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A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through each dip at a different speed.
For the different values given for the radius of curvature R and speed v, rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them.
A) R=60m v=16 m/s
B) R=15m v= 8m/s
C) R=30m v= 4m/s
D) R=45m v= 4m/s
E) R=30m v= 16 m/s
F) R= 15m v=12 m/s

Respuesta :

Ranking the magnitude of the forces from largest to smallest is;

N₆ > N₅ > N₁ = N₂ > N₃ > N₄

  • Since we want to find the magnitudes of the forces at the bottom of each dip, it means that at the bottom of the track there will be a force of gravity acting downwards and also a Normal force acting upwards.

Thus;

Force of gravity; F_g = Mg

Normal force = N

  • Also, because it is a circular motion, there will also be a centripetal force acting downwards as well.

Centripetal Force; F_c = Mv²/r

  • From equilibrium standpoint, we know that sum of upward forces is equal to sum of downward forces.

Thus;

N = Mg + Mv²/r

A) r = 60 m and v = 16 m/s

Thus; N = 9.8M + (16²/60)M

N₁ = 14.07M

B) r = 15 m and v = 8 m/s

Thus; N = 9.8M + (8²/15)M

N₂ = 14.07M

C) r = 30 m and v = 4 m/s

Thus; N = 9.8M + (4²/30)M

N₃ = 10.33M

D) r = 45 m and v = 4 m/s

Thus; N = 9.8M + (4²/45)M

N₄ = 10.16M

E) r = 30 m and v = 16 m/s

Thus; N = 9.8M + (16²/30)M

N₅ = 18.33M

F) r = 15 m and v = 12 m/s

Thus; N = 9.8M + (12²/15)M

N₆ = 19.4M

We want to rank the magnitude of the forces from largest to lowest. Thus;

N₆ > N₅ > N₁ = N₂ > N₃ > N₄

Read more at; https://brainly.com/question/24248463

snehashish65

The ranking for the forces acting at the bottom of dip is [tex]F_{6}>F_{5}>F_{1} =F_{2}>F_{3}>F_{4}[/tex].

The force acting at bottom of each dip is due to gravity force (downward direction) and Normal force (Upward) direction.

Then the net force at the bottom of dip is,

[tex]F=F_{c}+N\\F=\dfrac{mv^{2}}{r}+mg[/tex]

Here, g is the gravitational acceleration and m is the mass of coaster cart.

(A)

For r = 60 m and v = 16 m/s.

[tex]F_{1}=\dfrac{mv^{2}}{r}+mg\\F_{1}=\dfrac{m \times16^{2}}{60}+m(9.8)\\F_{1}=14.07m[/tex]

(B)

For r = 15 m and v = 8 m/s.

[tex]F_{2}=\dfrac{mv^{2}}{r}+mg\\F_{2}=\dfrac{m \times8^{2}}{15}+m(9.8)\\F_{2}=14.07m[/tex]

(C)

For r = 30 m and v = 4 m/s.

[tex]F_{3}=\dfrac{mv^{2}}{r}+mg\\F_{3}=\dfrac{m \times4^{2}}{30}+m(9.8)\\F_{3}=10.33m[/tex]

(D)

For r = 45 m and v = 4 m/s.

[tex]F_{4}=\dfrac{mv^{2}}{r}+mg\\F_{4}=\dfrac{m \times4^{2}}{45}+m(9.8)\\F_{4}=10.16m[/tex]

(E)

For r = 30 m and v = 16 m/s.

[tex]F_{5}=\dfrac{mv^{2}}{r}+mg\\F_{5}=\dfrac{m \times16^{2}}{30}+m(9.8)\\F_{5}=18.33m[/tex]

(F)

For r = 15 m and v = 12 m/s.

[tex]F_{6}=\dfrac{mv^{2}}{r}+mg\\F_{6}=\dfrac{m \times12^{2}}{15}+m(9.8)\\F_{6}=19.4m[/tex]

Thus we can conclude that the ranking for the forces acting at the bottom of dip is [tex]F_{6}>F_{5}>F_{1} =F_{2}>F_{3}>F_{4}[/tex].

Learn more about the centripetal force here:

https://brainly.com/question/14249440?referrer=searchResults

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