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musickatia [10]
3 years ago
11

Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot

ential lines around the two charges. The potential is defined to be zero at infinity. Identify the line of zero potential between the two charges, if there is one.
Physics
1 answer:
diamong [38]3 years ago
5 0

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

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Answer:

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Explanation:

The magnetic force is given by the expression

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where bold letters indicate vectors, the modulus of this expression is

       F = q v B sin θ

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the thumb indicates the direction of the speed, in this case from right to left

the palm the direction of the force, in our case upwards

the fingers extended the direction of the magnetic field, this case after fixing the other two components it points out of the blade

In short the magnetic field is leaving the sheet

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Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible l
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4 0
3 years ago
Help show how you got answers
lapo4ka [179]

The speed at which a wave of frequency of 14 Hz and a wavelength of 3 m will travel is  42 m / s

v = λ / T

f = 1 / T

v = f λ

v = Wave speed

λ = Wavelength

T = Time period

f = Frequency

f = 14 Hz

λ = 3 m

v = 14 * 3

v = 42 m / s

Frequency is the number of time an event occurs repeatedly in an unit amount of time. Its unit is hertz which in terms of SI unit is s^{-1}.

Therefore, the speed at which the wave will travel is 42 m / s

To know more about frequency

brainly.com/question/16861358

#SPJ1

4 0
1 year ago
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t
coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

               $=113.16 $  GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

                    $\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

                       = 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

   = 0.016 mm

7 0
3 years ago
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