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lorasvet [3.4K]
3 years ago
5

Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all

electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Physics
1 answer:
White raven [17]3 years ago
4 0

Answer:

C) experience a small induced magnetic moment when placed in an external magnetic field.

Explanation:

Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.

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Light hits a mirror at a 45° angle. It will be reflected at an angle _____.
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It will be equal to 45°
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What phenomenon causes colors of visible light to be separated by a prism?
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If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w
uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2

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V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL

So, the final volume of the sample is 200 mL.

5 0
2 years ago
A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder
Ratling [72]

Answer:

70509.8039216 N/C

Explanation:

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q = Charge = 2.00 µC

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r = Radius of cylinder = 10 cm

\lambda=\dfrac{q}{l}

Electric field is given by

E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C

The electric field at the surface of the cylinder is 70509.8039216 N/C

7 0
3 years ago
Read 2 more answers
Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
2 years ago
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