Radiation and Convection transfer energy from the core to the photosphere. This is because energy from the core moves through a thick layer called the radiative zone. Also in the convection zone energy moves mainly by convection. Rising currents of hot gas in the convection zone carry energy toward the Sun's surface.
Answer:
B) 71.5 [km]
Explanation:
To solve this problem we will decompose each of the directions in the x & y axes.
To solve this problem we will decompose each of the directions in the x & y axes. also for a greater understanding of the angles, you should look at the attached image, which contains the orientations for each angle (clockwise or counterclockwise).
<u>59.0 km in a direction 30.0° east of north</u>
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![d_{1x}= 59*sin(30) = 29.5[km]\\d_{1y}= 59*cos(30) = 51.09[km]](https://tex.z-dn.net/?f=d_%7B1x%7D%3D%2059%2Asin%2830%29%20%3D%2029.5%5Bkm%5D%5C%5Cd_%7B1y%7D%3D%2059%2Acos%2830%29%20%3D%2051.09%5Bkm%5D)
<u>58.0 km due south</u>
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![d_{2y} = - 58 [km]\\](https://tex.z-dn.net/?f=d_%7B2y%7D%20%3D%20-%2058%20%5Bkm%5D%5C%5C)
<u>It flies 100 km 30.0° north of west</u>
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<u />![d_{3x}= - 100*cos(30) = -86.6[km]\\d_{3y} = 100*sin(30)= 50 [km]](https://tex.z-dn.net/?f=d_%7B3x%7D%3D%20-%20100%2Acos%2830%29%20%3D%20-86.6%5Bkm%5D%5C%5Cd_%7B3y%7D%20%3D%20100%2Asin%2830%29%3D%2050%20%5Bkm%5D)
<u />
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Now we sum algebraically the components
![d_{x}=29.5-86.6 = -57.1[km]\\d_{y}=51.09 -58+50=43.09[km]\\\\](https://tex.z-dn.net/?f=d_%7Bx%7D%3D29.5-86.6%20%3D%20-57.1%5Bkm%5D%5C%5Cd_%7By%7D%3D51.09%20-58%2B50%3D43.09%5Bkm%5D%5C%5C%5C%5C)
Using the Pythagorean theorem we can find the magnitude of the displacement.
![d = \sqrt{(57.1)^{2} +(43.09)^{2} } \\d= 71.53[km]](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%2857.1%29%5E%7B2%7D%20%2B%2843.09%29%5E%7B2%7D%20%7D%20%5C%5Cd%3D%2071.53%5Bkm%5D)
1. Over-hunting
2. A disappearance of a primary food source
<span>3. Drought</span>
Answer:
0.0327 m
Explanation:
m = 2 kg
ω = 24 rad/s
A = 0.040 m
Let at position y, the potential energy is twice the kinetic energy.
The potential energy is given by
U = 1/2 m x ω² x y²
The kinetic energy is given by
K = 1/2 m x ω² x (A² - y²)
Equate both the energies as according to the question
1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)
y² = 2 A² - 2 y²
3y² = 2A²
y² = 2/3 A²
y = 0.82 A = 0.82 x 0.040 = 0.0327 m
Wavelength - the distance from one wave crest or trough to another wave crest or trough. Amplitude - the distance from the median point or "middle" of the wave straight up to a crest (a maximum) or straight down to a trough (or minimum), which is the peak amplitude; or the distance from a trough straight up to a crest, or a crest straight down to a trough, called peak-to-peak amplitude.
