Example of third class Lever first class Lever and second class Lever
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Use the following kinematic equation to solve for the final velocity:
In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:
Plug in the given acceleration and time:
Newtons 3.law: Action = Reaction
If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.
If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.
Answer:
θ = 62.72°
Explanation:
The projectile describes a parabolic path:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = x₀+ vx*t Formula (1)
vx = v₀x
Where:
x: horizontal position in meters (m)
x₀: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
v₀x: Initial speed in x in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)
vfy= v₀y -gt Formula (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 120 m/s , at an angle θ above the horizontal
v₀x= 55 m/s
x-y components of the initial velocity ( v₀)
v₀x = v₀*cosθ Equation (1)
v₀y = v₀*sinθ Equation (2)
Calculating of the angle θ
We replace data in the Equation (1)
55 = 120*cosθ
cosθ = 55 / 120
θ = 62.72°