I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
Answer:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
Answer:
L = 2.8 cm
Explanation:
Period T = 4 / 12 = 1/3 s
T = 2π√(L/g)
L = (T/2π)²g
L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm
True, the path of the ball, as observed from the train window, will be a horizontal straight line.
An object projected from a certain height has a parabolic path when observed from a fixed point.
However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.
When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.
Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.
Learn more about path of motion of objects here: brainly.com/question/82610
