THE GREEN HOSE: Define the (x,y) coordinates at a height of 4 feet, which corresponds to where Majra holds the green hose. This indicates the equation for the green hose takes the form y = a(x - h)² + 4. Water from the hose lands on the ground 10 feet away from Majra, thus y(10) = -4. Given that the curve passes through (0,0), this leads to ah² + 4 = 0; therefore, ah² = -4. To satisfy the previous equation, we find a(10 - h)² + 4 = -4, simplifying to a(10 - h)² = -8. Dividing (3) by (4) gives a ratio of h²/(10-h)² = 1/2, leading to 2h² = (10 - h)² = 100 - 20h + h², and resolving yields h² + 20h - 100 = 0. Applying the quadratic formula, we get x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142. We discard the negative solution. The vertex locates at (4.142, 4). From (3), we deduce a = -4/4.142² = -0.2332, leading to the equation for the green hose: y = 0.2332(x - 4.142)² + 4. THE RED HOSE: The vertex of the red hose is positioned at (3,7), represented by the equation y = -(x-3)² + 7. A graph depicting y(x) for both hoses is included in the attached figure. Answers: a. The red hose throws water higher. b. The green hose's equation is y = -0.2332(x - 4.124)² + 4, starting at a height of 4 feet. c. The feasible domain for the green hose is between 0 ≤ x ≤ 10 feet, with the corresponding range being -4 ≤ y ≤ 4 feet.
Thermal Power is approximately 460W. According to the Stephan-Boltzmann Law Formula: P = єσT⁴A, where: P = radiation energy, σ = Stefan-Boltzmann Constant, T = absolute temperature in Kelvin, є = emissivity of the material, and A = the surface area. Given that σ = 5.67 x 10^(-8), ε = 0.6, and T = 30°C which converts to Kelvin as 303K, with the human body dimensions of 2m length and 0.8m circumference leading to an area of 1.6m², so thermal power equals 0.6 x 5.67 x 10^(-8) x 303⁴ x 1.6 = 458.8W. Rounding gives about 460W.
Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
The velocity at the stagnation point is zero.
The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Thus, the answer is 2.2 psi
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
The moment of inertia for one blade
= 1/3 x m l²
where m denotes the mass of the blade
l represents the length of each blade.
Substituting the necessary values, the moment of inertia for one blade is
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Total moment of inertia for 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) The angular momentum 'L' is calculated using
L = x ω
where,
= the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²
ω= angular velocity =2π f
f represents the frequency of rotation of the blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolutions per second
ω = 2π f => 2π x 13 / 60 rad / s
L= x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹
