The headlights of a car are 1.4 m apart. What is the maximum distance (in km) at which the eye can resolve these two headlights? Take the pupil diameter to be 0.30 cm. (Assume the average wavelength of visible light is 555 nm.)

Answer:

a)   f₁ = 3.50 m ,  b)     f₂ = 0.84 m  

Explanation:

For this exercise we must use the constructor equation

          1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image

a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞

           1 / f₁ = 1 /∞ + 1 / 3.50

           f₁ = 3.50 m

b) in this case the image is at q = -0.600 m and the object p = 0.350 m

           1 / f₂ = 1 / 0.350 -1 / 0.600

the negative sign, is because the image is in front of the object

           1 / f₂ = 1,1905

            f₂ = 1 / 1,1905

            f₂ = 0.84 m

Answer:

electric field in the wide wire is

E₂ =[tex]\frac{E}{4}[/tex]

Explanation:

given

length of the copper wire = L

radius of the copper wire r₁ = b

length of the second copper wire = L

radius of the second copper wire r₂ = 2b

electric field in the narrow wire = E₁=E

recall

resistance R = ρL/A

where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.

Resistance of narrow wire, R₁

R₁ = ρL/A

where A  = πb²

R₁ = ρL/πb²---------- eqn 1

Resistance of wide wire, R₂

R₂ = ρL/A

where A = π(2b)²

R₂ = ρL/π(2b)²

R₂ = ρL/4πb²-------------- eqn 2

R₂ = ¹/₄(ρL/πb²)

comparing eqn 1 and 2

R₁ = 4R₂

calculating the current in the wire,

I = E/(R₁ + R₂)

recall

R₁ = 4R₂

∴ I = E/(4R₂ + R₂)

I = E/5R₂

calculating the potential difference across R₁ & R₂

V₁ = IR₁

I = E/5R₂

∴ V₁ = ER₁/5R₂

R₁ = 4R₂

V₁ = 4ER₂/5R₂

∴V₁  = ⁴/₅E

potential difference for R₂

V₂= IR₂

I = E/5R₂

∴ V₂ = ER₂/5R₂

V₂ = ER₂/5R₂

∴V₂  = ¹/₅E

so, electric field E = V/L

for narrow wire E₁ = V₁/L ----------- eqn 3

for wide wire, E₂ = V₂/L------------ eqn 4

compare eqn 3 and 4

E₂/E₁ = V₂/V₁( L is constant)

E₂/E₁ = ¹/₅E/⁴/₅E

E₂ = E₁/4

note E₁ = E

∴E₂ =[tex]\frac{E}{4}[/tex]

Answer:

-72.0°C

Explanation:

PV = nRT

Since n, number of moles, is constant:

PV / T = PV / T

(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T

T = 201.16 K

T = -72.0°C

Answer:

(a) The speed of the block after the collision is 0.0476v0.

(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Explanation:

Given;

mass of ball of clay, m₁ = 50 g = 0.05 kg

mass of brick, m₂ = 1 kg

initial velocity of the ball of clay, u₁ = v0

initial velocity of the brick, u₂ = 0

Since the clay ball sticks with the brick after collision, it is inelastic collision.

Therefore, let their final velocity = v

(a) What is the speed of the block after the collision?

Apply the principle of conservation linear momentum

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.05v₀ + 1(0) = v( 0.05 + 1)

0.05v₀ = 1.05v

v = 0.05v₀ / 1.05

v = 0.0476v₀

Thus, the speed of the block after the collision is 0.0476 of its initial velocity.

(b). What percentage of the ball's initial energy is "lost"?

Initial kinetic energy of the ball = ¹/₂mv₀²

                                                    = ¹/₂ x 0.05 x v₀²

                                                    = 0.025v₀²

Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²

                                                    = ¹/₂ x 1.05 x 0.0476v₀²

                                                    = 0.025v₀²

Change in kinetic energy = 0.025v₀² - 0.025v₀²

                                           = 0

percentage change in the initial kinetic energy of the ball;

= (0 / 0.025v₀²) x 100%

= 0 x 100%

= 0 %

Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Answer:

C) the rope tension is less than the objects weight

Explanation:

According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.

