Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3, 2, 1, 4 D. 3, 1, 2, 4

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:

The force is  [tex]F = 172 \ N[/tex]

Explanation:

From the question we are told that

    The  mass of the block is  [tex]m_b = 27.0 \ kg[/tex]

     The  coefficient of  static friction is  [tex]\mu_s = 0.65[/tex]

     The coefficient of kinetic friction is  [tex]\mu_k = 0.50[/tex]

The  normal force acting on the block is  

      [tex]N = m * g[/tex]

substituting values

     [tex]N = 27 * 9.8[/tex]

     [tex]N = 294.6 \ N[/tex]

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

        [tex]F_f = \mu_s * N[/tex]

substituting values

        [tex]F_f = 0.65 * 264.6[/tex]

        [tex]F_f = 172 \ N[/tex]

Now for this  block to move the force require is  equal to [tex]F_f[/tex] i.e

       [tex]F= F_f[/tex]

=>    [tex]F = 172 \ N[/tex]

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

Answer:

As a body moving upward

T=real weight + apparent weight

T=550+490

T=1040

hope u will get the answer:)

Explanation:

Answer:

c. 66 V

Explanation:

p =IV

I =P/V

P1/V1=P2/V2

60/110=36/V2

0.55 = 36/V2

V2 =36/0.55 = 65.5V

V2 = 66V

Answer:

The time taken is [tex]t = 0.0225 \ s[/tex]

Explanation:

From the question we are told that

    The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]

      The  refractive index is  [tex]n_f = 1.5[/tex]

The velocity of the signal is mathematically represented as

       [tex]v = \frac{c}{n_f }[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

 substituting values  

         [tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]

         [tex]v = 2.0*10^{8} \ m/s[/tex]

The time taken is mathematically evaluated as

      [tex]t = \frac{d}{v}[/tex]

substituting values  

      [tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]

      [tex]t = 0.0225 \ s[/tex]

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

 [tex]a_{max} = 0.00246 \ m/s^2[/tex]

b

   [tex]k =722.2 \ N/m[/tex]

Explanation:

From the question we are told that

     The  amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]

     The period is [tex]T = 17 \ s[/tex]

    The test weight is  [tex]W = 13 \ N[/tex]

Generally the radial acceleration is mathematically represented as

        [tex]a = w^2 r[/tex]

at maximum angular acceleration

       [tex]r = A[/tex]

So  

       [tex]a_{max} = w^2 A[/tex]

Now [tex]w[/tex] is the angular velocity which is mathematically represented as

      [tex]w = \frac{2 * \pi }{T}[/tex]

Therefore

       [tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]

substituting values

       [tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]

       [tex]a_{max} = 0.00246 \ m/s^2[/tex]

Generally this test weight is mathematically represented as

     [tex]W = k * A[/tex]

Where k is the spring constant

Therefore

        [tex]k = \frac{W}{A}[/tex]

substituting values        

      [tex]k = \frac{13}{0.018}[/tex]

     [tex]k =722.2 \ N/m[/tex]

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

Answer:

The correct answer is 88.84 mmHg.

Explanation:

The pressure differential between the brain and the heart while standing up will be 120 - rho × g (gravity) × h, here h is the distance from the brain to the heart. The h is 40 cm or 0.4 m.  

rho×g×h = 1060 kg/m³×9.8 m/s²×0.4m  

= 4155 Pa  

Now converting Pa to mmHg we get:  

4155 Pa × 760 mmHg / 1.01325 × 10⁵ Pa  

= 31.16 mmHg  

Thus, the pressure in the brain now is 120 - 31.16  

= 88.84 mmHg (hypotension)  

Answer:

A) 1.76U×10⁻³N

B) 2.716×10⁻³N

C) 264.5⁰

Explanation:

See detailed workings for (a), (b), (c) attached.

Answer:

Explanation:

v₁² = v₁ . v₁

= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )

= 12.25 + 8.41

= 20.66 m /s

a ) kinetic energy = 1/2 m v₁²

= 1/2 x .7 x 20.66

= 7.23 J

b )

changed velocity v₂ = v₂.v₂

= (6i - 5 j ) . (6i - 5 j )

= 36 + 25

= 61 m /s

kinetic energy = 1/2 m v₂²

= 1/2 x .7 x 61

= 21.35 J

Work done = change in energy

= 21.35 - 7.23

= 14.12 J .

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Answer:

C) Each successive stage lasts for approximately the same amount of time.

Explanation:

Nuclear burning is a series of nuclear processes through which a star gets its energy. The energy within a star is due to nuclear fusion of lighter elements (hydrogen) into more massive element (helium), with a release of a large amount of energy due to the conversion of some of the mass into energy. Each stage leads to a loss of some of the mass which is converted into energy (option A is valid).

The fusion of four hydrogen atoms into one helium atom means that there is a creation of element with a higher atomic weight (option D is valid), and the energy output of each stage exceeds its energy input, meaning that each stage will require a higher temperature than its previous stages (option B is valid).

A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.

Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.

Answer:

the smallest distance x  = 2.74 × 10⁻³ m or 2.74 mm

Explanation:

From the given information:

The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the  constructive interference is:

nΔy = mλ

where;

n = refractive index

Δy = path length (inside the glass)

So, from the diagram;

[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]

[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]

[tex]y = 5 \times 10^{-5} x[/tex]

Now;

Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])

Δy =1 ×  10⁻⁴[tex]x[/tex]

From nΔy = mλ

n( 1 ×  10⁻⁴[tex]x[/tex] ) = mλ

[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]

when the thickness is minimum then m = 1

Thus;

[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]

x =  0.00274 m

x = 2.74 × 10⁻³ m or 2.74 mm

Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.

Explanation: On Edge!!!!!!!!!!!!!!!!!!!!

Answer:

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

Refractive power is the measure of degree of convergence by a lens. The required refractive power for the given glasses is -4. 34 D.

Using lens formula  

[tex]\bold { \dfrac 1 v - \dfrac1 u = \dfrac {1}f}[/tex]

Where,

f-  focal point

v - distance of the image

u - distance of the object  

So,

[tex]\bold { \dfrac 1 {-17} - \dfrac1 {-65} = \dfrac {1}f}\\\\\bold { 0.434 = \dfrac {1}f}\\[/tex]

Since, [tex]\bold {power = \dfrac {- 100 }f}[/tex]

So,

[tex]\bold { power = - 100 \times 0.0434}}\\\\\bold { power = - 4.34\ D}[/tex]  

Therefore, the required refractive power for the given glasses is -4. 34 D.

To know more about  refractive power,

https://brainly.com/question/25164545

Answer:

a

Explanation:

The correct answer would be that the apparent depth of the object is less than the real depth.

The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.

Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.

The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.

Correct option: a

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

Answer:

Option A

Explanation:

From the graph, we came to know that Force and acceleration are in direct relationship.

Also,

Force = 0 when Acceleration = 0

Because Both are 0 at the origin.

Answer:

A. It will be 0 meters per second per second.

Explanation:

The force and acceleration is in a proportional relationship, that means the line goes through the origin.

On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.

Answer:

The  time interval is  [tex]t = 3 \ s[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = 4.0 \ rad/s^2[/tex]

     The  time taken is  [tex]t = 4.0 \ s[/tex]

      The angular displacement is  [tex]\theta = 80 \ radians[/tex]

The angular displacement can be represented by the second equation of motion as shown below

          [tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]

where  [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval

So substituting values

        [tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]

=>    [tex]w_i = 12 \ rad/s[/tex]

Now considering this motion starting from the start point (that is rest ) we have

       [tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]

Where  [tex]w__{0}}[/tex] is the angular velocity at rest which is zero  and  [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s

        [tex]12 = 0 + 4 t[/tex]

=>       [tex]t = 3 \ s[/tex]

Following are the response to the given question:

Given:

[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]

[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]

To find:

[tex]\to \omega=?\\\\\to t=?\\\\[/tex]

Solution:

Using formula:

[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]  

It would be the angle for rotation at the start of the 4-second interval.

This duration can be estimated by leveraging the fact that the wheel begins from rest.  

[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]

Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".

Learn more:

brainly.com/question/7464119

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

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.

Learn more about force here https://brainly.com/question/25239010

#SPJ2

Answer:

   R = r_protón / 2

Explanation:

The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement

      F = m a

Since the magnetic force is perpendicular, the acceleration is centripetal.

       a = v² / R

the magnetic force is

       F = q v x B = q v B sin θ

the field and the speed are perpendicular so the sin 90 = 1

we substitute

          qv B = m v² / R

          R = q v B / m v²

in the exercise they indicate

the charge  q = 2 e

the mass     m = 4 m_protón

        R = 2e v B / 4m_protón v²

we refer the result to the movement of the proton

         R = (e v B / m_proton) 1/2

the data in parentheses correspond to the radius of the proton's orbit

         R = r_protón / 2

Answer:

  B. matching your speed

Explanation:

To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*

_____

* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.

Answer:

It's C "signaling your intent"

Explanation:

The key thing to look at is they are asking the rest of the first step and that;s C

Answer:

Explanation:

Area of the loop = a b

current = I

magnetic moment of the loop M  = area x current

= ab I

Torque on the loop = MB sinθ

here θ = 90

Torque = MB

= abIB

In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .

Answer:

  x =  0.4654 cm

Explanation:

In this exercise we use the law of refraction

           n₁ sin θ₁ = n₂ sin θ₂

apply this formula to the first surface, where n₁ is the index of refraction of air (n₁ = 1) and n₂ is the index of refraction of glass (n₂ = n)

            θ₂ = sin⁻¹ (sin θ₁ / n)         (1)

having this angle we use trigonometry to find the value of the point where it comes out when we reach the other side

refracted ray

            tan θ₂ = x₂ / d

            x₂ = d tan θ₂

this value is the distance displaced by the refracted ray

now let's find the distance at which the incident beam should exit

           tan θ₁ = x₁ / d

           x₁ = d tan θ₁

the displacement of the ray is the difference between these two distances, we will call it x

           x = x₁ - x₂

            x = d tan θ₁ - d tan θ₂

           x = d (tan θ₁ - tan θ₂)        (2)

the easiest way to do the calculations is to find tea2 from the binding 1 and then perform the calculation with equation 2

calculate

            θ₂ = sin⁻¹ (sin 45 /1.5)

             θ₂ = 28.13º

             x = 1.0 (tan 45 - tan 28.13)

             x =  0.4654 cm

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

[tex]d (\frac{y}{L}) =[/tex] mλ

[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]

[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]

d = 1.0128×10⁻⁵m