Various Math #33 - Q&A

Question 1 At the origin, there is an A+10 nC charge. What is the electric field's strength at the coordinates (x,y)=(5.0 cm,0 cm)? Answer The field's strength at point (x,y), = (5 cm, 0 cm). Since there is a distance between the charge and that location. r = √(x² + y²) = √(0.05)² + (0)² = 0.05 m E = kq/r² = (9 × 10⁹) * (9 × 10⁻⁹) / (0.05)² = 3.6 × 10⁴ N/C Question 2 A small charged bead has a mass of 1.0 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 20 m/s². What is the charge on the bead? Answer 1 Because of the charge's upward acceleration in the example question, the charge's sign is positive. The following equation provides the electric force on the bed. F_e = qE = q * (200,000 N/C) Use Newton's second law F_Net = ma F_e - mg = ma qE - mg = ma So, q = (ma + mg)/E = (ma + mg)/E q = m(a + g)/E = (1.0 × 10⁻³kg) * (20 m/s² + 9.8 m/s²) / (200,000 N/C) q = 29.810⁻³ / 200,000 = 0.000149 C q = 0.149μC Answer 2 When two charged particles, one of which has a charge q₁ and the other has a charge q₂, are placed at a great distance from one another (r), they exert forces on one another that decrease with distance and rise with charge magnitude. This force is described by Coulomb's law and is represented by equation (22.2) in the following form: F_c = F_1on2 = F_2on1 = K|q₁||q₂|/r² (1) Where K is the electrostatic constant and it is given by 9.0 × 10^9 Nm²/C² The charge of the bead is given by 20 m/s². Convert the unit into Coulomb by q₁ = (m(a+g))/E = (1 × 10⁻³kg(20 + 9.8 m/s²)) / 200 N/C = 29.8×10⁻³/200 = 0.149μC Question 3 Which of the following are true statements? α: 2 ∈ {1,2,3} b: {2} ∈ {1,2,3} c: 2 ⊂ {1,2,3} d: {2} ⊂ {1,2,3} e: {2} ⊂ {{1}, {2}} f: {2} ∈ {{1}, {2}} Answer Definition used: Proper subset: A proper subset of a set A is a subset of A that is not equal to A. The element x belongs to the set A means the element in the set A. Consider statement a = 2 ∈ {1,2,3} Here, the element 2 belongs to the set {1, 2, 3} thus the statement is true consider statement b = {2} ∈ {1,2,3} Here the set {2} doesn't belong to the set {1, 2, 3} thus the statement is false consider statement c = 2 ⊂ {1,2,3} Here the number 2 is not the proper subset of {1, 2, 3}. Thus the statement is false consider he statement d = {2} ⊂ {1, 2, 3} Here the set {2} is proper subset of {1, 2, 3} Thus the statements is true consider the statement e = {2} ⊂ {{1}, {2}} Here, the set {2} is proper subset of {{1}, {2}} Thus, the statements is true consider statement f = {2} ∈ {{1}, {2}} Here the set {2} belongs tp {{1}, {2}}. Thus the statement is true Question 4 A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. Find the rate of growth after 3 hours. Answer From part(a), we know that P(t) = 100(4.2)^t Differentiate P'(t) = d/dt [100 * (4.2)^t] Recall that: d/dx(a^x) = a^x * lna P' (t) = 100 * (4.2)^t * ln4.2 P'(t) = P(t) * ln4.2 Substitute t=3 P' (3) = P(3) * ln4.2 From part (b), we know that P (3) = 7409 P' (3) = 7409 * ln4.2 ≈ 10633 Bacteria/hour Result: P' (3) = 7409 * ln4.2 ≈ 10633 Bacteria/hour Question 5 The tires on a new compact car have a diameter of 2.0 ft and are warranted for 60 000 miles. Determine the angle (in radians) through which one of these tires will rotate during the warranty period. Answer Given that the tire's diameter in this instance is d = 2.0 ft = 0.3048 x 2.0 m = 0.61 m Therefore the radius of the tire r = d/2 = 0.305 m Warranty of the tire l = 60000 mi = 1609.34 × 60000 m = 96560400 m a) As is common knowledge, an object moving in a circle with radius r will traverse an angle in radians given by θ = l/r = 96560400 m / 0.305 m = 316591475.4 rad = 3.2 × 10⁸ rad 28.8s