Pregunta 2 ( 20 puntos) Sean \( \vec{A}=3 \hat{\imath}+\hat{\jmath}+c \hat{k}, \vec{B}=-c \hat{\imath}-c \hat{\jmath}+2 \hat{k}, \vec{C}=\hat{\imath}-2 \hat{\jmath}+\hat{k} \) tres vectores en el plano \( x-y-z \), donde \( c \) es una constante. Responda las siguientes preguntas. (i) Calcule el vector \( \vec{D} \) tal que \( \vec{D}=2 \vec{A}-\vec{B}-\vec{C} \). (ii) Determine el valor de la constante \( c \) tal que el vector \( \vec{D} \) sea paralelo al plano \( x-y \). (iii) Determine el valor de la constante \( c \) tal que el vector \( \vec{D} \) sea perpendicular al eje de las \( x \).

9. Si dejamos caer al suelo una piedra cuya masa es de 4 kg desde una altura de 4 m , ¿Cuál es la energía cinética cuando ya ha recorrido la mitad de la caída a) hacia el suelo? b) CCual es la velocidad cuando alcanza el suelo?

A coil with an area of \( 0.6 \mathrm{~m}^{2} \) is held with its axis coinciding with the direction of a magnetic field (surface perpendicular to magnetic field), with a strength of 0.4 T . 8.3 Calculate the magnetic flux.

157. Una bombola a pareti rigide di 30 L contiene \( 0,5 \mathrm{moli} \) di un gas perfetto biatomico alla temperatura iniziale di \( 20^{\circ} \mathrm{C} \). Essa viene tenuta sotto il sole per un certo tem- po e si nota che la temperatura finale è di \( 80^{\circ} \mathrm{C} \). Trascu- rando la dilatazione termica della bombola, si calcoli: - il lavoro fatto dal gas; - la variazione di pressione del gas; - la variazione di energia interna del gas. [0 J; \( \left.8,3 \times 10^{3} \mathrm{~Pa} ; 6,2 \times 10^{2} \mathrm{~J}\right] \) (Esame di Fisica, Corso di laurea in Farmacia, Università La Sapienza di Roma, 2006/2007)

8.4 To produce an emf of 9 V , the area of the coil, with its axis coinciding with the direction of a magnetic field (area perpendicular to magnetic field), is halved in 2 minutes from \( 0.6 \mathrm{~m}^{2} \) to \( 0,3 \mathrm{~m}^{2} \). Calculate the number of turns around the coil.

QUESTION 6 (Start on a new page.) It has been established that the planet Omega has a moon and that the moon of Omega orbits the planet at a height, above the surface of Omega, which is equal to twice the radius of Omega. The mass of planet Omega is \( 3 \times 10^{15} \mathrm{~kg} \) and the mass of its moon is \( 1.5 \times 10^{13} \mathrm{~kg} \). The radius of the planet Omega is \( 8.5 \times 10^{6} \mathrm{~m} \) and the radius of its moon is \( 2.5 \times 10^{5} \mathrm{~m} \). 6.1 Determine the distance between the centre of the planet Omega and the centre of its moon.