at line D.

Learn more about velocity time graph here :brainly.com/question/769606

#SPJ1

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Read 2 more answers
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
2 years ago
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