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:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
Answer:
living things are found in air, water, and soil.
Explanation:
Answer:
50 N
4.2 N
Explanation:
i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.
ii) When the boom is resting on the end support, the normal force is:
∑τ = Iα
-W (0.50) + F (3.0) − N (6.0) = 0
-0.50 W + 3.0 F = 6.0 N
N = (-0.50 W + 3.0 F) / 6.0
N = (-0.50 × 2350 + 3.0 × 400) / 6.0
N ≈ 4.2
Answer:
Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.
