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Travka [436]
4 years ago
7

Why is there no electric field at the center of a charged spherical conductor?

Physics
1 answer:
Svetlanka [38]4 years ago
4 0
-- Since the sphere is a conductor, the charge on it will move around
until it's evenly distributed on the surface of the sphere.  When every
tiny smitch of charge is the same distance from the charge around it,
they won't need to move around any more.

-- At that point, the situation at the center of the sphere will be:
For every smitch of charge on the surface, causing en electric field
at the center, there is another smitch, of the same size and in exactly
the opposite direction, canceling out the field of the first one. 
Every smitch of charge on the surface causes a tiny bit of electric field
at the center, and they all cancel each other.

It turn out that if the sphere is hollow, then the electric field at EVERY
point inside it is zero, not only at the center.

It's exactly the same idea as a sphere with uniform, homogeneous mass.
For that sphere, the gravity at the center is zero.
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A seated cable row is an example of which level of training in the NASM OPT model?
Vera_Pavlovna [14]

Answer:

B. Strength

Explanation:

The OPT Model, or Optimum Performance Training Model, is a "fitness training system created by the NASM. The OPT Model is contructed with scientific evidence and principles that progresses an individual through five training phases: stabilization endurance, strength endurance, hypertrophy, maximal strength and power".

On the stabilization level we have the phase 1 called stabilization endurance.

For the level strength part we have 3 phases . Phase 2: Strength endurance , Phase 3: Hypertrophy, Phase 4: Maximal strength. And we can consider the case "A seated cable row" on this the level strength since we need to have some abilities to do this but not enough to stay on the power level since this one is the advancd level.

For the power level we have the last phase called power in order to mantain and conduct high training level programs.

8 0
4 years ago
What is momentum equal to?
Setler [38]

Answer:

D

Explanation:

mass times velocity

hope this help

8 0
3 years ago
What causes wind to follow patterns
denis-greek [22]

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I don't know if this was helpfull or not but I hope it was somewhat close to helpful? If it is not then please do not report this, I am sorry if this was the least bit helpful.

HAVE A NICE DAY! :)

6 0
3 years ago
A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil
NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ_{crss} = σ_{Y} ( cosФ cosλ)_{max}

now by putting values,

     = (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

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