The gravitational potential energy:
E p = m x g x h
where m is the mass and h is the height;
m = 50 kg, g = 9.81 m/s² , h = 40 m
E p = 50 kg x 9.81 m/s² x 40 m
Answer:
E p = 19,620 J = 19.62 kJ
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:
Explanation:
We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ
Applying law of conservation of momentum along direction of original motion
m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄
0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ
v cos θ = .8
Applying law of conservation of momentum along direction perpendicular to direction of original motion
1.03 sin 43 x .143 = .132 x v sinθ
v sinθ = .76
squaring and adding
v² = .76 ² + .8²
v = 1.1 m /s
Tan θ = .76 / .8
θ = 44°
