Answer:
Te ayudo con una de prueba $
Explanation:
Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
Answer:
9800 m
Explanation:
During acceleration, given:
v₀ = 0 m/s
a = 39.2 m/s²
t = 10.0 s
Find: v and Δy
v = at + v₀
v = (39.2 m/s²) (10.0 s) + 0 m/s
v = 392 m/s
Δy = v₀ t + ½ at²
Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s)²
Δy = 1960 m
During free fall, given:
v₀ = 392 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (392 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 7840 m
Therefore, h = 1960 m + 7840 m = 9800 m.
