Answer:
1. is the age group 35 and 44
2. is 2006 i think its 2006 i cant really tell in the picture but its the one before the last one!
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
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The movements of the tectonic plates
