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4 years ago

Describe how humans’ actions and natural processes have modified coastal regions in Louisiana and other locations

1 answer:

3 0

Coastal erosion has depleted a large portion of South Louisiana's wetlands along the coastline in swamps and marshes mainly due to storm surges. But other factors also contributed to this erosion. Canals and waterways dug through the marshes and swamps for the oil industry is one factor. Man-made levees erected to provide protection to residents living adjacent to the river is another major cause. Large scale logging especially in the early 1900's also damaged the wetlands.

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The pulse site located at the point where the upper leg bends is called the

VLD [36.1K]

The pulse site located at the point where the upper leg bends is called the femoral. It is an artery found in the thigh. It is large and is deemed as the main arterial supply for the lower part of the body. It is known as the second artery that is the largest. It is being used as the catheter access artery. From it, diagnostics for the heart, brain, arms, kidney and other parts can be directed to the other arterial system. It can also be used as a source to draw blood that is from the arteries when there is low blood pressure.

3 0

3 years ago

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

5 0

4 years ago

A coin is made of ______

lys-0071 [83]

Answer:

coins made with the combination of copper zinc and Nickel

A cloth is made of fabrics

A towel is made from cotton

an umbrella is made up of fiberglass

a sweater is made up of wool

hope it helps

please mark me as the brainliest

5 0

3 years ago

A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou

prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

$tan\theta =\frac{v^{2}}{rg}$

$tan\theta =\frac{18.61^{2}}{53.1\times 9.8}$

tan θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

6 0

3 years ago

An airplane has an effective wing surface area of 17.0 m2 that is generating the lift force. In level flight the air speed over

olga_2 [115]

To solve this problem it is necessary to apply the equations given from Bernoulli's principle, which describes the behavior of a liquid moving along a streamline. Mathematically this expression can be given as,

$P_1 + \frac{1}{2}\rho*v_1^2 + P_2 + \frac{1}{2}*\rho*v_2^2=0$

Where,

$P_i =$ Pressure at each state

$\rho$ = Density

$v_i =$ Velocity

Re-organizing the expression we can get that

$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Our values are given as

$v_1 = 40m/s$

$v_2 = 55m/s$

$\rho_{water} = 1.2kg/m^3 \rightarrow$ Normal Conditions

Replacing we have,

$P_1 -P_2 = \frac{1}{2}*1.2*(55^2-40^2)$

$P_1 - P_2 = 855Pa$

If we consider that there is a balance between the two states, the Force provided by gravity is equivalent to the Support Force, therefore

$F_l = F_g$

Here the lift force is the product between the pressure difference previously found by the effective area of the aircraft, while the Force of gravity represents the weight. There,

$F_g = W$

$F_l = (P_2-P_1)A$

Equating,

$(P_1 - P_2)*A = W$

$W = 855*17$

$W = 14535 N$

Therefore the weight of the plane is 14535N

3 0

3 years ago