A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest

Answer:

The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.

Explanation:

Here’s the breakdown:

The electric field from the positive sheets E₁ = б/2E₀

E₂ is from the negative sheet = -б/2E₀

At points 1, 4, and 5, the electric fields created by the sheets oppose each other.

At point 1, the total field is calculated as -E₁ + E₂ = 0.

Similarly, at point A, the total field results in -E₁ - E₂ = 0.

However, at any point in between the plates, the electric field is directed consistently in one way.

At points 2 and 3, the field is directed to the right.

Thus, we have:

E net = E₁ + E₂

= б/2E₀ + -б/2E₀

=б/E₀

Note: Please refer to the attached document for the full question accompanying this solution.

According to Newton's second law, provided F is directed horizontally,

• the net horizontal force acting on the larger block is

F - µmg = 3mA

where µmg represents the friction experienced by the larger block from contact with the smaller block, µ is the static friction coefficient between both blocks, and A indicates the acceleration of the block;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg denotes the magnitude of the normal force exerted by the surface on the combined mass of both blocks, 3mg corresponds to the weight of the larger block, and mg indicates the weight of the smaller block;

• the net horizontal force acting on the smaller block is

µmg = ma

where µmg again signifies the friction between both blocks; however, it is important to note that this force aligns in the same direction as F. It is the sole force influencing the smaller block in the horizontal direction, thus (b) static friction causes the smaller block's acceleration;

• the net vertical force on the smaller block is

mg - mg = 0

where mg represents the force of both the normal force from the larger block pushing up against the smaller one, and the weight of the smaller block.

(You should be able to create your own free-body diagrams based on the forces discussed.)

(c) Solve the equations stated above to find A and a:

A = (F - µmg) / (3m)

a = µg

Answer:

≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + k

The normal vector N is computed as

Where:

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u

Thus

≈ 3077.34

Answer:

d

Explanation: