action and reaction are equal in magnitude and opposite in direction.Then why do they not balance each other.

Given that,

A particle to move with uniform circular motion in a uniform magnetic field.

Suppose, The conditions are,

(I). The charged particle has to be positive and it should be moving in a direction opposite to a uniform magnetic field.

(II). The charged particle  should be moving parallel to the magnetic force and perpendicular to the magnetic field.

(III). The magnetic field  should be uniform and charge particle should be moving perpendicular to the magnetic field.

We know that,

An particle to move with uniform circular motion.

Here, electric force is perpendicular to velocity of particle.

The electric field is defined as,

[tex]F_{c}=\dfrac{mv^2}{r}[/tex].....(I)

Suppose, there is magnetic field, if a charge moving with velocity and the magnetic field exerts a field.

The magnetic force is defined as,

[tex]F_{m}=qvB[/tex].....(II)

We need to find the magnetic field

Using equation (I) and (II)

[tex]F_{c}=F_{m}[/tex]

[tex]\dfrac{mv^2}{r}=qvB[/tex]

[tex]B=\dfrac{mv}{qr}[/tex]

Hence, The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.

(III) is correct option.

Answer:

Power = Force × Distance/time

Power = Force × Velocity

Power = 2,300.0 N × 28.0 m/s²

Power = 64400 Nm/s

Explanation:

First show the formula of Power

Re-arrange formula and used to work out Power

Pretty simple stuff!

Hope this Helps!!

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

Answer:

The  amount by which the second filter reduces the intensity of light emerging from the first filter is

     z =  0.60

Explanation:

From the question we are told that

    The angle between the axes is  [tex]\theta = 41^o[/tex]

The intensity of polarized light that emerges from the second filter is  mathematically represented as

         [tex]I= I_o cos^2 \theta[/tex]

 Where [tex]I_o[/tex] is the intensity of light emerging  from the first filter

        [tex]I = I_o [cos(41.0)]^2[/tex]

      [tex]I =0.60 I_o[/tex]

This means that the second filter reduced the intensity by z =  0.60

Answer:

The impulse is 2145 kg-m/s

The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.

Explanation:

Force on the rail = 3900 N

Elapsed time of impact = 0.55 s

Impulse is the product of force and the time elapsed on impact

I = Ft

I is the impulse

F is force

t is time

For this case,

Impulse = 3900 x 0.55 = 2145 kg-m/s

If the initial velocity was 2.95 m/s

and mass of car plus driver is 190 kg

neglecting friction, the initial momentum of the car is given as

P = mv1

where P is the momentum

m is the mass of the car and driver

v1 is the initial velocity of the car

initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s

We know that impulse is equal to the change of momentum, and

change of momentum is initial momentum minus final momentum.

The final momentum = mv2

where v2 is the final momentum of the car.

The problem translates into the equation below

I = mv1 - mv2

imputing values, we have

2145 = 560.5 - 190v2

solving, we have

2145 - 560.5 = -190v2

1584.5 = -190v2

v2 = -1584.5/190 = -8.34 m/s

Answer:

Explanation:

For current in ideal transformer the formula is

I₁ / I₂ = N₂ / N₁

I₁  and I₂ are current in primary and secondary coil respectively and N₁ and N₂ are no of turns in primary and secondary coil .

Putting the given values

100 / I₂ = 200 / 100 = 2

I₂ = 50 A .

output current = 50 A .

Answer:

the greatest intensity is obtained from   c

Explanation:

An electromagnetic wave stagnant by the expression

           E = E₀ sin (kx -wt)

when two waves meet their electric fields add up

           E_total = E₁ + E₂

the intensity is

           I = E_total . E_total

           I = E₁² + E₂² + 2E₁ E₂ cos θ

where θ  is the phase angle between the two rays

Let's examine the two waves

in this case E₁ = E₂ = E₀

          I = Eo2 + Eo2 + 2 E₀ E₀ coasts

         I = E₀² (2 + 2 cos θ )

         I = 2 I₀ (1 + cos θ )

     let's apply this expression to different cases

a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀

b) cos π = -1     this implies that     I_total = 0

c) the cosine is  1,

         I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ

         I = E₀² (5 +4 cos θ)

         I = E₀² 9

         I = 9 Io

d) in this case the cos pi = -1

          I = E₀² (5 -4)

          I = I₀

e) we rewrite the equation

         I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ

         I = Eo2 (10 + 6 cos θ)

         cos π = -1

         I = E₀² (10-6)

         I = 4 I₀

the greatest intensity is obtained from   c

The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.

