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Alex787 [66]
3 years ago
6

A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?

Physics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

It will become a temporary magnet because the domains will easily realign.

Explanation:

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3 years ago
The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine th
Rzqust [24]
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis 
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is 
T = 1/2 I₀w² + 1/2mAVA²

= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation  is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
5 0
3 years ago
Along hot spots, vast amounts of basaltic lava can flow from fissures, forming:
Fynjy0 [20]

The answer would be flood basalt. This is the outcome of a huge volcanic eruption or sequence of eruptions that covers large expanses of land or the ocean floor with basalt lava. The development and effects of a flood basalt hinge on a variety of factors, like latitude, continental configuration, rate, volume, period of eruption, the preexisting climate state, style and location, and the biota flexibility to alteration.

8 0
3 years ago
The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per
Natasha_Volkova [10]

Answer:

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Explanation:

Given that:

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The radius of the thick wire  r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0  × 10¹⁸ electrons

Electron mobility  μ =  6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

I = \frac{q}{t}

I = \frac{Ne}{t}

I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}

I = 0.96 \ A

The magnitude of the electric field is:

E = \frac{I}{ \mu n eA}

E = \frac{I}{ \mu n e(\pi r^2)}

E = \frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}

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8 0
3 years ago
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Fofino [41]

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Explanation:

3 0
2 years ago
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