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Bumek [7]
3 years ago
6

A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of

the resultant force in kN on the 80 kg driver of this car?
Physics
1 answer:
djverab [1.8K]3 years ago
3 0

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

F_c=ma_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN

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Kaylis [27]

Answer:

 Δt'/ T% = 90.3%

Explanation:

Simple harmonic movement is described by the expression

         x = A cos (wt)

we find the time for the two points of motion

x = - 0.3 A

        -0.3 A = A cos (w t₁)

         w t₁ = cos -1 (-0.3)

         

remember that angles are in radians

        w t₁ = 1.875 rad

x = 0.3 A

        0.3 A = A cos w t₂

        w t₂ = cos -1 (0.3)

         w t₂ = 1,266 rad

         

Now let's calculate the time of a complete period

x= -A

        w t₃ = cos⁻¹ (-1)

        w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

         T = 2 t₃

         T = 2π / w     s

now we can calculate the fraction of time in the given time interval

        Δt / T = (t₁ -t₂) / T

        Δt / T = (1,875 - 1,266) / 2pi

        Δt / T = 0.0969

 

This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is

         Δt'/ T = 1 - 0.0969

         Δt '/ T = 0.903

         Δt'/ T% = 90.3%

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