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3 years ago

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An electron is placed in a magnetic field that is directed along a z axis. The energy difference between parallel and antiparall

el alignments of the z component of the electron's spin magnetic moment with is 7.50 × 10-25 J. What is the magnitude of ?

1 answer:

Luda [366]3 years ago

8 0

Answer:

B = 0.0404 т

Explanation:

Given

ΔE = 7.50 x 10⁻²⁵ J

μb = 927.4 x 10⁻²⁶ J/т

ΔE = 2 * μb * B

B = ΔE / 2 μb

B = 7.50 x 10⁻²⁵ / 2 * 927.4 x 10⁻²⁶

B = 7.50 / 1854.8 x 10⁻²⁵⁻⁽⁻²⁶⁾

B = 0.00404 x 10¹

B = 0.0404 т

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A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F

EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B = 1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ) eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0

3 years ago

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, $F_c=\frac{mv^2}{R}$

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

$f=\mu N$

As there is no vertical motion, therefore, $N=mg$ . So,

$f=\mu mg$

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816$

Therefore, option (c) is correct.

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3 years ago

What are the principals of electric arc welding

Wewaii [24]

Principles<span> of </span>arc welding<span>. </span>Arc welding<span> is a </span>welding<span> process, in which heat is generated by an </span>electric arc<span> struck between an electrode and the work piece. </span>Electric arc<span> is luminous</span>electrical<span> discharge between two electrodes through ionized gas.</span>

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3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

Chemical weathering is greatest under conditions of

Pavel [41]

<h2>Answer: higher mean annual rainfall and temperatures. </h2>

Explanation:

Chemical weathering is the set of destructive processes through which rocky materials go trhough. These processes cause changes in the color, texture, composition, firmness and shape of the material.

It should be noted that this happens when the rocks come into contact with atmospheric agents such as oxygen and carbon dioxide.

Another important aspect is that rocks are able to break up more easily thanks to this type of weathering, since <u>the mineral grains within the rock lose adherence and dissolve better under the action of some physical agents</u>, such as <u>humidity (rainfall included) and temperature</u>.

Therefore:

Chemical weathering is greatest under conditions of <u>higher mean annual rainfall and temperatures. </u>

5 0

4 years ago