It's important to match your exercise shoes with the type of exercise in which you will be participating.

Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x--1 m and has a surface charge density

MariettaO [177]

Answer:

Ea = 3.2 10⁵ N / C

Explanation:

To calculate the electric field of each plane we will use Gauss's law, we create a Gaussian surface that is a cylinder that has the axis perpendicular to the plane, in this case the flow line between the cylinder walls and the surface is zero and all the flow is perpendicular to the base of the cylinder.

We apply the law of gauss flow to each side is the value of the electric field (E) for the area of the cylinder (A); whereby the flow in the two directions is 2 E A

Φ = 2E A = ρₙt / εo

Where ρₙ is the charge inside the cylinder, as the charge density gives us, σ = Q / A

ρₙ = σ A

By which we can clear the electric field

E = σ A / 2εo

where it is worth 8.85 10⁻¹² C² / N² m². Let's calculate with this equation in the field for each plane

1 plane σ= -2.0 pC / m²

E1 = -2. 10⁻¹² / 2 8.85 10⁻¹²

E1 = -0.113 N / C

The field line is directed to the plane

2 Plane σ = 5.8 mC / m2

E2 = 5.8 10-6 / 2 8.85 10-12

E2 = 3,277 10 5 N / A

Field lines leave the plane

As we have the values of each field in the whole space. Let's calculate in the field at the point x = 1 m

To do this we must add the fields at the selected point vectorally, for this distance the point is between the two planes, so the field of plane 1 points to the left and the point of plane 2 also points to the left, consequently field adds

Ea = E1 + E2

Ea = 0.113 + 3.277 10⁵

Ea = 3.2 10⁵ N / C

At point X = -1 m in this case the point is on plane 1, so this plane does not generate any field, it is an equipotential surface, the total field is equal to field 2

Ea = E1 = 3.2 10⁵ N / C

Note that there is no difference in numbers values by the difference between the load between each plane

5 0

3 years ago

A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration

Julli [10]

Answer:

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) The maximum height attained by the rocket from the ground is 348.65 m

(c) The time taken by the rocket to maximum height after lift off is 8.85 s.

(d) The total time taken by the rocket in air is 17.3 second.

Explanation:

u = 57 m/s, a = 3 m/s^2, h = 140 m

let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.

Use III equation of motion

V^2 = u^2 + 2a h

v^2 = 57^2 + 2 x 3 x 140

v = 63.95 m/s

Now use I equation of motion

v = u + at

t = (63.95 - 57) / 3 = 2.32 s

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) Let H be the maximum height reached by the rocket after the engine stops.

Use III equation of motion

v^2 = u^2 + 2aH

here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2

0 = 63.95^2 - 2 x 9.8 x H

H = 208.65 m

The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m

(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.

Use I equation of motion

v = u + a t'

0 = 63.95 - 9.8 x t'

t' = 6.53 s

The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.

(d) let t'' be the time taken by the rocket to fall freely

Use II equation of motion

H' = ut'' + 1/2 gt''^2

Here, H' = 348.65 m, u = 0

348.65 = 0 + 0.5 x 9.8 x t''^2

t''^ = 8.44 s

The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.

4 0

3 years ago

In a thunderstorm at 32.0°C, Reginald sees a bolt of lightning and hears the thunderclap 2.00s later. How far from Reginald did

ratelena [41]

Answer:

dont have the answer but you cute

Explanation:

5 0

2 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

3 0

3 years ago

Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,

White raven [17]

The net force acting on the car is 65 N to the left

The net force acting on an object is simply defined as the resultant force acting on the object.

From the question given, we obtained the following data:

Force applied to the right (Fᵣ) = 250 N
Force applied to the left (Fₗ) = 315 N
Net force (Fₙ) =?

The net force acting on the car can be obtained as follow:

Fₙ = Fₗ – Fᵣ

Fₙ = 315 – 250

Therefore, the net force acting on the car is 65 N to the left

Learn more on net force: brainly.com/question/19549734

8 0

2 years ago

Read 2 more answers