Physics, 20.07.2019 13:00 pnicklin00

The radius of curvature of a loop-the-loop for a roller coaster is 12.4 m. at the top of the loop (with the car inside the loop), the force that the seat exerts on a passenger of mass m is 0.80mg. find the speed of the roller coaster at the top of the loop.

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Answer from: Ideen

14.79 m/s At the top of the loop, there's 2 opposing forces. The centripetal force that's attempting to push the roller coaster away and the gravitational attraction. These 2 forces are in opposite directions and their sum is 0.80 mg where m = mass and g = gravitational attraction. So let's calculate the amount of centripetal force we need. 0.80 = F - 1.00 1.80 = F So we need to have a centripetal force that's 1.8 times the local gravitational attraction which is 9.8 m/s^2. So 1.8 * 9.8 m/s^2 = 17.64 m/s^2 The formula for centripetal force is F = mv^2/r where F = force m = mass v = velocity r = radius We can eliminate mass from the equation since the same mass is being affected by both the centripetal force and gravity. So: F = v^2/r 17.64 m/s^2 = v^2/12.4 m 218.736 m^2/s^2 = v^2 14.78972616 m/s = v So the velocity at the top of the loop (rounded to 2 decimal places) is 14.79 m/s.

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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The radius of curvature of a loop-the-loop for a roller coaster is 12.4 m. at the top of the loop (w...

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