Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
a. be sure to hold expansion cards by the edge connectors
Explanation:
Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.
But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question
The half-life equation 
in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>
<em>
</em>
<em>
</em>
Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.
<em>2 * 5700 = </em>11400 yr
Answer: Neither Sandra nor Marissa will be in her THR zone.
Explanation:
1) Actual pulse of both Sandra and Marissa : 144 bpm
2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm
3) Sandra's TRH is in the range 135 - 172 bpm.
Since 124 < 135, she will be below the range.
4) Marissa's TRH range is 143 - 176 bpm.
Since, 124 < 143, she is below the range
In conlusion, neither Sandra nor Marissa will be in her THR zone.
Answer:
f_D = =3.24 N/m
Explanation:
data given
properties of air
k = 0.0288 W/m.K
WE KNOW THAT
Reynold's number is given as
= 1.941 *10^4
drag coffecient is given as
solving for f_D
Drag coffecient for smooth circular cylinder is 1.1
therefore Drag force is
f_D = =3.24 N/m
