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Oksi-84 [34.3K]
3 years ago
15

The pressure inside a champagne bottle can be quite high and can launch a cork explosively. Suppose you open a bottle at sea lev

el. The absolute pressure inside a champagne bottle is 6 times atmospheric pressure; the cork has a mass of 7.5 g and a diameter of 18 m. Assume that once the cork starts to move, the only force that matters is the pressure force.What is the acceleration of the cork? Express your answer with the appropriate units.
Physics
1 answer:
valina [46]3 years ago
4 0

Answer:

Explanation:

Given that,

The absolute pressure inside the champagne is 6 times atmospheric pressure

Atmospheric pressure is 101,325 N/m²

So, the absolute pressure in the champagne is

P_abs = 6 × P_atm

P_abs = 6 × 101,325

P_abs = 607,950 N/m²

If the cork of the champagne has a mass of

M = 7.5g

M = 7.5 / 1000 = 0.0075 kg

The diameter of the cork is 18mm, It is no possible for the diameter to be 18m, so I assumed it is 18mm

d = 18mm = 0.018m

The area of the circle will be

Area = πd² / 4

Area = π × 0.018² / 4

Area = 2.545 × 10^-4 m²

We know that,

Pressure = Force / Area

Then,

Force = Pressure × Area

Force = 607,950 × 2.545 × 10^-4

Force = 154.723 N

Then, from newton second law,

F = ma

a = F / m

a = 154.723 / 0.0075

a = 20629.77 m/s²

a ≈ 20,630 m/s²

The acceleration of the cork is 20,630 m/s²

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<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

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In this case, the object is the Earth. Therefore,

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<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

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<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

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(ii) The energy (rotational energy) E is given by;

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E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

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