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3 years ago

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What is a magnitute?

2 answers:

ValentinkaMS [17]3 years ago

5 0

Answer:

Magnitude is a specific type of norm. Magnitude is what is known as the Euclidean norm. There are other norms, such as the Leibniz and the Chebychev norm.

Explanation:

erik [133]3 years ago

5 0

Magnitude refers to that the maximum or greater extent of size and it's direction of an object

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A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

3 years ago

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III,

Rom4ik [11]

Answer:

See explanations for step by step procedures to get answer.

Explanation:

Given that;

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.

4 0

4 years ago

Steam enters a turbine in a Rankine cycle power plant at 200 psia and 500 °F. a) Calculate the isentropic thermal efficiency if

Aleks04 [339]

Answer:

η=0.19=19% for p=14.7psi

η=0.3=30% for p=1psi

Explanation:

enthalpy before the turbine, state: superheated steam

h1(p=200psi,t=500F)=2951.9KJ/kg

s1=6.8kJ/kgK

Entalpy after the turbine

h2(p=14.7psia, s=6.8)=2469KJ/Kg

Entalpy before the boiler

h3=(p=14.7psia,x=0)=419KJ/Kg

Learn to pronounce

the efficiency for a simple rankine cycle is

η= $\frac{h1-h2}{h1-h3}$

η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)

η=0.19=19%

second part

h2(p=1psia, s=6.8)=2110

h3(p=1psia, x=0)=162.1

η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)

η=0.3=30%

7 0

4 years ago

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given :

$PV^{1.3}=C$

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

$\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}$

where $r_{c}$ is compression ratio = $\frac{v_{c}+v_{s}}{v_{c}}$

$r_{c}$ = $1+\frac{v_{s}}{v_{c}}$

where $v_{c}$ is compression volume

$v_{s}$ is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, $v_{30}=v_{c}+0.7v_{s}$

and at 70% compression, 30% of the swept volume remains

∴ $v_{70}=v_{c}+0.3v_{s}$

We know,

$\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}$

$\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}$

$\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\$

$1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}$

$v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}$

$0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}$

$0.2218v_{s} = 0.594v_{c}$

$v_{c}=0.3734 v_{s}$

∴ $r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}$

Therefore, compression ratio is $r_{c}$ = 3.678

Now efficiency, $\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}$

$\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}$

$\eta =0.32342$ , this is the ideal efficiency

Therefore actual efficiency, $\eta_{act} =0.5\times \eta _{ideal}$

$\eta_{act} =0.5\times 0.32342$

$\eta_{act} =0.1617$

Therefore total power required = 1 kW x 3600 J

= 3600 kJ

∴ we know efficiency, $\eta=\frac{W_{net}}{Q_{supply}}$

$Q_{supply}=\frac{W_{net}}{\eta _{act}}$

$Q_{supply}=\frac{3600}{0.1617}$

$Q_{supply}=22261.78 kJ$

Therefore fuel required = $\frac{22261.78}{45000}$

= 0.4947 kg/hr

5 0

3 years ago

The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are

IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum = $9.93*10^{-8}$

Number of vacant atoms = $5.854 * 10^{15} \ vacancies/cm^3$

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = $6.022*10^{23$

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

$5.854*10^{15}$ = $9.93*10^{-8}$ × Total number of sites(N)

Total number of sites (N) = $\dfrac{5.854*10^{15}}{9.93*10^{-8}}$

Total number of sites (N) = $5.895*10^{22}$

From the expression of the total number of sites; we can determine the density of the electrum;

$N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}$

where ;

$N_A$ = Avogadro's Number

$\rho_{electrum} =$ density of the electrum

$A_{electrum}$ = Atomic mass

$5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}$

$5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}$

$8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}$

$\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}$

$\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}$

Thus; the density of the electrum is 14.30 g/cm³

7 0

3 years ago