It would float outward as all space is expanding from the center constantly. of by some chance it for caught in the orbit of another celestial body with huge mass it may orbit that, bur other wise it would just push outward from the center of the universe theoretically the big bang
Answer: 9cm
Explanation:
Refractive index can also be defined as the ratio of the real depth to the apparent depth.
Given that the
Real depth = 12 m
Refractive index of water = 1.33
Refractive index of air = 1.00
nair/nwater = real depth/apparent depth
Substitute all the parameters into the formula
1.33/1 = 12/ apparent depth
Cross multiply
1.33 Apparent depth = 12
Apparent depth = 12/1.33
Apparent depth = 9.02 cm
Therefore, A submerged swimmer in the pool sees the tip of the triangle at 9cm approximately distance above the water.
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)
Answer:
step bro was stuck on the elevator
Explanation:
