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AleksandrR [38]
3 years ago
13

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

s 13.2 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car. Group of answer choices 10.1 m/s 11.4 m/s 14.9 m/s 12.5 m/s
Physics
1 answer:
Mila [183]3 years ago
8 0

Answer:

11.4 m/s

Explanation:

The expression for the Centripetal acceleration is :

a=\frac{v^2}{R}

Where, a is the accleration

v is the velocity around circumference of circle

R is radius of circle

In the given question,

a = g = Acceleration due to gravity as the car is at top = 9.81\ m/s^2

v = ?

R = 13.2 m

So,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

<u>v = 11.4 m/s</u>

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What happens to the average kinetic energy of water molecules as water freezes? It decreases as the water absorbs energy from it
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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

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3 years ago
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