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3 years ago

In your opinion, is "avoiding" an effective conflict management strategy? Why or why not?

2 answers:

6 0

I don't think avoiding would be an effective conflict management strategy, because if it's something serious or even something small you shouldn't really ever avoid anything especially if it's between you and another person because it could damage something between you two for awhile or maybe forever. So no i don't think that is a good why to handle something.

soldi70 [24.7K]3 years ago

4 0

In my personal opinion, avoiding in not an effective conflict management strategy. When you are avoiding a conflict, you are really just pushing it to the side and it will need to be dealt with sooner or later. Avoiding a conflict is also ammeter thing to do as well, you should take responsibility and handle any situation.

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Which three characteristics make iron a good blackbody radiator?

victus00 [196]

Answer:

A, C, and D

Explanation:

Just did the quiz

4 0

3 years ago

Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of

julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b) E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center

E₁ = k q₁ / d²

shell 2

the point is on the outside,d>ro

E₂ = k q₂ / d²

the total camp is

E_total = -E₁ + E₂

E_total = k ( $\frac{-q_1 + q_2}{d^2}$ )

E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

E₂ = 0

E_total = E₁

E_total = k q₁ / d²

E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point is inside the shell d< ro

as there are no charges inside

E₁ = 0

shell 2

point is inside the radius of the shell d <ro

E₂ = 0

the total field is

E_total = 0

3 0

3 years ago

Explain why not everyone can become a bodybuilder, even if they train hard enough.

Anni [7]

People have diffrent body builds and bone structure

5 0

4 years ago

Help with question 16 and 17

storchak [24]

Your answer is correct.

___

The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.

R = R₀×(1 + α×ΔT)

R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω

5 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago