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Taya2010 [7]
3 years ago
13

When groups interact with one another, sometimes there is conflict. What is the BEST example of this type of group interaction?

Physics
1 answer:
Degger [83]3 years ago
4 0

Answer: (D) Two countries go to war with one another.

Explanation: I checked.

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If south-east Texas is where H then it is C, but if not the answer is A.

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Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the
Leya [2.2K]

Answer: 1.298\times 10^{-5}\ N/m

Explanation:

Given

Current in the first wire I_1=2.2\ A

Current in the second wire I_2=5.9\ A

wires are 20\ cm apart

Force per unit length between the current-carrying wires is

\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}

Force exerted  by the wires is the same

Put the values

\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m

This force will be repulsive in nature as the current is flowing opposite

4 0
3 years ago
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4 years ago
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Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a
victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

  1. F=\frac{GMm}{R^{2} }

Where:

  • F => Gravitational force or force of attraction between two masses
  • M => Mass of asteroid 1
  • m => Mass of asteroid 2
  • R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

  • M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

  • R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

  • m=ρV

This is going to be helpful because the volume of a sphere is:

  • \frac{4}{3}\pi r^{3}

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

  • F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }
  • F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }
  • F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2}  }{36}

Now let´s plug in the values we know:

  1. F = 1 lb     mutual gravitational attraction force
  2. G = 6.67(10)^{-11}     gravitational constant
  3. ρ_{iron} =491.5 \frac{lb}{ft^{3} }

  • 1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}

Solve for r and multiply by 2 because 2r = diameter

  • d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }

Result is d = 21.81 Feet

6 0
3 years ago
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