The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
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Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
And the distance traveled downwards is:
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
Replacing into the first equation:
Rationalizing:
Let's call v1 the final speed of the package dropped from a height H. Thus:
Let v2 be the final speed of the package dropped from a height 4H. Thus:
Taking out the square root of 4:
Dividing v2/v1 we can compare the final speeds:
Simplifying:
The final speed of the second package is twice as much as the final speed of the first package.
There are different forces acting on an object like nuclear force , gravitational force...plus external forces like friction and other..
net sum of all these is resultnat
Answer:
2.2 s
Explanation:
Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)
So, T = 2π√(I/mgh)
T = 2π√(mL²/3 /mgL/2)
T = 2π√(2L/3g)
substituting the values of the variables into the equation, we have
T = 2π√(2L/3g)
T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))
T = 2π√(3.66 m/(29.4 m/s² ))
T = 2π√(0.1245 s² ))
T = 2π(0.353 s)
T = 2.22 s
T ≅ 2.2 s
So, the period of the man's leg is 2.2 s
As the rollercoaster goes up. kinetic energy changes to gravitational potential energy. When it moves back down, gpe changes back to ke.
