Answer:
a. The moment of the 4 N force is 16 N·m clockwise
b. The moment of the 6 N force is 12 N·m anticlockwise
Explanation:
In the figure, we have;
The distance from the point 'O', to the 6 N force = 2 m
The position of the 6 N force relative to the point 'O' = To the left of 'O'
The distance from the point 'O', to the 4 N force = 4 m
The position of the 4 N force relative to the point 'O' = To the right of 'O'
a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point
a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise
b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.
Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
Answer:
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.)
Show step by step please.
Note: The answer is given it's should be 1000 m ??
This what i can up with so see what it is kid
Explanation:
Liquid would like wet the pages.
Hope this helps !
Brainliest if it pleases you :D
