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Gemiola [76]
4 years ago
13

The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h. (a) If the train goes

around a curve at that speed and the acceleration experienced by the passengers is to be limited to 0.060g, what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a 1.80 km radius to be at the acceleration limit?
Physics
1 answer:
faust18 [17]4 years ago
6 0

Answer:

a)  r = 6122 m and b) v = 32.5 m / s

Explanation:

a) The train in the curve is subject to centripetal acceleration

         a = v2 / r

Where v is The speed and r the radius of the curve

They indicate that the maximum acceleration of the person is 0.060g,

        a = 0.060 g

        a = 0.060 9.8

        a = 0.588 m /s²

Let's calculate the radius

        v = 216 km / h (1000m / 1km) (1 h / 3600 s =

        v = 60 m / s

        r = v² / a

        r = 60² /0.588

        r = 6122 m

b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m

        v = √a r

        v = √( 0.588 1800)

        v = 32.5 m / s

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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s
Nezavi [6.7K]

Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:  

x-component:  

x=V_{o}cos\theta t   (1)  

Where:  

V_{o}=2.80 m/s is the initial speed  

\theta=39\° is the angle at which the venom was shot

t is the time since the venom is shot until it hits the ground  

y-component:  

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)  

Where:  

y_{o}=0.4 m  is the initial height of the venom

y=0  is the final height of the venom (when it finally hits the ground)  

g=-9.8m/s^{2}  is the acceleration due gravity

Knowing this, let's begin:

First we have to find t from (2):

0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}   (3)

Rearranging (3):  

-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0   (4)  

This is a <u>quadratic equation</u> (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:  

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)  

Where:  

a=-4.9  

b=1.762  

c=0.4  

Substituting the known values:  

t=\frac{-1.762 \pm \sqrt{(1.762)^{2} - 4(-4.9)(0.4)}}{2(-4.9)} (6)  

Solving (6) we find the positive result is:  

t=0.517 s (7)

Substituting (7) in (1):

x=2.8 m/s cos(39\°) (0.517 s)   (8)

Finally:

x=1.124 m   (9)

4 0
4 years ago
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