University:
University of North Georgia
Course:
MATH 1450 | Calculus I
Academic year:

2024
Views:

265

Pages:

3
Author:

Autumn Cox

Question 1 A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.35 m/s², how far will it travel before becoming airborne? (b) How long does this take? Answer Given: v_i = 0 m/s a = 0.35 m/s² v_f = 6 m/s Required: a) x =? b) t =? a) According to kinematic equation and given information conclude that x = (v_f² - v_i²)/(2a) = (6²)/(2 * 0.35) = 51.4 m b) According to kinematic equation and given information conclude that t = (v_f - v_i)/a = 6 / 0.35 = 17.1 s Question 2 The compressive strength of concrete is normally distributed with μ = 2500 psi and σ = 50 psi. Find the probability that a random sample of n = 5 specimens will have a sample mean diameter that falls in the interval from 2499 psi to 2510 psi. Answer Data given for compressive strength of concrete: μ = 2500 psi, σ = 50 psi and n=5 Let us calculate the probability that the sample mean diameter that falls in the interval from 2499 to 2510 psi P[2499 V_i = m_c * v_c / (m_c + m_t) Now, consider the momentum conservation along the "j" direction: m_c * v_c + m_t * v_t = (m_c + m_t) * V_j => V_j = m_t * v_t / (m_c + m_t) Therefore, the velocity of the vehicles after the collision is: V = (m_c/(m_c + m_t)) * v_c * i + (m_t/(m_c + m_t)) * v_t * j By substituting the corresponding values, we get the velocity of the vehciles after the collision as: V = -(1047/(1047+1593)) * 15i - (1593/(1047+1593)) * 10.8j => V = -6.04i - 6.45j |V| = √( (-6.04)² + (-6.45)² ) = 8.836 m/s After the collision the kinetic energy of the both vehicles combined will be utilized to do work against the friction force (given µ_k = 0.5), The friction force both the vehicles is given by: f = µ_k(m_c + m_t)g = 0.5*(1074 + 1593)*9.81 = 13081.635 N Now, according to work energy theorem the we can write: (1/2)(m_c + m_t)V² = fd Where "d" is the distance travelled by the wreckage after the collision. Substituting the corresponding values, we get: d = (1/2) * (1047+1593)(8.836)² / 13081.635 = 7.958 m Hence, the distance travelled by the wreckage after the collsion is 7.958 m. Question 5 Calculate the nuclear binding energy of ⁵⁵Mn(A = 55, z = 25) in joules. The atomic mass of ⁵⁵₂₅Mn is 54.938 amu The speed of light is c = 3.00 × 10⁸ m/s The mass of a proton is 1.0073 amu The mass of a neutron is 1.0087 amu The mass of an electron is 5.4858 × 10⁻⁴ amu 1 kg = 6.022 × 10²⁶ amu 1 J = 1 kg * m²/s² 1 MeV = 1.602 × 10⁻¹³ J Answer The number of the neutrons of the substance is N = A - Z = 55 - 25 = 30 The expression for the final mass of the substance is mf = Z(mp) + N(mn) = 25(1.0073 amu) + 30(1.0087 amu) = 55.4435 amu The difference in mass is Δm = m_f - m_i Δm = 55.4435 amu - 54.938 amu = 0.5055 amu Δm = 0.5055 amu * (1.66 × 10⁻²⁷ kg / 1 amu) = 8.39 × 10⁻²⁸ kg The binding energy is BE = Δmc² BE = (8.39 × 10⁻²⁸kg) * (3 × 10⁸ m/s)² = 7.55 × 10⁻¹¹ J