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Dmitry_Shevchenko [17]

3 years ago

5

A machine puts out 6,000 J of work. To produce that much work the machine

1 answer:

topjm [15]3 years ago

5 0

Answer:

Efficiency of the machine = 75%

Explanation:

Given:

Input work = 8,000 J

Output work = 6,000 J

Find:

Efficiency of the machine

Computation:

Efficiency of the machine = [Output work / Input work]100

Efficiency of the machine = [6,000 / 8,000]100

Efficiency of the machine = 75%

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3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

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4 years ago

an airplane of mass 50 tonnes is flying at a speed of 5 km\s at a height of 50 km above the ground. what is the total energy it

Sonbull [250]

Answer:

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3 years ago

(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th

KonstantinChe [14]

Answer:

Part A:

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=3.03*10^{-3} m^2/V$

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

$N-Au=\frac{Density*Avogadro Number}{atomic weight}$

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$Atomic weight=196.97g/mol$

So:

$n=1.5*\frac{Density*Avogadro Number}{atomic weight}$

$n=1.5*\frac{19.32*6.02*10^{23}}{196.97}$

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}$

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

$Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}$

$Electron Mobility=3.03*10^{-3} m^2/V$

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4 years ago