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Nezavi [6.7K]
2 years ago
13

Sodium is a metal why​

Physics
2 answers:
kirill [66]2 years ago
7 0

Explanation:

Sodium is a chemical element with the symbol Na (from Latin "natrium") and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table. ... The free metal does not occur in nature, and must be prepared from compounds.

Fofino [41]2 years ago
4 0

Explanation:

Sodium is an alkali metal, being in group 1 of the periodic table.

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A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc
vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

6 0
3 years ago
How did mount st. Helen get its name
steposvetlana [31]

The modern name, Mount St. Helen's, was given to the volcanic peak in 1792 by seafarer and explorer Captain George Vancouver of the British Royal Navy. He named it in honor of fellow countryman Alleyne Fitzherbert, who held the title 'Baron St. Helen's.

5 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453
Drupady [299]

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

0.0001\times \frac{100}{0.4539}=0.022%

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

\frac{1}{0.022}\times {100}=4545.5\ kg

Mass in pounds would be

\frac{4545.5}{0.4539}=10014.32\ lb

Mass in pound-mass is 10014.32 lb

8 0
3 years ago
The bullet's horizontal velocity and vertical velocity vectors do not affect each other and are known as
lisabon 2012 [21]
Values in physics that do not affect each other are considered Independent values
8 0
2 years ago
Read 2 more answers
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