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3 years ago

Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three

Engineering

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

Explanation:

We have evaluate $\frac{(204mm\times 0.00457kg)}{34.6N}$

We know that 1 mm $=10^{-3}m$

So 240 mm $=204\times 10^{-3}m$

Newton can be written as $kgm/sec^2$

So $\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}$

So the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

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It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

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3 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

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3 years ago

An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel

Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

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3 years ago

Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u

Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × $10^{-2}$ N/m

so average radius = $\frac{diameter}{2}$ = 100 µm = 100 × $10^{-6}$ m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r² .....................1

here r is radius and Δp pressure difference and σ surface tension

Δp = $\frac{2 \sigma }{r}$

put here value

Δp = $\frac{2*2.69*10^{-2}}{100*10^{-6}}$

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

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4 years ago

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elena55 [62]

Answer:

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3 years ago