In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.

Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:

C) the rope tension is less than the objects weight

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

Explanation:

Let the first wave is :

[tex]y_1=A\sin\omega t[/tex]

And another wave is :

[tex]y_2=A\sin (\omega t+\phi)[/tex]

[tex]\phi[/tex] is phase difference between waves

Let y is the resultant of these two waves. So,

[tex]y =y_1+y_2[/tex]

The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,

[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]

So, the phase difference between the two waves is [tex]\pi[/tex].

Answer:

θ = 26.19 x 10³ radians

Explanation:

FOR ACCELERATED MOTION:

we use 2nd equation of motion for accelerated motion:

θ₁ = ωi t + (1/2)αt²

Where,

θ₁ = Angular Displacement covered during accelerated motion = ?

ωi = Initial Angular Speed = 0 rad/s

t = Time Taken = 2.8 x 10³ s

α = Angular Acceleration = 2.24 x 10⁻³ rad/s²

Therefore,

θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²

θ₁ = 8.78 x 10³ radians

Now we find final angular velocity (ωf) by using 1st equation of motion:

ωf = ωi + αt

ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)

ωf = 6.272 rad/s

FOR UNIFORM ANGULAR SPEED:

For uniform angular speed we use following equation:

θ₂ = ωt

where,

θ₂ = Angular Displacement during uniform motion = ?

ω = Uniform Angular Speed = ωf = 6.272 rad/s

t = Time Taken = 1.57 x 10³ s

Therefore,

θ₂ = (6.272 rad/s)(1.57 x 10³ s)

θ₂ = 9.85 x 10³ radians

FOR DECELERATED MOTION:

Now, we use 3rd equation of motion for decelerated motion:

2αθ₃ = ωf² - ωi²

where,

α = Angular deceleration = - 2.01 x 10⁻³ rad/s²

θ₃ = Angular Displacement during decelerated motion = ?

ωf = Final Angular Speed = 2.99 rad/s

ωi = Initial Angular Speed = 6.272 rad/s

Therefore,

2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²

θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)

θ₃ = 7.56 x 10³ radians

FOR TOTAL ANGULAR DISPLACEMENT:

Total Angular Displacement = θ = θ₁ + θ₂ + θ₃

θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians

θ = 26.19 x 10³ radians

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:

3.867 μm

Explanation:

The index of refraction, μ = 1.6

Wavelength of the light, λ = 580 nm

N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have

(N2 - N1) λ = 2L (n2 - n1)

L = [(N2 - N1) * λ] / 2(n2 - n1)

L = (8 * 580) / 2(1.6 - 1.0)

L = 4640 nm / 1.2

L = 3867 nm or 3.867 μm

Therefore we can come to the conclusion that the thickness of the film is 3.867 nm

Answer:

The  voltage is [tex]V_c = 9.92 \ V[/tex]

Explanation:

From the question we are told that

     The voltage of the battery is  [tex]V_b = 24 \ V[/tex]

     The capacitance of the capacitor is  [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]

     The  resistance of the resistor is [tex]R = 100\ \Omega[/tex]

     The time taken is  [tex]t = 0.16 \ s[/tex]  

Generally the voltage of a charging charging capacitor after time t is mathematically represented as

       [tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]

Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so  

      [tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]

      [tex]V_c = 9.92 \ V[/tex]

Hey There!!

Your answer will be A. It get longer.

Because, The kinetic energy of its atoms increase. They vibrate faster. This means that each atom will take up more space due to its movement so the material will expand. Some metals expand more than others due to differences in the forces between the atoms / molecules. . . .

Hope This Helps <3!!

Answer:

here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.

Explanation:

Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current

            P = Vi = i² R

           i = √ P / R

            i = √ (5/10)

           i = 0.707 A

This is the current that must circulate in the circuit.

Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel

The equivalent resistance is

  1 /[tex]R_{equi}[/tex] = 1/10 + 1/10 + 1/10 = 3/10

   Requi = 10/3

   Requi = 3.3 Ω

The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors

current in each residence in parallel

              i_P = 0.707 / 3

              I_p = 0.2357 A

now let's look at the power dissipated in each resistor

            P = R i²

            P = 10  0.2357²

            P = 0.56 W

the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate

             P_total = 9 P

             P = total = 9 0.56

             P_total = 5 W

we see that this combination meets the specifications of the problem.

Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = 14691.456 J   is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

[tex]P = \frac{w}{t}[/tex]

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

[tex] I = \frac{q}{t} [/tex]

Where:

q: is the electric charge transferred through the surface

t: is the time      

The charge, q, is:

[tex] q = n*e [/tex]

Where:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

[tex] q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C [/tex]

Hence, the current in the wire is:

[tex] I = \frac{126.88 C}{3.97 s} = 31.96 A [/tex]

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

Answer:

Explanation:

 This problem bothers on  application of the expression for motion emf which is expressed as

[tex]E= Blv[/tex]

where B= magnetic field in Tesla

           l= length of the conductor

           v= speed of conductor

Given data

l= 2 meters

v= 6 m/s

B= 5 Tesla

Applying the formula we have

[tex]E=5*2*6= 60mV[/tex]

Answer:

D) VL(t) leads IL(t) and VC(t) lags IC(t)

Explanation:

This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,

( I(t), V(t)), is zero.

Answer:

7.68×10^25kg

Explanation:

The formula for energy used per year is calculated as

Energy used per year =12 x Energy used per month

By substituting Energy used per month in the above formula, we get

Energy used per year =12 x 1600kWh

= 19200kWh

Conversion:

From kWh to J:

1 kWh=3.6 x 10^6 J

Therefore, it is converted to J as

19200 kWh =19200 x 3.6 x 10^6 J

= 6.912×10^10 J

Hence, energy used per year is 6.912×10^10 J

To find the mass that is converted to energy per year.

E = MC^2 ............1

E is the energy used per year

C is the speed of light = 3.0× 10^8m/s

Where E= 6.912×10^10 J

Substituting the values into equation 1

6.912×10^10 J = M × 3.0× 10^8m/s

M = 6.912×10^10 J / (3.0× 10^8m/s)^2

M = 6.912×10^10 J/9×10^16

M = 7.68×10^25kg

Hence the mass to be converted is

7.68×10^25kg

Answer:

C is the best answer for this question

Answer:

C) proton acceptor

Explanation:

A proton acceptor is another name for a base,

Answer:

Explanation:

A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.

At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.

Answer:

The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s

Explanation:

I'll assume that the correct question is

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?

mass of box = 51 kg

for the first 12 m, it is pulled with a constant force of 240 N

The acceleration of the box for this first 12 m will be

from F = ma

a = F/m

where F is the pulling force

m is the mass of the box

a is the acceleration of the box

a = 240/51 = 4.71 m/s^2

Since the body started from rest, the initial velocity u = 0

applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have

[tex]v^{2}= u^{2}+2as[/tex]

where v is the final velocity

u is the initial velocity which is zero

a is the acceleration of 4.71 m/s^2

s is the distance covered which is 12 m

substituting value, we have

[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)

[tex]v^{2}[/tex]  = 113.04

[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s

For the final 10.5 m, coefficient of friction is 0.21

from  f = μF

where f is the frictional force,

μ is the coefficient of friction = 0.21

and F is the pulling force of the box 240 N

f = 0.21 x 240 = 50.4 N

Net force on the box = 240 - 50.4 = 189.6 N

acceleration = F/m = 189.6/51 = 3.72 m/s^2

Applying newton's equation of motion

[tex]v^{2}= u^{2}+2as[/tex]

u is initial velocity, which in this case =  10.63 m/s

a = 3.72 m/s^2

s = 10.5 m

v = ?

substituting values, we have

[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)

[tex]v^{2}[/tex]  = 112.9 + 78.12

v  = [tex]\sqrt{191.02}[/tex]  = 13.82 m/s

Answer:

The  kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]

Explanation:

