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Ad libitum [116K]
3 years ago
13

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

he roadway is frictionless?
Physics
1 answer:
kari74 [83]3 years ago
8 0

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

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

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0

Therefore, the road bank angle is 16.38⁰.

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a car traveling at 30m/s notices an accident about 65m up the road. The car's brakes are capable of decelerating at a rate of 6.
uysha [10]
Yes because you would have at least 3 car spaces
4 0
3 years ago
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
erica [24]

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0
3 years ago
a conducting rod whose length is 25 cm is placed on a u-shaped metal wire that has a resistance of 8.0 ω . the wire and the rod
andreyandreev [35.5K]

Answer:

The current is I  =  1 A

The direction is anti-clockwise

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

     the length of the conducting rod is L  = 25 \ cm =  \frac{25}{100}  =  2.5 \ m

     The resistance is  R = 8  \Omega

      The magnetic field is  B = 0.40\ T

        The speed of the rod is v = 6.0 m/s

The emf induced is

          \epsilon  = BLv

 substituting values we have

           \epsilon = 0.40 * 2.5 * 8

           \epsilon = 8V

From ohm law the induced current would be

      I  =  \frac{\epsilon}{R }

 substituting values we have

       I  =  \frac{8}{8 }      

       I  =  1 A

The direction anticlockwise this because according to lenze law the current due to change in magnetic field will act in the opposite direction of the  force causing the magnetic field to change

4 0
4 years ago
Estimate the average rate of change in elevation from d to points i, ii, and iii, assuming the distance from d to i is 3500 m, t
Anit [1.1K]
Ummm i am not going to be able say i am high

8 0
3 years ago
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
4 years ago
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