(a) Check all of the following that are correct statements, where E stands for γmc2. Read each statement very carefully to make
sure that it is exactly correct.
The definition K = E - mc2 is valid even for speeds near the speed of light.
The energy principle can be written Esys,f = Esys,i + Wsurr.
The definition of work is W = Fxdeltax + Fydeltay + Fzdeltaz.
At speeds close to the speed of light, kinetic energy is approximately equal to (1/2)mv2.
The definition of work is W = < Fxdeltax, Fydeltay, Fzdeltaz >.
The definition of work is W = |< Fxdeltax, Fydeltay, Fzdeltaz >|.
The energy principle can be written deltaEsys = Wsurr.
(b) An object with mass 70 kg moved in outer space. When it was at location < 14, -32, -7 > its speed was 11.5 m/s. A single constant force < 240, 400, -120 > N acted on the object while the object moved to location < 19, -40, -12 > m. What is the speed of the object at this final location?
The definition K = E - mc2 is valid even for speeds near the speed of light.
The energy principle can be written Esys,f = Esys,i + Wsurr.
The definition of work is W = Fxdeltax + Fydeltay + Fzdeltaz.
At speeds close to the speed of light, kinetic energy is approximately equal to (1/2)mv2.
The definition of work is W = < Fxdeltax, Fydeltay, Fzdeltaz >.
The definition of work is W = |< Fxdeltax, Fydeltay, Fzdeltaz >|.
The energy principle can be written deltaEsys = Wsurr.
(b) An object with mass 70 kg moved in outer space. When it was at location < 14, -32, -7 > its speed was 11.5 m/s. A single constant force < 240, 400, -120 > N acted on the object while the object moved to location < 19, -40, -12 > m. What is the speed of the object at this final location?
1 answer:
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Answer:
Explanation:
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Potential difference
New Capacitor
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Answer:
18 Ω
Explanation:
As K and F are at the same voltage, we can redraw the diagram as in figure 2
Series resistances add directly, so we get figure 3
Adding parallel resistances gets us to figure 4
Now we can move two 6Ω resistances for clarification in figure 5
As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6
Add series resistances to get to figure 7
Add parallel resistances to get to figure 8
Add series resistances to get to figure 9
