Answer:
v = 54.2 m/s
Explanation:
We can utilize conservation of energy to solve this issue.
Initial condition Higher
Em₀ = U = m g h
Final condition. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ (2gh)
Now let's perform the calculation
v = √ (2 * 9.8 * 150)
v = 54.2 m/s
Kindly use English in your post so that I or someone else can assist you.
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.
Kinetic energy refers to the energy an object possesses while in motion, whereas thermal energy corresponds to heat energy. In situations where heat increases in materials, such as a solid turning into a liquid, the molecules start to move more rapidly, resulting in a rise in kinetic energy.
