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3 years ago

Which of the following energy resources produces greenhouse gases?

1 answer:

4 0

Coal is burned to get energy. By burning coal, green houses gases like carbon dioxide, nitrogen oxides and suphur oxides are produced.

Rest of the energy sources given in the option are clean energies.

Answer is A.

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The displacement is a straight line from the starting point to the finish in a ____ direction

Inessa05 [86]

To answer this question, we need to see the picture or description that goes along with it.

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4 years ago

A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft

rjkz [21]

$4.2*9.8\\41.16$

41.16 meters

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3 years ago

Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

mote1985 [20]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is $W_c = P_o V_o ln (R_v)$

Explanation:

Generally smallest workdone done by a gas is mathematically represented as

$dW = PdV$

Generally for an isothermal process

$PV = nRT = constant$

=> $P = \frac{nRT}{V}$

Generally the total workdone is mathematically represented as

$W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV$

=> $W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV$

=> $nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.$

=> $W_c = nRT [ln(V_f) - ln(V_o)]$

=> $W_c = nRT ln \frac{V_f}{V_o}$

From the question $\frac{V_f}{V_o } = R_v$

=> $W_c = P Vln (R_v)$

at initial state

$W_c = P_o V_o ln (R_v)$

7 0

3 years ago

A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?

Reika [66]

Answer:

$0.542*10^{23}\ Aluminum\ Atoms$

Explanation:

$We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms$

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3 years ago

What types of EM waves have you seen or used today? Describe them.

Mrrafil [7]

Answer:

none

Explanation:

3 0

3 years ago

Read 2 more answers