Au Pair

da SDH. Na SDH, a informação está organizada em módulos síncronos de transporte (STM), os quais contêm três elementos básicos: 1. SOH (cabeçalho de seção): cumpre funções de sincronismo de quadro, canais de serviço, funções de controle, etc. 2. AU - Pointer (ponteiro da unidade administrativa): indica como está estruturada a informação na área da carga útil, e indica como localizar os “virtual container”, onde está a informação dos tributários. 39 3. “Payload” (área de…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

B = (4.00i j j " " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector…

⋅ C = (7.22)(−3.0) + (9.58)(−5.20) = −71.5 m 2 1.51: a) From Eq.(1.21), r r A ⋅ B = (4.00 )(5.00 ) + (3.00 )(− 2.00 ) = 14.00. b) A ⋅ B = AB cos θ , so θ = arccos [(14.00) (5.00 × 5.39 )] = arccos(.5195) = 58.7 o. 1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle φ as r r  A ⋅ B  Ax Bx + Ay B y   . φ = arccos     AB  = arccos AB     In the intermediate calculations given here, the…

" " A⋅ B 14.0 = = 0.519; φ = 58.7°. (b) cos φ = AB (5.00)(5.39) " " " EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to " " ⎛ A⋅ B ⎞ ⎛ Ax Bx + Ay By ⎞ give the angle φ as φ = arccos ⎜ ⎟ = arccos ⎜ ⎟. AB ⎝ ⎠ ⎝ AB ⎠ SET UP: Eq.(1.14) shows how to obtain the components for a vector

⋅ C = (7.22)(−3.0) + (9.58)(−5.20) = −71.5 m 2 1.51: a) From Eq.(1.21), r r A ⋅ B = (4.00 )(5.00 ) + (3.00 )(− 2.00 ) = 14.00. b) A ⋅ B = AB cos θ , so θ = arccos [(14.00) (5.00 × 5.39 )] = arccos(.5195) = 58.7 o. 1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle φ as r r  A ⋅ B  Ax Bx + Ay B y   . φ = arccos     AB  = arccos AB     In the intermediate calculations given here, the…