Hence the expression of ω in terms of m and k is
Given the expressions;
Equating both expressions we will have;
Divide both equations by 2π
Square both sides
Take the square root of both sides
Hence the expression of ω in terms of m and k is
Answer:
22m/s
Explanation:
lowest part on the graph (closest to x-axis)
Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Answer:
R=2F
Explanation:
As the forces are in same direction so the resultant force will be:
R=F+F
R=2F
