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levacccp [35]
3 years ago
11

Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volum

e of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?
Physics
1 answer:
s2008m [1.1K]3 years ago
8 0

there would be atleast 24 moles of air not counting the irreplacable ones

hope this helped

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
What is gibbs paradox?
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What is the resistance of a nichrome wire at 0.0 âc if its resistance is 100.00 Ï at 11.5 âc? express your answer using four sig
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30 years times the number of dogs
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A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?
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Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
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