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Gre4nikov [31]
3 years ago
12

The viscosity of all liquids decreases as the temperature is a) Increased b) decreased c) maintained constant d) fluctuating e)

none of these
Engineering
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

b) decreased

Explanation:

Viscosity is the measure of the resistance between the layers of the liquid. On increasing temperature, the kinetic energy of the molecules of the liquid increases which in turn increase in the random motion of the molecules. Thus,  the relative motion of  the molecules in the liquid become easier and hence, viscosity decreases.

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loris [4]

Answer: 9,719

Explanation:

5 0
3 years ago
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

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8 0
3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

z1= \frac{\left(150-137\right)}{27.7}=0.4693

z2=\frac{\left(201-137\right)}{27.7}=2.3105

P(0.4693

B.)

z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289

\\P(2.9309

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0
4 years ago
Read 2 more answers
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
Which lens is wide-angle?
PIT_PIT [208]

A wide-angle lens has a focal length of 35mm or shorter, which gives you a wide field of view. The wider your field of view, the more of the scene you'll be able to see in the frame. These lenses are ideal for many scenarios, and most photographers have at least one trusty wide-angle lens in their kit.

BRAINLIEST PLSSS

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