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melamori03 [73]
3 years ago
15

A curve of radius 55.1 m is banked so that a car of mass 1.6 Mg traveling with uniform speed 61 km/hr can round the curve withou

t relying on friction to keep it from slipping on the surface. 1.6 Mg µ ≈ 0 θ At what angle is the curve banked? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of deg.
Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

θ = 28°

Explanation:

For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.

X axis

      Fₓ = m a

      Nₓ = m a

Where the acceleration is centripetal

      a = v² / r

The only force that we must decompose is normal, let's use trigonometry

      sin θ = Nₓ / N

      cos θ = N_{y} / N

      Nₓ = N sin θ

      N_{y} = N cos θ

Let's replace

     N sin θ = m v² / r

Y Axis

     N_{y} - W = 0

     N cos θ = mg

Let's divide the two equations of Newton's second law

     Sin θ / cos θ = v² / g r

     tan θ = v² / g r

     θ = tan⁻¹ (v² / g r)

We reduce the speed to the SI system

      v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s

Let's calculate

     θ = tan⁻¹ (16.94 2 / (9.8 55.1)

     θ = tan⁻¹ (0.5317)

     θ = 28°

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Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

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    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

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A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe
Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

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We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

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Now, substituting the given information we have:

alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

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Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

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Rounding to the next integrer: 3.

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