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Vanyuwa [196]
2 years ago
9

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-hi

gh hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify
Physics
1 answer:
Andrew [12]2 years ago
3 0

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -  

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is  

 k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

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3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
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Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

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comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

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     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

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        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

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  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

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  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

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Angelina_Jolie [31]

Answer: B

Explanation: the teacher just told us the answer

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As we know that the combination is maintained at rest position

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Answer:

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3 years ago
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