A 0.435 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.435 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed>

off by an adjustable plunger, and comes to rest some distance above the plunger. The plastic tube>

has an inner radius of 6.90 mm, and is frictionless. Neither the plunger nor the metal cylinder>

allow any air to flow around them. If the plunger is suddenly pushed upwards, increasing the>

pressure between the plunger and the metal cylinder by a factor of 2.79, what is the initial>

acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.style="font-family: 'droid sans', georgia, 'times new roman', serif; font-size: 16px; text-align: -webkit-center;">

Explanation / Answer

given Mass m = 0.435 kg;

inner dia di= 6.9 * 10^-3 m;

pressure inside pi= 2.79 atm;

pressure outside po =1 atm;

Change in pressure = F/A;

delta p = 1.79 *10^5 Pa ;

1.79 * 10^5 * (3.14 *(6.9)^2 * 10^-6)/4 = m*a;

F=m*a;

so we get acceleration a = 6.68/m;

so a= 6.68/0.435 = 15.379 m/s2.

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