Various Math #23 - Q&A

Question 1 A -3.0 nC charge is on the x-axis at x = -9 cm and a +4.0 nC charge is on the x-axis at x = 15 cm. At what point or points on the y-axis is the electric potential zero. y = ____ cm Answer The charge q1 = -3.0 nC The charge q2 = 4.0 nC r1 = √(y² + (9 cm)²) r2 = √(y² + (15 cm)²) the net potential at given point on y-axis = 0? kq1/r1 + kq2/r2 = 0 kq1/r1 = - kq2/r2 -3.0 nC/√(y² + (9 cm)²) = -4.0 nC/√(y² + (15 cm)²) By equating both sides 9[y² + (15 cm)²] = 16[y² + (9 cm)²] 7y² = 729 cm² y = 10.205 cm Question 2 How many 3d electrons does an Fe3+ ion have? Answer Electronic configuration of Fe = [Ar]3d⁶4s² ion=[Ar]3d⁵ so the ion will have 5 d-electrons Question 3 The drawing shows a circuit that contains a battery, two resistors, and a switch. What is the equivalent resistance of the circuit when the switch is (a) open and (b) closed? What is the total power delivered to the resistors when the switch is (c) open and (d) closed Answer Switch open Req = R1 = 65Ω Total power delivered to Resistor = V²/R = 9²/65 = 1.246W Switch is Close Req = R1R2/(R1+R2) = 65*96/(65+96) = 38.75Ω Total Power delivered to Resistors = V²/Req = 9²/38.75 = 2.098W Question 4 A delivery truck travels 18 blocks north, 10 blocks east, and 16 blocks south. What is its final displacement from the origin?Assume blocks are equal length. Answer 1 initial position: <0, 0> After traveling 18 blocks north: <0, 18> After traveling 10 blocks east: <10, 18> After traveling 16 blocks south: <10, 2> Final position: <10, 2> |<10, 2>| = √10² + 2² =10.2 blocks Answer 2 Explanation: Distance is a scalar quantity that gives a measure of the total length of the path travelled by an object, regardless of its direction Displacement is a vector quantity which connects the initial position to the final position of motion of the object. Being a vector, it has both magnitude and direction: the magnitude is the shortest distance between initial and final position, while the direction goes from the initial to the final position. In this problem, the motion is: 18 blocks north 10 blocks east 16 blocks south So, the distance is the sum of the distances covered in each part: distance = 18+ 10+ 16 = 44 blocks The displacement can be calculated by evaluating the initial and the final position. Along the north-south direction, the displacement is dy = 18(n) - 16(s) = 2 blocks north Along the east-west direction, dx = 10 blocks east So, the magnitude of the displacement is given by Pythagorean's theorem d = √dx²+dy² = √2²+ 10² = 10.2 blocks And the direction is given by θ = tan⁻¹(dy/dx) = tan⁻¹(2/10) = 11.3° measured as north of east. Answer 3 The north movement contributes a displacement of 18 blocks in the positive y-direction (+ y). The east movement contributes a displacement of 10 blocks in the positive x-direction (+x). The south movement contributes a displacement of 16 blocks in the negative y-direction (-y). To find the final displacement, we need to add these vector components. We can represent each component as a vector in two dimensions using the i and j unit vectors. The north component can be represented as 18j, the east component as 10i, and the south component as -16j. The final displacement vector, D, is given by the sum of these components: D = 18j + 10i – 16j. Now, let's simplify this expression. Since i and j are orthogonal unit vectors, we can combine the components separately. For the x-component: 10i = 10*1i = 10i. For the y-component: 18j – 16j = (18-16)j = 2j. Now we have the simplified expression for the displacement vector: D = 10i + 2j. To find the magnitude of the displacement, we can use the Pythagorean theorem. The magnitude of a vector D can be calculated as: |D| = √(Dx)² + (Dy)², where Dx and Dy are the x and y components of the vector, respectively. In this case, Dx = 10 and Dy = 2. Substituting these values, we get: |D| = √10² + 2² = √100 + 4 = √104. Finally, evaluating the square root, we find: |D| = √104 ≈ 11.3. Therefore, the final displacement from the origin is approximately 11.3 blocks. Answer 4 To find the final displacement of the delivery truck from the origin, we can treat the blocks traveled in each direction as vectors and add them up. Let's denote the vectors as v1, v2, and v3, representing the north, east, and south directions, respectively. The magnitude of each vector can be calculated using the Pythagorean theorem: |v1| = 18 blocks (north) |v2| = 10 blocks (east) |v3| = 16 blocks (south) To find the final displacement, we can add these vectors together: D = v1 + v2 + v3 Now, let's calculate the x-component and y-component of D. The x-component can be calculated by subtracting the magnitude of the vector in the west direction (which is equal to the magnitude of the vector in the east direction) from the magnitude of the vector in the east direction: x = |v2| - 0 = 10 The y-component can be calculated by subtracting the magnitude of the vector in the south direction from the magnitude of the vector in the north direction: y = |v1| - |v3| = 18-16 = 2 Now we can calculate the magnitude of the final displacement using the Pythagorean theorem: |D| = √x² + y² = √10² + 2² = √100 + 4 = √104 ≈ 10.2 blocks. Therefore, the final displacement of the delivery truck from the origin is approximately 10.2blocks. Answer 5 Solution: Let's assume that moving north corresponds to increasing the y-coordinate, moving east corresponds to increasing the x-coordinate, and moving south corresponds to decreasing the y-coordinate. The origin is at (0,0). We can break down the truck's movements into two components: the x-component (east/west) and the y-component (north/south). The truck travels 18 blocks north, so the y-component of its displacement is +18 blocks ( Δy = + 18). The truck then travels 10 blocks east, so the x-component of its displacement is +10 blocks ( Δx = + 10). Finally, the truck travels 16 blocks south, so the y-component of its displacement is -16 blocks (Δy = - 16). To find the final displacement from the origin, we need to add up the x and y components. Let's call the final displacement vector d. The x-component of d is the sum of the x-components of the individual displacements: Δx = + 10. The y-component of d is the sum of the y-components of the individual displacements: Δy = +18+(-16) = + 2. Therefore, the final displacement vector d is given by d = (Δx,Δy) = (+10, + 2). To find the magnitude of the final displacement, we can use the Pythagorean theorem. The magnitude |d| is given by: |d| = √Δx² + Δy² = √ (10)² + (2)² = √104 = 2√26. Hence, the final displacement from the origin is 2√26 blocks. Question 5 A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 N. The coefficients of static friction between hishands and the book is .40. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert. Answer The frictional force on both sides of the book are equal to μsN So the weight of the book must be equal to thesum of these forces. 31 = 2μsN then solve for normal force N = 31/2μs N = 31/(2*0.40) = 38.75 N ≈ 39 N