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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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Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium
ivolga24 [154]

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

5 0
3 years ago
A lava flow is an example of what igneous rock
nydimaria [60]
The answer is extrusive. 
7 0
3 years ago
Read 2 more answers
If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
2 years ago
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

8 0
3 years ago
A jet stream generally diverges above a low-pressure (warm) center. However, at Earth's surface, air converges at a low-pressure
Gemiola [76]

The correct answer is that the surface winds will get stronger.

6 0
3 years ago
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