13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

Answer: 24.24 m

Explanation:

A player launches a football 50.0 m at an angle of 61° to the north of west. We will break this down into vertical and horizontal elements.

Horizontal component: 50 cos 61° = 24.24 m directed westward

Vertical component: 50 sin 61° = 43.73 m directed toward the north.

Refer to the diagram below.

Therefore, the westward displacement of the football corresponds to the horizontal component of the displacement, which is 24.24 m.

Response:

a) 80 V

b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

Solution

(a) Utilizing the given values for VA, and q, we derive a relationship among the three parameters and VB to compute VB.

                                   .........................................(1)                                          

Where = 0, and the potential energy U of the charge is defined as U = q V

In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:

                                  0+qVA= +qVB                    (Dividing by q)

                                         VA= /q + VB                  (Restructuring for VB)

                                         VB=VA- /q.......................................(2)

We now have the relation between VB, VA, and , allowing us to substitute our values for VA, , and q into equation (2) to obtain VB

                                         VB=VA- /q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)

                                              =80 V

(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points

                                   VA-VB =...................................(a)

                                   VA-VB=E                      (Restructuring for E)

                                            E= VA-VB/..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

                                            E= VA-VB/

                                              =-100 N/C

The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.

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.

Answer:

57.94°

Explanation:

We understand that the formula for flux is

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78

                          E=

                          S=

 =  

 =0.5306

 θ=57.94°