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3 years ago

Which method should be used to determine which type of natural event produces the greatest number of sand dunes?

Physics

1 answer:

jeka943 years ago

5 0

Answer:

Stabilizing dunes involves multiple actions. Planting vegetation reduces the impact of wind and water. Wooden sand fences can help retain sand and other material needed for a healthy sand dune ecosystem. Footpaths protect dunes from damage from foot traffic.

Explanation:

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A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

zhannawk [14.2K]

Answer:

$y = 17.04 ft$

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

$y = 180 ft$

horizontal distance of the tower from mid point is given as

$x = \frac{1300}{2} = 650 ft$

now from the equation of parabola we have

$y = k x^2$

$180 = k(650^2)$

$k = 4.26 \times 10^{-4}$

now we have

$y = (4.26 \times 10^{-4})x^2$

now we need to find the height at distance of 200 ft from center

so we have

$y = (4.26 \times 10^{-4})(200^2)$

$y = 17.04 ft$

8 0

2 years ago

Aluminum oxide can be produced during rocket launches. Show that the sum of positive and negative charges in a unit of Al2O3 equ

Triss [41]

Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.

Al203 has 2 Al atoms and 3 O atoms.

Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)

= 2(3) + 3(-2)

= 6 - 6

= 0

Explanation:

Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.

Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.

4 0

3 years ago

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

Ksju [112]

g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:

R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s

the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)

6 0

2 years ago

Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to

boyakko [2]

B. Elastic potential to kinetic energy

The elastic potential energy in the slingshot will be transferred to the stone as kinetic energy as the stone is launched.

6 0

2 years ago

Read 2 more answers

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

8 0

3 years ago