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soldier1979 [14.2K]
3 years ago
5

Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo

ok?
She does work on the book at no time during this process.
She does work only while she carries the book across the room.
She does work from the moment she touches the book until she lets it go.
She does work as she picks up the book and as she lifts it to the shelf.
Physics
2 answers:
noname [10]3 years ago
5 0

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

ZanzabumX [31]3 years ago
3 0

option d

took test on edu

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Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy
erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

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5 0
3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Zarrin [17]

Answer:

\rm 9.186\times 10^{-7}\ C.

Explanation:

<u>Given:</u>

  • Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
  • Distance of separation between the plates, d = 1.0 cm = 0.01 m.
  • Minimum value of electric field that produces spark, \rm E=3\times 10^6\ N/C.

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

\rm E=\dfrac{\sigma}{\epsilon_o}.

where,

  • \rm \sigma = surface charge density of the plate of the capacitor = \dfrac qA.
  • \rm q = magnitude of the charge on each of the plate.
  • \rm A = surface area of each of the plate =\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.
  • \epsilon_o = electrical permittivity of free space, having value = 8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.

For the minimum value of electric field that produces spark,

\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0
3 years ago
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