In moving out of a dormitory at the end of the semester, a student does 2.17 x 104 j of work. in the process, his internal energy decreases by 5.03 x 104 j. determine each of the following quantities (including the algebraic sign): (a)w, (b) u, and (c)q.

a. W = 2.17 x 10 ⁴ J

b. ΔU = - 5.03 x 10⁴ J

c. Q = 2.86 x 10 ⁻³ J

Explanation:

a.

The student work does 2.17 x 10⁴ J

So W = 2.17 x 10 ⁴ J  

b.

The internal energy decreases 5.03 x 10⁴ J

ΔU = U₁ - U₂

ΔU = 0 - 5.03 x 10⁴ J = - 5.03 x 10⁴ J

c.

The formula heat can use:

Q = W + ΔU

Q = 2.17 x 10⁴ J + ( - 5.03 x 10⁴ J )

Q = 2.86 x 10 ⁻³ J

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

need to see circuit to get time constant