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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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If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?
UkoKoshka [18]
Ninguém sabe exatamente
4 0
2 years ago
Describe an example of newton’s 3rd law of motion
professor190 [17]

Answer:

For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

Explanation:

8 0
3 years ago
2.08
natka813 [3]

Calculate its average speed in meters per second

Answer:

5.77 m/s

Explanation:

Speed= Distance/Time

Distance= 40+ half of 40= 40+20= 60 m

Time= 8.8+1.6=10.4 s

Average speed= 60/10.4=5.769230769  m/s

Approximately, the average speed is 5.77 m/s

5 0
2 years ago
A cycle track is 500 metres long. Amy completes 10 laps. She travelled at an average speed of 12.5 metres per second. She puts o
yanalaym [24]
The fast lap is irrelevant to the question, because it didn't happen
until after the 9 laps that you're interested in.

To be perfectly technical about it, we don't actually have enough
information to answer the question.  You told us her average speed
for 10 laps, but we don't know anything about how her speed may
have changed during the whole 10 laps.  For all we know, maybe
she took a nap first, and then got up and drove 10 laps at the speed
of 125 metres per second.  That would produce the average speed
of 12.5 metres per second and we would never know it  Why not ?
That's only 280 miles per hour.  Bikes can do that, can't they ?

IF we can assume that Amy maintained a totally steady pace through
the entire 10 laps, then we could say that her average for 9 laps was
also 12.5 metres per second.
5 0
3 years ago
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