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melamori03 [73]
2 years ago
14

An object which starts at the origin is traveling in the positive direction with a constant velocity V. After traveling a distan

ce D with this constant velocity, the object begins to slow down at a constant rate until it stops after traveling an addition distance of 2D. In terms of D and V, determine the average velocity of the entire trip.
Physics
2 answers:
zepelin [54]2 years ago
7 0

With the concept of kinematics we find the average speed is 0.6V

given parameters

   * initial velocity V

   * the distances in each part of the movements x₁ = D and x₂ = 2D

to find

  The average speed of the entire trajectory

In kinematics we analyze the motion of bodies, the average velocity is defined by

        V_{avg} = \frac{\Delta x}{\Delta t}           1

where V_{avg} is the average velocity, Δx and Δt are the variation of displacement and time in the interval

For this exercise we have two types of movement in the first part a uniform movement and in the second part an accelerated movement  let's solve each onaccelerated movement,e separately

1 Part. uniform motion

we look for the time

          V = \frac{x_1}{t_1}

          t₁ = x₁ / V

          t₁ = \frac{D}{V}

2 Part. Accelerated movement, let's start by looking for acceleration, they tell us that the body stops at the end of the interval so its velocity is zero v_f=0, the initial velocity is v₀ = V

           v_f^2 = v_o^2 - 2 \ a \ x_2

           0 = v₀² - 2 a x₂

           a = \frac{v_o^2}{2x_2}

we substitute

           a = \frac{V^2}{D}

now we look for the time it takes to stop

          v = v₀ - a t₂

          0 = v₀ - a t₂

          t₂ = \frac{v_o}{a}

we substitute

          t₂ = \frac{V}{\frac{V^2}{4D} }

          t₂ = \frac{4D}{V}4D / V

for the entire movement, the total displacement is

          Δx = x₁ + x₂

          Δx = D + 2D

          Δx = 3D

the total time is  

          Δt = t₁ + t₂

          Δt = \frac{D}{V} + \frac{4D}{V}

          Δt = \frac{5D}{V}

we substitute in equation 1 of average velocity

         V_{avg} = \frac{3D}{\frac{5D}{V} }

         V_{avg} = \frac{3}{5} \ V

         V_{avg } = 0.6 V

With the use of the kinematics equations with constant acceleration we can find the average velocity of the body throughout the journey is 0.6V

                                                               

learn more about average speed here:

brainly.com/question/11265533

34kurt2 years ago
3 0

Answer:

Explanation:

time to travel first leg

D = Vt₁

t₁ = D/V

time to travel second leg.

at constant deceleration, the average velocity will be half of the original.

2D = (V/2)t₂

t₂ = 4D/V

t = t₁ + t₂ = 5D/V

Vavg = d/t = (D + 2D)/(5D/V) = (3D/5D)V = 0.6V

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Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

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Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
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Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
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  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

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r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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The answer is the red sidelight on a powerboat should be visible from the front and from the left (port side).

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  • All-Round white light: This light is the major light that is used to join the masthead light and the stern light. This single light is visible to all vessels from all directions.
  • Thus, as a rule for a boat rider, he should show the vision of red light and it should be visible from the front and from the left (port side).

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