The 1st one goes two added sodoes the second one then the third goes to removed then the fourth goes to added and the rest go to removed
1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.
The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is
a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration
Since it started from rest, v₀ = 0. Using the formula:
0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a

⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s