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Montano1993 [528]

3 years ago

9

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Container C has 1 mol of gas in 22.4 L. Container D has 2 mol of gas in 11.2 L. Which containers have equal pressures?

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

Container A and C

Explanation:

ideal gas equation gives P=nRT/V

so at constant Temperature and pressure, P=n/T

Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

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OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

<u>F = 85696.5 N = 85.69 KN</u>

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What is the density of the block if it has a mass of 45.5 grams and a volume of 2.4 cm³?

rjkz [21]

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3 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

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3 years ago