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Radda [10]
3 years ago
10

A 72.9 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of

1.21 m/s in 0.80 s. It travels with this constant speed for 4.97 s, undergoes a uniform downward acceleration for 1.52 s, and comes to rest. What does the spring scale register (a) before the elevator starts to move?
Physics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

(a) 72.9 kg

Explanation:

Before the elevator starts to move, only gravitational force exerts on the man, this force is generated by the man mass and the gravitational acceleration, which in turn register in the scale. So the scale would probably indicate the man mass, which is 72.9 kg.

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A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green. the te
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<u>Answer:</u>

a) Time spend by ball in air = 4.368 seconds

b)   Longest hole that golfer can make = 93.59 meter

<u>Explanation:</u>

  Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

a) We have golf ball travels maximum distance, so range is maximum.

                 Maximum range is when, sin 2θ =1

                             =>  θ = 45⁰

    Now we have travel time of projectile, t =  2u sin θ/g  

          Initial velocity = 30.3 m/s and  θ = 45⁰

                     So time spend in air, t = \frac{2*30.3*sin45}{9.81} =4.368 seconds

 b) Longest hole that golfer can make = Range of projectile = \frac{u^2sin2\theta}{g}

      Longest hole that golfer can make = \frac{30.3^2sin(2*45)}{9.81}=93.59 meter

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P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the
timofeeve [1]

Answer:

a

The total friction drag for the long side of the plate is 107 N

b

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image  

Where the U is the initial velocity = 6 m/s

    So the equation we will be working with is

               F = \frac{1}{2} \rho C_fAU^2

    Where \rho is the density of SAE 10W = 870\ kg/m^3 This is obtained from the table of density at 20° C

                C_f is the friction drag coefficient

   This coefficient is dependent on the Reynolds number if the Reynolds number is less than 5*10^5 then the flow is of laminar type and

          C_f  = \frac{1.328}{\sqrt{Re} }

But if the Reynolds number is greater than 5*10^5 the flow would be of Turbulent type and

         C_f = \frac{0.074}{Re_E^{0.2}}

Where Re is the Reynolds number

   To obtain the  Reynolds number  

                                      Re = \frac{\rho UL}{\mu}

          where L is the length of the long side = 110 cm = 1.1 m

 and \mu is the Dynamic viscosity of SAE 10W oil = 1.04*10^{-1} kg /m.s

  This is gotten from the table of Dynamic viscosity of oil

  So        

                    Re = \frac{870 *6*1.1}{1.04*10^{-1}}

                          = 55211.54

Since            55211.54 < 5.0*10^5

Hence

                    C_f = \frac{1.328}{\sqrt{55211.54} }

                          = 0.00565

                 A is the area of the plate  = \frac{ (110cm)(55cm)}{10000} =0.55m^2

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

                F = \frac{1}{2} \rho C_f(2A)U^2

                F  =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2

                 F = 107N

Considering the short side

            To obtain the Reynolds number

                      Re = \frac{\rho U  b}{\mu}

Here b is the short side

                        Re =\frac{870*6*0,55}{1.04*10^{-1}}

                              =27606

Since the value obtained is not greater than 5*10^5 then the flow is laminar

   And

              C_f = \frac{1.328}{\sqrt{Re} }

                    = \frac{1.328}{\sqrt{27606} }

                   = 0.00799

The next thing to do is to obtain the total friction drag

             F = \frac{1}{2} \rho C_f(2A)U^2

      Substituting values

           F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2

                = 151.4 N

4 0
3 years ago
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