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Dimas [21]
3 years ago
7

A fluid flows along the x axis with a velocity given by V = ( x / t ) ˆ i , where x is in feet and t in seconds. (a) Plot the sp

eed for 0 ≤ x ≤ 10 ft and t = 3 s . (b) Plot the speed for x = 7 ft and 2 ≤ t ≤ 4 s . (c) Determine the local and convective acceleration. (d) Show that the acceleration of any fluid particle in the flow is zero. (e) Explain physically how the velocity of a particle in this unsteady flow remains constant throughout its motion.

Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

c)

 V_local = -x/t^2

 V_convec = x/t^2

d)

a =  V_local +  V_convec = 0

e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases

Explanation:

Hi!

You can see plots for a) and b) attached on this document

c)

The local acceleration is just teh aprtial derivative of the velocity with respect to t:

\frac{dV}{dt} = \frac{d}{dt} \frac{x}{t}=- \frac{x}{t^2}

And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:

\vec{v} \cdot \nabla \vec{v} = v ( \frac{dv}{dx} ) =\frac{x}{t} \frac{1}{t} = \frac{x}{t^2}

d)

Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign

e)

This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases,  the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero

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GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

3 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

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Answer:

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A. through a relatively short distance.

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A gallow

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