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:
5.82812 rad/s
Explanation:
L = Length of meter stick = 1 m = 100 cm
= The center of mass of the stick = 
= Angular velocity
Moment of inertia of the system is given by

As the energy in the system is conserved

The maximum angular velocity is 5.82812 rad/s
I dont know from option
Because SI Unit of acceleration is m/s^2
Figure B. The second option.