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WINSTONCH [101]
2 years ago
12

A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in

Physics
1 answer:
KatRina [158]2 years ago
6 0

Answer:

37.5 seconds

Explanation:

The given parameters are;

The mass of the block on the spring, m = 2.50 kg

The force with which the loaded spring launches the block, F = 450 N

The change in momentum of the block, Δp = 12.0 kg·m/s

We have;

Let the force with which the block was launched = The net force, F_{NET}

By Newton's second law of motion, we have;

F = F_{NET} = Δp × Δt

Where;

Δt = The time the block is in contact with the spring

Therefore;

\Delta t = \dfrac{F_{NET}}{\Delta p}

By plugging in the values for F_{NET} and Δp, we have;

\Delta t = \dfrac{450 \ N}{12.0 \ kg \cdot m/s} = 37.5 \ s

The time duration the block is in contact with the spring, Δt = 37.5 seconds.

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Answer:

Explanation:

1) Moc to wskaźnik czasu pracy w stosunku do czasu.

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czas = 0,5 min = 0,5 * 60 = 30 sekund

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Moc = 25 watów

3) Moc = Gotowe / czas

Czas potrzebny = praca / moc

Biorąc pod uwagę pracę wykonaną przez ciągnik = 10 MJ = 10 * 10 ^ 6 Dżuli

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