Answer :
Answer:
Explanation:
Given that,
Diameter of pipe
d = 20cm
Then, radius =d/2 =20/2
r = 10cm =0.1m
The speed at the bottom is
Vi = 3m/s
Speed at the top Vf?
At the bottom the cube is at a height of 0m
Then, y1 = 0m
At the top the cube is at a height which is the same as the diameter of the pipe
y2 = 0.2m
Now, let us consider, the energy conservation equation , which is the sum of kinetic energy and gravitational potential energy, given by,
K2 + U2 = K1 + U1
½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1
Divide all through by m
½•Vf² + g•y2 = ½•Vi² + g•y1
Since y1 = 0
So we have,
½•Vf² + g•y2 = ½•Vi²
½•Vf² = ½•Vi² — g•y2
Multiply through by 2
Vf² = Vi² —2g•y2
Vf = √(Vi²—2g•y2)
g is a constant =9.81m/s2
Vf = √(3²—2×9.81×0.2)
Vf = √(9—0.981)
Vf = √8.019
Vf = 2.83m/s
The speed of the ice cubes at the top of the pipe is 2.83m/s
The final velocity of the sugar cube at the top is 2.25 m/s.
The given parameters;
- initial position, h₁ = 20 cm = 0.2 m
- initial speed, v₁ = 3 m/s
The final velocity of the sugar cube at the top is calculated by applying the principle of conservation of energy;
[tex]mgh_1 + \frac{1}{2} mv_1^2 = mgh_2 + \frac{1}{2} mv_2^2\\\\gh_1 + \frac{1}{2} v_1^2 = gh_2 + \frac{1}{2} v_2^2\\\\\frac{1}{2} v_2^2 = gh_1 - gh_2 + \frac{1}{2} v_1^2\\\\v_2^2 = 2g(h_1 - h_2) + v_1^2\\\\v_2 = \sqrt{ 2g(h_1 - h_2) + v_1^2} \\\\v_2 = \sqrt{ 2\times 9.8(0-0.2) + 3^2} \\\\v_2 =2.25 \ m/s[/tex]
Thus, the final velocity of the sugar cube at the top is 2.25 m/s.
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