John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. He is pulling horizontally wi

Response:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.Please review the attached document for further explanations

Answer:

Part A) Electric fields at the designated point due to charges q₁ and q₂:

E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) The overall electric field at P (Ep)

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Explanation:

Conceptual analysis

The electric field at point P caused by a point charge is calculated as:

E = k*q/d²

E: Electric field measured in N/C

q: charge magnitude in Newtons (N)

k: electric constant measured in N*m²/C²

d: distance from the charge q to point P in meters (m)

Equivalence:

1 nC = 10⁻⁹ C

1 cm = 10⁻² m

Data:

k = 9 * 10⁹ N*m²/C²

q₁ = -6.00 nC = -6 * 10⁻⁹ C

q₂ = +3.00 nC = +3 * 10⁻⁹ C

d₁ = 4 cm = 4 * 10⁻² m

d₂ = 5 * 10⁻² m

Part A) Calculation for electric fields at point from q₁ and q₂:

Refer to the attached illustration:

E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.

E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from the charge.

E₁ = k*q₁/d₁² = 9 * 10⁹ * 6 * 10⁻⁹ / (4 * 10⁻²)² = 33.75 * 10³ N/C

E₂ = k*q₂/d₂²= 9 * 10⁹ * 3 * 10⁻⁹ / (5 * 10⁻²)² = 10.8 * 10³ N/C

E₁ = 33.75 * 10³ N/C (-j)

E₂x = E₂cosβ = 10.8 * (3/5) = 6.48 * 10³ N/C

E₂y = E₂sinβ = 10.8 * (4/5) = 8.64 * 10³ N/C

E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) Calculation for net electric field at P (Ep)

The electric field at point P from multiple point charges is the vector sum of the individual electric fields.

Ep = Epx (i) + Epy (j)

Epx = E₂x = 6.48 * 10³ N/C (-i)

Epy = E₁y + E₂y = (33.75 * 10³ (-j) + 8.64 * 10³ (+j)) N/C = 25.11 * 10³ (-j) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Answer: The result to the query is 0.25 ohms

Explanation:

R = u x/A.......1

where u represents the resistivity of the

rod, A is the cross-sectional area, and x denotes

the length of the rod.

Let R* represent the resistance across the adjacent sections of the rod.

Then, R* = u1/4.......2

By comparing equation 1 with equation 2, we find that

R* = 1/4

which equals 0.25 ohms.

Answer:

50.2 cm

Explanation:

We have the following data:

Height, h=3.5 m

Initial horizontal velocity,

Time, t=0.32 s

We need to determine how far the ball is from the ground after 0.32 s.

Initial vertical velocity,

Where

a)

i) 120 s

ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c)

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.

c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force , leading us to the equation:

where

represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

- The weight (gravitational force) of the book is

Upon comparing these two equations, we conclude:

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.