Answer:
Radiative transport
Explanation:
Radiative transportation takes place extensively in the core of the sun. It produces solar energy which are emitted outward from the core of the sun, and it comprises more than 70% of the sun's heat energy. At the core of the sun, the process of nuclear fusion takes where the hydrogen atoms combines with one another, and forms helium, and during this process, it releases a large amount of energy. The photons continuously interacts with the ions (or atoms) and releases this energy to the particles and the excited particles converts the excess amount of energy into another photon which are carried outward by the process of radiative transport.
The angle of refraction is 15.6 deg.
When a ray of light passes from medium 1 to medium 2, the refractive index of medium 2 with respect to 1 is the ratio of the absolute refractive index of medium 2 and medium 1
When light passes from air into diamond, with the given values of refractive indices,
According to Snell's law,
The angle of refraction is 15.6 deg
Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
I think its B because it never has a the turn over to any other side except for the west.
Answer:
charge Qint = 7.17 10⁻⁴ C
Explanation:
For this problem we must use Gauss's law
F = ∫ E. dA = Qint / εₙ
let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.
Φ = E A = Qint / ε₀
A = 1 km² (1000 m / 1km)² = 1 10⁶ m²
We can calculate the charge
Qint = E A ε₀
Qint = 81 1 10⁶ 8.85 10⁻¹²
Qint = 7.17 10⁻⁴ C
