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3 years ago

Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its

Engineering

1 answer:

kirill115 [55]3 years ago

3 0

<h2>Answer:</h2>

<h3>Required Answer is as follows :-</h3>

$\sf \dfrac{P _{i}}{ P_{f}} = \dfrac{3}{4}$

3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²

¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²

¾ = m₁ × (V)²/m₂ × (V²/4)

¾ = m₁/m₂ × 4

m₂ = 4 × 4/3 = 16/3m₁

$\bold{ }$

∆m = m₂ - m₁
∆m = 16m₂/3 - m₁ = 13m₁/3
Ratio = (13m₁/3)/ m₁
Ratio = 13/3

Ratio = 13:3

$\bold{ }$

Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg. $\bold{ }$

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An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th

photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

${W_{u}}= \eta {W_{pump}}$ .............1

put here value

${W_{u}}$ = (0.80)(20)

${W_{u}}$ = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

$\Delta{E_{mech}} = {m} \Delta pe$ ..............2

$\Delta {E_{mech}} = {m} g \Delta z$

and that can be express as

$\Delta {E_{mech}} = \rho Q g \Delta z$ ..................3

$\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]$ ......4

solve it we get

$\Delta {E_{mech}} = 13.614$ hp

so here

due to frictional effects, mechanical power lost in piping

we get here

${W_{frict}} = {W_{u}}-\Delta {E_{mech}}$

put here value

${W_{frict}}$ = 16 -13.614

${W_{frict}}$ = 2.37 hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0

3 years ago

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

yKpoI14uk [10]

Explanation:

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Therefore,

$-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}$

$Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}$

5 0

3 years ago

W²-5w+14

alexira [117]

Answer:w²-5w+14

4x²+11x+16

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Explanation:

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3 years ago

The statement that is NOT true about the concept of a boundary layer on an object is: a. the Reynolds number is greater than uni

Ilya [14]

Answer:

Option E

Explanation:

All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.

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3 years ago

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o

grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

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3 years ago