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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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Testing shows that a sample of wood from an artifact contains 25% of the

daser333 [38]

Answer:

The artifact is 11,460 years old.

<u>Explanation</u>:

Given that,

The half life of the carbon-14 is 5730 years and we are left with 255 of the sample of wood from an arti-fact.

So it takes 5730 years for the sample to reduce into half

Initially there will be 100% of the sample so

after first 5730 years, the sample reduces into 50% percent

Now the left 50% sample will take another 5730 years to decay into half of its amount.

after next 5730 years the sample reduces into 25% percent

So totally after 2 half-life the sample reduces into 25%

That is (5730 +5730) years = 11460 years

5 0

4 years ago

Read 2 more answers

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago

Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?

mixer [17]

Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2

For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg

Input the value of bullet in the formulae;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J

Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J

Which means that th Energy of the ball is more than the bullet.

7 0

4 years ago

What is the density to the object g/cm3

alexdok [17]

Should be 1.4, I hope this helps you out

6 0

3 years ago

Is it possible for surfers to use dynamite to produce water waves that are suitable for surfing

hammer [34]

if they had a suitable amount to cause an interruption in the waves so huge and vast that it makes waves..... it depends because you can have any amount and get different results any day though

hope this helps plz mark me brainliest

7 0

3 years ago