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kotykmax [81]
3 years ago
9

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)
If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

What torque does this force exert on the tree, measured about the point where the trunk meets the ground?
Physics
1 answer:
frez [133]3 years ago
3 0

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

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Answer:

6.4\cdot 10^{-15} N

Explanation:

The electric force exerted on the electron is given by:

F=qE

where

q=1.6\cdot 10^{-19}C is the magnitude of the electron charge

E = 40000 V/m is the electric field

Substituting,

F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N

By comparison, the gravitational force exerted on the electron is:

F=mg

where

m=9.10939\cdot 10^{-31} kg is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N

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Differences between work against friction and work against gravity.
Goshia [24]

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Calculate the speed of a bus that travels a distance of 55 miles in 0.75 hours.
Sonja [21]
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Florida law applies the same protections given to pedestrians to any person riding or leading an animal upon a roadway
dmitriy555 [2]

Answer:

True

Explanation:

The Florida law states that a person driving an animal-drawn vehicle has the same protections that are applicable to a person driving a vehicle and the protections that are applicable to pedestrians also apply to a person riding or leading an animal upon a roadway or the shoulder of it. According to this, the answer is that the statement is true.

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The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4
scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

        v = v₀ + a t

        y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

  • Positive slope the body is accelerating and the speed is increasing.
  • Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b) The acceleration is the first 8 s

          v = v₀ + a t

          a = \frac{v-v_o}{\Delta t}  

          a = \frac{2-0}{8-0}  

          a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

          a = \frac{6-2}{ 14-10}  

          a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

         d_{total } = d_1 +d_2 + d_3 +d_4  

We look for every displacement.

       d₁ = v₀ + ½ a₁ Δt²

       d₁ = 0 + ½ 0.25 8²

       d₁ = 8 m

In the second section the velocity is constant

         d₂ = v₂ Δt₂

         d₂ = 2 (10-8)

         d₂ = 4 m

The third section.

    d₃ = v₀ + ½ a t²

    d₃ = 2 + ½ 1 (14-10) ²

    d₃ = 10 m

The distance of the fourth section.

       

we look for acceleration

          a₄ = \frac{v-v_o}{\Delta t}  

          a₄ = \frac{0-6}{18-14}  

          a₄ = -1.5 m / s²

     

          d₄ = 6 + ½ (-1.5) (1814) ²

          d₄ = -6 m

The total displacement is;

          d_{total} = 8 + 4 + 10 -6

          d_{total} = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

          v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

  a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0
2 years ago
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