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jek_recluse [69]
2 years ago
15

A rollercoaster loop has a radius of 22.7 m. what is the minimum speed the coaster must have at the top of the loop to not fall

off the track? (unit = m/s)
Physics
1 answer:
Softa [21]2 years ago
4 0

Answer:

v = 14.92 m/s

Explanation:

First, make a <u>free body diagram</u> and see the <u>forces in the y-direction</u>.

  • ∑F_y = F_N - F_g

Use Newton's 2nd Law F = ma to replace ∑F_y with m * a_y.

  • m * a_y = F_N - F_g

The acceleration in the y-direction is the centripetal acceleration, a_c = v^2/r.

  • m * v^2/r = F_N - mg

The <u>normal force is 0</u> because this is where the rollercoaster is not falling off the track yet not touching the track.

  • m * v^2/r = - mg

The masses cancel out.

  • v^2/r = -g

<u>Solve for v</u> to find the speed of the rollercoaster at the top of the loop.

  • v^2 = -(-9.81) * r
  • v = √(9.81 * 22.7)
  • v = 14.9227

The minimum speed the coaster must have at the top of the loop to not fall off the track is <u>14.92 m/s</u>.

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A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s
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Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a =  5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down  are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

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