1 Free Energy

1- Situação de mercado Tamanho: Red Bull é uma bebida energética sem álcool em sua composição. Tornou-se um sucesso de vendas em mais de 160 países, devido, em grande parte, a uma ótima campanha de marketing, uma excelente distribuição e um bom projeto de embalagem. O Red Bull foi criado no ano de 1984, quando se iniciou a pesquisa de sabor e sua comercialização legalizada somente em 1987. O austríaco Dietrich Mateschitz, durante uma viagem de negócios a Tailândia, enquanto executivo de…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…

You are free: to copy, distribute and transmit the work; to adapt the work. You must attribute the work in the manner specified by the author or licensor OBTENÇÃO DE BIODIESEL EMPREGANDO ROTA ETÍLICA E CATALISADOR ALCALINO Cesar A. da Silva1 RESUMO Os insumos potencialmente capazes de promover o abastecimento da demanda de geração energética têm originado diversas pesquisas científicas com o objetivo de amenizar os impactos ambientais. Dentre as soluções encontradas destaca-se o…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= (26, 10, 4) = (0.92, 0.36, 0.14) |(26, 10, 4)| b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2.

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= (26, 10, 4) = (0.92, 0.36, 0.14) |(26, 10, 4)| b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2.

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,…