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krek1111 [17]
9 months ago
9

a 1500 kg vehicle is traveling on a curved, icy road. the road is banked at an angle of 10.0 degrees and has a radius of curvatu

re of 400 m. the velocity of the car necessary to travel on the icy road without sliding is .
Physics
1 answer:
Monica [59]9 months ago
5 0

The vehicle speed required to drive on an icy road without sliding is 28.3 m/s.

The weight of the car is m = 1500 kg

The angle at which the road is inclined is θ = 10

The radius of curvature is r = 400m

The expression for the speed of the car required to travel on the road without sliding is V =\sqrt{rgtan}

V =\sqrt{400*9.8*tan10}

V = 28.3 m/s

Velocity is the rate of change in direction of an object in motion measured by a specific time standard and observed from a specific reference point (for example, 60 km/h north). A key idea in kinematics, the branch of classical mechanics that studies the motion of bodies, is velocity.

The definition of velocity requires both its magnitude and its direction, since it is a physical vector quantity. Velocity is a coherently derived unit that is measured in the SI (metric system) as meters per second (m/s or m/s1). Velocity is a scalar absolute value (magnitude) of speed.

Learn more about Velocity here:

brainly.com/question/18084516

#SPJ4

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Answer:

a

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b

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Explanation:

From the question we are told that  

   The  wavelength of the light is \lambda  = 550 \ nm  =  550 *10^{-9} \ m

    The  distance of the slit separation is  d = 0.500 \ mm = 5.0 *10^{-4} \ m

 

Generally the condition for two slit interference  is  

     dsin \theta =  m \lambda

Where m is the order which is given from the question as  m = 2

=>    \theta  =  sin ^{-1} [\frac{m \lambda}{d} ]

 substituting values  

      \theta  =  0.0022 rad

Now on the second question  

   The distance of separation of the slit is  

       d =  0.300 \ mm  =  3.0 *10^{-4} \ m

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      I  =  I_o  [\frac{sin \beta}{\beta} ]^2

Where  \beta is mathematically evaluated as

       \beta  =  \frac{\pi *  d  *  sin(\theta )}{\lambda }

  substituting values

     \beta  =  \frac{3.142  *  3*10^{-4}  *  sin(0.0022 )}{550 *10^{-9} }

    \beta  = 0.06581

So the intensity is  

    I  =  I_o  [\frac{sin (0.06581)}{0.06581} ]^2

   I  =  0.000304 I_o

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
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Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

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for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

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where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

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