From the question we are told that

       The  radius of the orbit is  [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]

       The gravitational force is  [tex]F_g = 6600 \ N[/tex]

The kinetic energy of the satellite is mathematically represented as

       [tex]KE = \frac{1}{2} * mv^2[/tex]

where v is the speed of the satellite which is mathematically represented as

     [tex]v = \sqrt{\frac{G M}{r^2} }[/tex]

=>  [tex]v^2 = \frac{GM }{r}[/tex]

substituting this into the equation

      [tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]

Now the gravitational force of the planet is mathematically represented as

      [tex]F_g = \frac{GMm}{r^2}[/tex]

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     [tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]

=>    [tex]KE = \frac{ 1}{2} *F_g * r[/tex]

substituting values

       [tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]

         [tex]KE = 7.59 *10^{10} \ J[/tex]

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.  

Find out more information about voltage here:

https://brainly.com/question/4389563

Answer:

The hall  voltage is [tex]\epsilon =1.45 *10^{-6} \ V[/tex]

Explanation:

From the question we are told that

     The magnetic field is  [tex]B = 2.00 \ T[/tex]

     The diameter is  [tex]d = 2.588 \ mm = 2.588 *10^{-3} \ m[/tex]

       The  current is  [tex]I = 20 \ A[/tex]

The radius can be evaluated as

        [tex]r = \frac{d}{2}[/tex]

substituting values

       [tex]r = \frac{2.588 * 10^{-3}}{2}[/tex]

       [tex]r = 1.294 *10^{-3} \ m[/tex]

The hall voltage is mathematically represented as

      [tex]\episilon = B * d * v_d[/tex]

where[tex]v_d[/tex] is the  drift velocity of the electrons on the current carrying conductor which  is mathematically evaluated as

       [tex]v_d = \frac{I}{n * A * q }[/tex]

Where n is the number of electron per cubic meter which for copper is  

        [tex]n = 8.5*10^{28} \ electrons[/tex]

  A is the cross - area of the wire  which is mathematically represented as

           [tex]A = \pi r^2[/tex]

substituting values

          [tex]A = 3.142 * [ 1.294 *10^{-3}]^2[/tex]

          [tex]A = 5.2611 *10^{-6} \ m^2[/tex]

so the drift velocity is  

       [tex]v_d = \frac{20 }{ 8.5*10^{28} * 5.26 *10^{-6} * 1.60 *10^{-19} }[/tex]

       [tex]v_d = 2.7 *10^{-4 } \ m/s[/tex]

Thus the hall voltage is

     [tex]\epsilon = 2.0 * 2.588*10^{-3} * 2.8 *10^{-4}[/tex]

     [tex]\epsilon =1.45 *10^{-6} \ V[/tex]

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

Answer: The width is 1.25692 μm

Explanation:

The data that we have here is:

Distance between the single slit to the screen = L = 43cm

λ = 636 nm

Distance from the central maximum to the first minimum = Z =  3.8mm

We know that the angle for the destructive diffraction is:

θ = pλ/a

where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,

then we have:

θ = (636nm/a)

And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:

Tg(θ) =  Z/L

Tan(636nm/a) = 3.8cm/43cm

First, we need to use the same units in the right side:

3.8mm = 0.38cm

Tg(636nm/a)  = 0.38cm/43cm

636nm/a = Atg(  0.38cm/43cm ) = 0.506

a = 636nm/0.506 = 1,256.92 nm

1 μm = 1000nm

then:

a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm

Answer:

amplitudes

Explanation:

In everyday physics we define the amplitude of a wave as the maximum (this can also be called the highest)displacement or distance moved by a point on a given vibrating body or wave as measured from its equilibrium position. The key idea in defining the amplitude of a wave motion is the idea of a 'maximum displacement from the position of equilibrium'.

Given the equations;

E= EoSin(kx - ωt)y

B= Bosin(kx- ωt)z

Both Eo and Bo refer to the maximum displacement of the electric and magnetic field components of the electromagnetic wave. This maximum displacement is known as the amplitude of the electric and magnetic components of the electromagnetic wave.