Answer:
d_{b} = 2 d_{a}
Explanation:
The electrical resistance for a cylindrical wire is described by the expression
R = ρ L / A
The area of a circle is
A = π r²
r = d / 2
A = π d²/4
We substitute
R = ρ L 4 /π d²
Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B
= 4 R_{b}
We substitute
ρ 4/π 
² = 4 (ρ 4/π d_{b}²)
1 / d_{a}² = 4 / d_{b}²
d_{a} = d_{b} / 2
Answer:
The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)
Explanation:
The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.
a) Outer semi-sphere:
A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²
b) Inner semi-sphere:
A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²
c) Edge (Ring):
A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²
Therefore, the total surface area of the bowl is given by:
A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)
Changing units to m², as required in the problem, we get:
A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)
It wouldn't indicate the same source if the paint was from a different place. If the paint was the same brand and color it would be but if not then no
I think it's C, longer wave length.
2 because that’s correct I I I I I I I I I I I I I I I I I I I