An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.

In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.

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Answer:effective

Explanation:

Answer:

Explanation:

Change in entropy is expressed as ΔS = ΔQ/T where;

ΔQ is the total heat change during conversion of ice to water.

T is the room temperature

First we need to calculate the total change in heat using the conversion formulae;

ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)

m is the mass of ice/water = 900g = 0.9kg

L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg

c is the specific heat capacity of water = 4200J/kgK

Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K

Substituting the given values into ΔQ;

ΔQ = 0.9(333000)+0.9(4200)(283)

ΔQ = 299700 + 1069740

ΔQ = 1,369,440 Joules

Since Change in entropy ΔS = ΔQ/T

ΔS =  1,369,440/30+273

ΔS = 1,369,440/303

ΔS = 4519.60 J/K

Hence, the change in entropy of the universe is 4519.60 J/K

Answer:

1) c. Helium

2) Iron

3) False.

Explanation:

1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.

2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.

3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

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Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

Answer:

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Explanation:

The electric field is given by the following formula:

E = F/q

E= W/q

E = mg/q

where,

E = magnitude of electric field = ?

m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

= charge = 3 nC = 3 x 10⁻⁹ C

Therefore,

E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

Hope this helps ;) ❤❤❤

Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

Thanks!!!!❤❤❤

Answer:

Explanation:

he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is:  m L²/ 3  where m is mass and L is length of rod

If the axis of rotation passes through the center of the rod, then the moment of inertia is:   m L² / 12

So for the former case , moment of inertia is higher that that in the later case .

In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .

In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .

Since range of distance from axis is less , moment of inertia too will be less because

Moment of inertia = Σ m r² where r is distance of mass m from axis .

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

Answer:

Explanation:

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Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

[tex]\mu_{sB}=0.126[/tex]

[tex]\mu_{sC}=0.168[/tex]

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

[tex]\sum \tau_{A}=0[/tex]

[tex]N(3m)-W(1.5m)=0[/tex]

When solving for N we get:

[tex]N=\frac{W(1.5m)}{3m}[/tex]

[tex]N=\frac{(1962N)(1.5m)}{3m}[/tex]

[tex]N=981N[/tex]

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

[tex]\sum F_{y}=0[/tex]

[tex]-F_{By}+N_{c}=0[/tex]

[tex]F_{By}=N_{c}[/tex]

Next, the forces in x.

[tex]\sum F_{x}=0[/tex]

[tex]-f_{sB}-f_{sC}+P_{x}=0[/tex]

We can find the x-component of force P like this:

[tex]P_{x}=360N(\frac{4}{5})=288N[/tex]

and finally the torques about C.

[tex]\sum \tau_{C}=0[/tex]

[tex]f_{sB}(1.75m)-P_{x}(0.75m)=0[/tex]

[tex]f_{sB}=\frac{288N(0.75m)}{1.75m}[/tex]

[tex]f_{sB}=123.43N[/tex]

With the static friction force in point B we can find the coefficient of static friction in B:

[tex]\mu_{sB}=\frac{f_{sB}}{N}[/tex]

[tex]\mu_{sB}=\frac{123.43N}{981N}[/tex]

[tex]\mu_{sB}=0.126[/tex]

And now we can find the friction force in C.

[tex]f_{sC}=P_{x}-f_{xB}[/tex]

[tex]f_{sC}=288N-123.43N=164.57N[/tex]

[tex]f_{sC}=N_{c}\mu_{sC}[/tex]

and now we can use this to find static friction coefficient in point C.

[tex]\mu_{sC}=\frac{f_{sC}}{N}[/tex]

[tex]\mu_{sC}=\frac{164.57N}{981N}[/tex]

[tex]\mu_{sB}=0.168[/tex]

Answer:

its supposed to be (a) 1W

Answer:

Explanation:

The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.

In compound microscopes, the distance between the two lenses is expressed as L = v0+ue

v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.

Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.

Given L = 18.0cm

Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm

1/f0 = 1/u0+1/v0

f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.

