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Juliette [100K]
3 years ago
5

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

or compressed. The ball is then released from rest and it falls to a height y - h before moving upward. Consider the following quantities: translational kinetic energy, gravitational potential energy, elastic potential energy. When the ball was at a height y - (h/2), which of the listed quantities has (have) values other than zero joules
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

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When switched on, the grinding machine accelerates from rest to its operating speed of 3550 rev/min in 10 seconds. When switched
Paraphin [41]

Answer:

During start total turns

N_1 = 296 turn

After half of the time total turns

N_3 = \frac{29.6 + 0}{2}(5) = 74 turns

Total number of turns during it stop

N_2 = \frac{59.17 + 0}{2}(31) = 917 turn

After half of the time total turns

N_4 = 688 turns

Explanation:

Initially the machine is at rest and then starts rotating with speed 3550 rev/min

now we will have

f = \frac{3550}{60} = 59.17 rev/s

now we know that it took 10 s to reach the speed

so angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2

now it stops in 31 s so the angular deceleration is given as

\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2

now initially number of turn to reach the given speed

N_1 = \frac{59.17 + 0}{2}(10) = 296 turn

number of turns during it stop

N_2 = \frac{59.17 + 0}{2}(31) = 917 turn

Now during startup speed after t = 5 s is given as

\omega_1 = (5.917)(5) = 29.6 rev/s

N_3 = \frac{29.6 + 0}{2}(5) = 74 turns

now during it stop the speed after half the time is given as

\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s

now the number of turns is given as

N_4 = \frac{59.17 + 29.56}{2}(15.5)

N_4 = 688 turns

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Answer:

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