Because one side of the spoon is convex and the other side is concave, the spoon’s different sides reflect differently.
Red is the lowest because it has the shortest wavelengths
Answer: h = 0.52m
Explanation:
Using the equation of out flow;
A1 × V1 = A2 ×V2
Where A1 = area of the first nozzle
A2 = area of the second nozzle
V1= velocity of flow out from the first nozzle
V2 = velocity of flow out from 2nd nozzle
But AV= area of nozzle × velocity of water = volume of water per second(m³/s).
Now we can set A×V = Area of nozzle × height of rise.
Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)
D1 = 20mm= 0.02m; h1 = 0.13m
D2 = 10mm = 0.01m; h2= ?
h2 = π(D1/2)²× h1 /π(D2/2)²
h2 = (0.02/2)² × 0.13/(0.01/2)²
= (0.01)² ×0.13 /(0.005)²
= 1.3 × 10^-5/(5 × 10^-3)²
= 1.3 × 10^-5/25 × 10^-6
= (1.3/25) 10^-5 × 10^6
= 0.052× 10
= 0.52m
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
