In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ
1 answer:
Answer:
(a) W=1.20×10⁴J
(b) U= -5.46×10⁴J
(c) Q= -4.26×10⁴J
Explanation:
Given that student does 1.20×10⁴J work
(a) W=1.20×10⁴J
Work done by student,so positive sign
During the process, his internal energy decreases by 5.46×10⁴J.
(b) U= -5.46×10⁴J.
As the Energy decreases therefore negative sign
For (c) Q
We know the formula
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Answer: 313920
Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.
Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.
The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.
The depth below the water surface of each strip is,
di = 8 − yi
and that in turn gives us the pressure on the strip,
Pi =ρgdi = 9810 (8−yi)
The area of each strip is,
Ai = 2√4− (yi) 2Δy
The hydrostatic force on each strip is,
Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy
The total force on the plate is found on the attached image.
Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR
ΔV=0.484mV