The answer is:
Transverse dune
The explanation:
Transverse dune : is abundant barchan dunes It may merge into barchanoid ridges, which then grade into linear .
The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.
It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.
Transverse dunes have a gently sloping windward side and a steeply sloping leeward side.
They general form in areas of sparse vegetation and abundant sand are transverse dunes.
Answer:
The answer is the 3rd option!
C. element only one substance
Answer:
Er = 231.76 V/m, 27.23° to the left of E1
Explanation:
To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:
E1:
![E_1_x = 0V/m\\E_1_y=100V/m](https://tex.z-dn.net/?f=E_1_x%20%3D%200V%2Fm%5C%5CE_1_y%3D100V%2Fm)
E2:
Keep in mind that the x component of electric field E2 is directed to the left.
![E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m](https://tex.z-dn.net/?f=E_2_x%3D%20150V%2Fm%2A-sin%2845%29%20%3D%20106.07%20V%2Fm%5C%5CE_2_y%3D150V%2Fm%2Acos%2845%29%20%3D%20106.07V%2Fm)
∑x: ![E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m](https://tex.z-dn.net/?f=E_1_x%2BE_2_x%20%3D%200V%2Fm%20-%20106.07V%2Fm%20%3D%20-106.07V%2Fm)
∑y: ![E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m](https://tex.z-dn.net/?f=E_1_y%20%2B%20E_2_y%20%3D%20100V%2Fm%20%2B%20106.07V%2Fm%20%3D%20206.07V%2Fm)
The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.
or 27.23° to the left of E1.