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Gekata [30.6K]
3 years ago
6

A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. the lid of a

pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. a separate metal piece, the petcock, sits on top of this opening and prevents steam from escaping until the pressure force overcomes the weight of the petcock. the periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value. determine the mass of the petcock of a pressure cooker whose operation pressure is 100 kpa gauge and has an opening cross-sectional area of 4 mm2. assume an atmospheric pressure of 101 kpa.
Physics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

m=40.816\ gm

Explanation:

Given;

The operational pressure of the cooker, P=10^5\ Pa (gauge)

area of the opening, a=4\ mm^2

  • We know that the gauge pressure is measured with respect to the atmospheric pressure. As the atmospheric pressure acts on all the components of the system being observed so, we only need to counter the gauge pressure.

As the petcock sits on the opening of the cross-section so, it balances the built-in pressure due to its weight.

<u>Hence:</u>

m.g=P\times a

where:

m= mass of the petcock

g= acceleration due to gravity

m\times 9.8=10^5\times 4\times 10^{-6}

m=40.816\ gm

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Answer:

Explanation:

There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.

Going with the physics version first, here's what we know:

a = -9.8 m/s/s

v₀ = 3.75 m/s

t = ??

That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:

v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:

v = v₀ +at and filling in:

0 = 3.75 + (-9.8)t and

-3.75 = -9.8t so

t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:

2(.38) = .76 seconds. Now onto the calculus way:

The position function is

s(t)=-4.9t^2+3.75t The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:

v(t) = -9.8t + 3.75 and

0 = -9.8t + 3.75 and

-3.75 = -9.8t so

t = ,38 and multiply that by 2 to find the time the whole trip took:

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Answer:

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Explanation:

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F2 = G M_e m2 / r²

where

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Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

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a1 = v² / r ⇒ v² = r a1    and

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F1 = F1c

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a1 = G M_e / r²

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Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

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