<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
Answer:
It is direct proportionality. The greater the mass, the greater is the gravitational potential energy. The equation for GPE is : GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground. As you can see GPE is directly proportional to mass, and height. KT.
Explanation:
Gravitational potential energy is a function of both the mass of your system and the mass of the thing generating the gravity field around your system.
The relationship is linear, which means that if you multiply or divide one of the masses by some number but leave everything else the same, you multiply or divide the potential energy by the same number. A 3kg mass has three times the gravitation potential energy of a 1kg mass, if placed in the same location.
B. Refraction because objects refract in water
Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,
the diameter decreases 86% so
Thus, speed in smaller section is 48.6 m/s
