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Molodets [167]
3 years ago
6

Initially, 0.05kg of air is contained in a piston cylinder device at 200 oc and 1.6 mpa. the air then expands at constant temper

ature to a pressure of 0.4 mpa. assume the process occurs slowly enough that the acceleration of the piston can be neglected. the ambient pressure is 101.35kpa. (i) determine the work (kj) performed bt the air in the cylinder on the piston. (ii) determine the work (kj) performed by the piston on the ambient environment. neglect the cross-sectional area of the connecting rod. (iii) determine the work transfer (kj) from the piston to the connecting rod. neglect friction between the piston and cylinder. assume that no heat transfer to or from the piston occurs and that the energy of the piston does not change solution
Physics
1 answer:
myrzilka [38]3 years ago
6 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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Discuss how friction is reduced in oce skating
belka [17]

Answer:

below

Explanation:

Ice melts, meaning it has a watery layer upon its surface. This allows things to be moving like they are on a liquid but it has the solidity of a solid. The thin metal of the ice skates also decrease the surface area meaning it exerts more force but in turn, it allows you to move faster and further reduces friction.

7 0
2 years ago
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0
3 years ago
i just said something super cringey to someone and now i wish i had a time machine because i cant forget it :((( SOME HELP PLEAS
Llana [10]

Answer:

what did you say???

Explanation:

6 0
2 years ago
Read 2 more answers
I need to know the right answer to that question
ad-work [718]
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation

v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.
8 0
3 years ago
How do you know which magnitude is higher or how do you compare them?
Paul [167]
Magnitudes are measured by intensity so a 3.4 earthquake is much less stronger than a 4.5 earthquake it’s very literally when measuring them the higher the number the stronger it is
7 0
3 years ago
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