How is a good slide into a base performed?

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

il63 [147K]

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.

3 0

3 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

2 years ago

The phrase terminal velocity can best be defined as

Anna007 [38]

None of these are a good definition; a good definition would be "the maximum velocity that an object can fall at." however the best answer out go those is c. the constant velocity of some falling objects.

8 0

2 years ago

Read 2 more answers

A factory worker pushes a 32.0 kgkg crate a distance of 4.0 mm along a level floor at constant velocity by pushing horizontally

Luda [366]

Answer:

a) The force that the worker must apply has a magnitude of 75.317 newtons.

b) The external force does a work of 0.301 joules.

c) The friction force does a work of -0.301 joules.

d) Both normal force and gravity have done a work of 0 joules.

e) The total work done on the crate is 0 joules.

Explanation:

a) As the crate is moving at constant velocity, we know that magnitude of the force done on the crate must be equal to the friction force. Hence, we must use the following formula:

$F = \mu\cdot m\cdot g$ (1)

Where:

$F$ - External force, in newtons.

$\mu$ - Coefficient of kinetic friction, no unit.

$m$ - Mass, in kilograms.

$g$ - Gravity acceleration, in meters per square second.

If we know that $\mu = 0.24$ , $m= 32\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the external force is:

$F = (0.24)\cdot (32\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 75.317\,N$

The force that the worker must apply has a magnitude of 75.317 newtons.

b) The direction of the force is parallel to the direction of motion. The work done by this force ( $W_{F}$ ), in joules, is determined by this formula:

$W_{F} = F\cdot \Delta s$ (2)

Where $\Delta s$ is the distance travelled by the crate, in meters.

If we know that $F = 75.317\,N$ and $\Delta s = 0.004\,m$ , then the work done by the force is:

$W_{F} = (75.317\,N)\cdot (0.004\,m)$

$W_{F} = 0.301\,J$

The external force does a work of 0.301 joules.

c) The direction of the friction force is antiparallel to the direction of motion. The work done by this force ( $W_{f}$ ), in joules, is determined by this formula:

$W_{f} = -W_{F}$ (3)

$W_{f} = -0.301\,J$

The friction force does a work of -0.301 joules.

d) The direction of the normal force is perpendicular to the direction of motion. Therefore, no work is done due to normal force.

$W_{N} = 0\,J$

Likewise, no work is done by gravity.

$W_{g} = 0\,J$

Both normal force and gravity have done a work of 0 joules.

e) The total work is the sum of the works done by the external force and the friction force:

$W = W_{F}+W_{f}$

$W = 0.301\,J - 0.301\,J$

$W = 0\,J$

The total work done on the crate is 0 joules.

5 0

2 years ago

Where does the potential energy of rocket goes when it escape from the influence of the earth

Nonamiya [84]

The potential energy of a rocket increases when its escapes from the influence of the earth. It increases when it escapes the earth's boundary.

<h3>What is escape speed?</h3>

Escape velocity or escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.

It is commonly expressed as an ideal speed, neglecting atmospheric friction.

If a spacecraft is launched from the moon at the escape speed of the earth then the value of escape speed will be the difference in both the velocities.

An object's height above the zero position directly relates to its gravitational potential energy.

The kinetic energy of motion is converted into gravitational potential power as your rocket ascends.

The rocket possesses gravitational potential energy as it climbs because it is becoming farther away from the Earth's surface, much as a positive and negative charge moving apart.

The potential energy of a rocket increases when its escapes from the influence of the earth. It increases when it escapes the earth's boundary.

To learn more about the escape speed refer to the link;

brainly.com/question/14178880

#SPJ1

6 0

1 year ago