Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
F = 351×10³lb
Explanation:
Given the density
ρg = 64.6lb/ft³
Diameter d = 12ft
The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft
The pressure in the tank is
P = ρgh = 64.6 × 12 = 775.2lb/ft²
The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²
F = 351×10³lb.
Answer:
The musculoskeletal system involves the complex interactions of muscles, bones, and connective tissues.
When a muscle contracts, a bone will move. When a bone contracts, a muscle will move. When a ligament contracts, a tendon will move.
Due to friction from sliding against the ground, the player decelerates in 1 direction. Thus his momentum decreases.