Answer:
66.35m/s
Explanation:
Para resolver el ejercicio es necesario la aplicación de las ecuaciones de continuidad, que expresan que
From our given data we can lower than:
So using the continuity equation we have
Therefore the velocity at the exit end is 66.35m/s
I think answer is A. The Brightness of a Star
Explanation:
Given:
Final speed of mass A = Va
Final speed of mass B = Vb
Mass of A = Ma
Mass of B = Mb
Ma = 2 × Mb
By conservation of linear momentum,
0 = Ma × Va + Mb × Vb
0 = 2 × Mb × Va + Mb × Vb
Vb = - 2 × Va
Energy of the spring, U = 1/2 × k × x^2
1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
Ma × Va² + Mb × Vb² = 70
2 × Mb (-Vb/2)² + Mb × Vb² = 70
1/2 × Mb × Vb² + Mb × Vb² = 70
3/2 × Mb × Vb² = 70
Mb × Vb² = 140/3
= 46.7 J
Ma = 2 × Mb and Vb = - 2 × Va
Ma/2 × (4 × Va²) = 140/3
Ma × Va² = 70/3
Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J
Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J
Correct Answer is B.
In a parallel circuit, the voltage is same across all the branches however the current in each branch is different and depends on the resistance of that branch.
Points to Remember:
1) In series circuit current remains the same and voltage varies
2) In parallel circuit voltage remains the same and current varies
