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serious [3.7K]
3 years ago
9

Sort the following forces as relevant or not relevant to this situation. The symbols are defined as follows: normal force = n⃗ ,

tension force = T⃗ , horizontal force of the road on the car = F⃗ rc, horizontal force of the road on the truck = F⃗ rt, weight = w⃗ , force of the car pushing on the truck = F⃗ ct, and force of the truck pushing on the car = F⃗ tc.
Physics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

it would be least to graetest

Explanation:

10-84

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The curvature of the helix r​(t)equals(a cosine t )iplus(a sine t )jplusbt k​ (a,bgreater than or equals​0) is kappaequalsStartF
4vir4ik [10]

Answer:

\kappa = \frac{1}{2 b}

Explanation:

The equation for kappa ( κ) is

\kappa = \frac{a}{a^2 + b^2}

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

\kappa (a) = \frac{a}{a^2 + b^2}

Now, the conditions to find a maximum at a_0 are:

\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0

\frac{d^2\kappa(a)}{da^2}  \left | _{a=a_0} < 0

Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 -  2 a^2)

\frac{d}{da} \kappa = \frac{b^2 -  a^2}{(a^2 + b^2)^2}

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 -  a^2}{(a^2 + b^2)^2} )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 -  a^2 ) + (b^2 -  a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2}  )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2  a ) + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

\frac{d^2}{da^2} \kappa = \frac{-2  a}{(a^2 + b^2)^2} + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

We dcan skip solving the equation noting that, if a=b, then

b^2 -  a^2 = 0

at this point, this give us only the first term

\frac{d^2}{da^2} \kappa = \frac{- 2  a}{(a^2 + a^2)^2}

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

\kappa = \frac{b}{b^2 + b^2}

\kappa = \frac{b}{2* b^2}

\kappa = \frac{1}{2 b}

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