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Irina18 [472]
3 years ago
14

0.250 moles of gas are in a piston.The gas does 109 J of work while 240 J of heat are added. What is the change in internal ener

gy?
(Be careful with + and - signs. +W = expansion, +Q =added, +ΔU = temp goes up)


Thank you!
Physics
1 answer:
irina [24]3 years ago
6 0

Answer:

+131Joules

Explanation:

Energy can be expressed using below expresion.

ΔE = (q + w).........eqn(1)

q will be + be if heat is gained hence, q= 240 J

work "w" will be - ve if work is done by the system, hence w= -109 J

Then substitute into eqn(1)

Change in Internal energy=

= (240 -109 )

= +131J

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<u>Given the following data;</u>

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A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t
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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
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Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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2 years ago
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