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The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 40-m-diameter loop-the-loop. What is the car's speed at the top?

Respuesta :

Answer:

Explanation:

Given

Diameter [tex]d=40\ m[/tex]

radius [tex]r=20\ m[/tex]

From diagram, at top point

If Normal force is equal to Gravitational force

[tex]N=mg[/tex]

where N=normal reaction

m=mass of car

Normal reaction will provide centripetal force

[tex]N=\frac{mv^2}{r}[/tex]

thus

[tex]\frac{mv^2}{r}=mg[/tex]

[tex]v=\sqrt{2gr}[/tex]

[tex]v=\sqrt{2\times 9.8\times 20}[/tex]

[tex]v=19.79\ m/s[/tex]              

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The car's speed at the top if the  normal force equals the magnitude of the gravitational force is 14m/s

According to the question, we are told that the normal force is equal to the magnitude of the gravitational force as a roller coaster.

N = G

The formula for calculating the normal force is expressed as:

N = mg

The formula for calculating the gravitational force is expressed as:

G = [tex]\frac{mv^2}{r}[/tex]

If they are both equal, hence;

[tex]mg=\frac{mv^2}{r}\\g=\frac{v^2}{r}[/tex]

Given the following parameters

diamater = 40m

radius = 20m

acceleration due to gravity g = 9.8m/s^2

Substitute the given parameters into the formula to get the car's speed

[tex]9.8=\frac{v^2}{20}\\v^2 = 9.8 \times 20\\v^2 = 196\\v=\sqrt{196}\\v = 14m/s\\[/tex]

Hence the car's speed at the top if the  normal force equals the magnitude of the gravitational force is 14m/s

Learn more here: https://brainly.com/question/23792224

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