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Lady_Fox [76]
3 years ago
14

A race car drives around a turn on a flat race track at constant speed. Identify the force that allows the race car to travel ar

ound the turn?
A)
Air resistance applies the force needed.


B)
No force is required to go around the turn.


C)
The weight of the car applies the force needed.


D)
The friction between the tires and the track applies the force needed
Physics
1 answer:
lidiya [134]3 years ago
7 0
D) The friction between the tires and the track applies the force needed.
Without the friction, the car wouldn't handle the curve and would fly off the track.
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Answer:

The actual elevation angle is 12.87 degrees

Explanation:

In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that \theta_2=90-43=47.

Using Snell's Law we can write:

n_1\sin(\theta_1)=n_2\sin(\theta_2)

\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2),

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)

\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13. Calling \alpha the actual angle of elevation, we get from the picture that \alpha=90-77.13=12.97

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