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3 years ago

Describe two methods by a magnet can be demagnetized

Physics

2 answers:

Mekhanik [1.2K]3 years ago

8 0

Answer:

Demagnetizing a magnet :

By hitting with a hammer
By heating with a high temperature
By dropping the magnet

leva [86]3 years ago

5 0

Being hit with a hammer
Being dropped

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Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

Lelechka [254]

Answer:

The distance is $D = 0.000712 \ m$

Explanation:

From the question we are told that

The wavelength of the light source is $\lambda = 700 \ nm = 700 *10^{-9} \ m$

The distance from a pin hole is $x = 9\ m$

The diameter of the pin hole is $d = 1.2 \ mm = 0.0012 \ m$

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

$D = \frac{1.22 \lambda }{d }$

substituting values

$D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }$

$D = 0.000712 \ m$

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3 years ago

What type of Literary Devices is this passage and provide an explanation

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3 years ago

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

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3 years ago

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Answer:

Explanation:

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This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

Tcosθ = mw²R

Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = $\frac{g}{\omega^2R}$

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3 years ago

Describe two methods by a magnet can be demagnetized​

Describe two methods by a magnet can be demagnetized