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3 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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3. If I run 150m in 30 seconds, what speed will I have been running at?

Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

you were running fast .....5m/s is a good speed

7 0

2 years ago

What is the transition from a gas to a liquid?

postnew [5]

The transition from gas to liquid is called condensation. An example would be water droplets forming on an ice cold glass placed in room temperature.

6 0

3 years ago

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago

How can Newton's third law describe the forces affecting a rocket as it

rusak2 [61]

Answer:D

Explanation:

Apex

7 0

3 years ago