1/v0 = 1/7.4-1/12

1/v0 = 0.1351-0.0833

1/v0 = 0.0518

v0 = 1/0.2184

v0 = 19.31cm

Note that v0 = ue = 19.31cm

To get ve, we will use the lens formula 1/fe = 1/ue+1/ve

1/ve = 1/fe-1/ue

Given ue = 19.31cm and fe = -7.00cm

1/ve = -1/7.0-1/19.31

1/ve = -0.1429-0.0518

1/ve = -0.1947

ve = 1/-0.1947

ve = -5.14cm

Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens

b) Magnification of the final image M = ve/ue

M = 5.14/19.31

M = 0.27

Magnification of the final image is 0.27

Answer:

The inventors  claim is not real

a)  No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

     The  power input is  [tex]P_i = 0.25 kW = 0.25 *10^{3} \ W[/tex]

      The  rate of heat transfer [tex]J = 3050 J/s[/tex]

       The temperature of the freezer content is [tex]T = 270 \ K[/tex]

       The  ambient temperature is  [tex]T_a = 293 \ K[/tex]

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

      [tex]COP = \frac{T }{Ta - T}[/tex]

substituting values

     [tex]COP = \frac{270 }{293 - 270}[/tex]

     [tex]COP =11.7[/tex]

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

       [tex]COP = \frac{J}{P_i}[/tex]

substituting values

       [tex]COP = \frac{3050}{0.25 *10^{3}}[/tex]

       [tex]COP = 12.2[/tex]

Now given that the COP  of an ideal refrigerator is  less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

Answer:

 I = E/R   e^{-t/RC}

Explanation:

In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,

                  E -q / c-IR = 0

we replace the current by its expression and divide by the resistance

                   I = dq / dt

                   dq / dt = E / R  -q / RC

                   dq / dt = (CE -q) / RC

we solve the equation

                   dq / (Ce-q) = -dt / RC

we integrate and evaluate for the charge between 0 and q and for the time 0 and t

                   ln (q-CE / -CE) = -1 /RC   (t -0)

eliminate the logarithm

              q - CE = CE [tex]e^{-t/RC}[/tex]

               q = CE (1 + 1/RC  e^{-t/RC} )

In general the teams measure the current therefore we take the derivative to find the current

               i = CE (e^{-t/RC} / RC)

               I = E/R   e^{-t/RC}

This expression is the one that describes the charge of a condensate in a DC circuit

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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Answer:

1) 14 kW

2) 4.67 x 10^-5 N

Explanation:

Area of solar panel = 10 m^2

Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

P = 1.4 kW/m^2  x  10 m^2 = 14 kW

b) For a perfect absorbing surface, the radiation pressure is given as

p = I/c

where p is the radiation pressure

I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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Answer:

(a) [tex]\vec D = 2\,i - 2\,j[/tex], (b) [tex]\vec E = -6\,i + 12\,j[/tex]

Explanation:

Let be [tex]\vec A = 3\,i - 3\,j\,[m][/tex], [tex]\vec B = i - 4\,j\,[m][/tex] and [tex]\vec C = -2\,i + 5\,j \,[m][/tex], each resultant is found by using the component method:

(a) [tex]\vec D = \vec A + \vec B + \vec C[/tex]

[tex]\vec D = (3\,i - 3\,j) + (i-4\,j) + (-2\,i+5\,j)\,[m][/tex]

[tex]\vec D = (3\,i + i -2\,i)+(-3\,j-4\,j+5\,j)\,[m][/tex]

[tex]\vec D = (3 + 1 -2)\,i + (-3-4+5)\,j\,[m][/tex]

[tex]\vec D = 2\,i - 2\,j[/tex]

(b) [tex]\vec E = -\vec A - \vec B + \vec C[/tex]

[tex]\vec E = -(3\,i-3\,j)-(i - 4\,j)+(-2\,i+5\,j)[/tex]

[tex]\vec E = (-3\,i + 3\,j) +(-i+4\,j) + (-2\,i + 5\,j)[/tex]

[tex]\vec E = (-3\,i-i-2\,i) + (3\,j+4\,j+5\,j)[/tex]

[tex]\vec E = (-3-1-2)\,i + (3+4+5)\,j[/tex]

[tex]\vec E = -6\,i + 12\,j[/tex]