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Sergeeva-Olga [200]
3 years ago
7

If a magnet can cut into a small piece , each piece will remain a magnet ? Explain this phenomenon

Physics
1 answer:
Hitman42 [59]3 years ago
8 0

yes ,each piece will remain a magnet because the small pieces will cut into north and south

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Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0
NikAS [45]
The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J

For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
6 0
3 years ago
Read 2 more answers
In avarage,How many times do a child breathe in a
ollegr [7]
On an approximate scale, A child breaths 20 times a minute as compared to only 12 to 16 in resting phase of an Adult.

So, In 60 minutes (1 hour), They breathe = 20 * 60 = 1200
In 24 hours (1 day), They breathe = 1200 * 24 = 28,800

In short, Your Answer would be: 28,800

Hope this helps!
5 0
3 years ago
Read 2 more answers
Which of the following statements is FALSE regarding the law of conservation of energy?
Jobisdone [24]

Answer:

D

Explanation:

because it doesn't agree with the law of definite proportion

and energy can niether be created nor destroyed but can be transformed from one form to another

5 0
3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
I just need the answer
Archy [21]

Answer:

I'll try to help which grade are you?

7 0
3 years ago
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