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erastovalidia [21]
3 years ago
8

If a bar magnet is cut in half, what will happen?

Physics
2 answers:
-BARSIC- [3]3 years ago
5 0

Explanation:

if a bar magnet is cut in half it's new sides will come that are North , South and if you'll cut the bar magnet as much as you can get new sides of magnetic attraction appear...

stiv31 [10]3 years ago
3 0
You will have 2 bar magnets.
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A nonconducting ring of radius 10.0 cm is uniformly charged with a total positive charge 10.0μC. The ring rotates at a constant
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The magnitude of the magnetic field on the axis of the ring 5 cm from its center is 143 pT.

The radius of the nonconducting ring is R = 10 cm.

The ring is uniformly charged q = 10 μC.

The angular speed of the ring, ω = 20 rad/s

The ring is x = 5 cm from the center of the ring.

Now,

R = 10 cm = 0.1 m

q = 10.0 μC = 10 × 10⁻⁶ C

x = 5 cm = 0.05 m

The magnetic field on the axis of a current loop is given as:

B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]

Now, I = q / [2π/ω]

So, the magnitude of the magnetic field which is directed away from the center is:

B =  [ μ₀ ωqR² ] / [4π(x² + R²)^{3/2} ]

B = [ μ₀ (200) (10 × 10⁻⁶) (0.1)² ] / [4π((0.05)² + (0.1)²)^{3/2} ]

B = 1.43 × 10⁻¹⁰ T

B = 143 pT

Learn more about the magnetic field here:

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2 years ago
Balance the equation: ___ SiO2 + ___ HF  ___ SiF4 + ___ H2O
olga_2 [115]
Let's start from the Fluorine (F), which has 4 atoms on the right side: this means we must put a factor 4 in front of HF to balance the number of atoms of Fluorine. 
Nowe we have 4 atoms of hydrogen, H, on the left side. Since we have a H_2 on the right side (2 atoms), we must put a factor 2 in front of H_2 O, to balance the number of atoms of hydrogen.
So now the atoms of oxygen are also balanced (2 on both sides), as well as the atoms of Silicon (Si), so the balanced reaction is

Si O_2 + 4 HF \rightarrow SiF_4 + 2 H_2 O
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3 years ago
If an element has 7 electrons how many of them will be in the 2nd energy level
inna [77]
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8 0
4 years ago
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Molds belong to a group of plants called <br> a.yeast<br> b.fungi<br> c.viruses
Marat540 [252]
B. fungi

I hope this helps!
5 0
4 years ago
Calculate the vapor pressure for a mist of spherical water droplets of radius 3.70×10−8m surrounded by water vapor at 298 K. The
poizon [28]

Answer:

The vapor pressure for a mist is P= 25.92\ Torr

Explanation:

From the question we are given that

        The radius is  r = 3.70 *10^{-8} m

        The temperature is T  = 298K

       The vapor pressure of water P_o = 25.2\ Torr

      The density of water is  \rho = 998 kg.m^{-3}

      The surface tension of water is \sigma  = 71.99 m N \cdot m^{-1}

Generally the equation of that is mathematically represented as

                            ln (\frac{P}{P_0} )  = \frac{2 \sigma M}{r \rho RT}

 Where  P is the vapor pressure for mist

               R is the  ideal gas constant = 8.31

      making P the subject in the formula

    P = e^ {\frac{2 \sigma M}{r \rho RT}} * P_0

        = e^{\frac{2 *(0.07199)(0.018)}{(3.70*10^{-8})(998)(8.31)(298)} } * 25.2

        P= 25.92Torr

               

4 0
3 years ago
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