Fermat's principle is of least time rather than of least distance. Would least distance apply for reflection? For refraction? Why are your answers different?

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

[tex] E = \frac{Kq}{d^{2}} [/tex]

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]

The electric field due to charge 2 is:

[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]

[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].

(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.

The answer can be explained as follows.

Given that the two charges are;

(a) At the midpoint; [tex]r = 0.5\,m[/tex].

We know that the electric field due to charge [tex]q_1[/tex].

Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]

The electric field due to charge [tex]q_2[/tex] is given by;

Therefore, the net electric field in the midpoint is given by;

The direction is towards the right side.

(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.

So the force on the charge is ;

Find out more about electrostatic force and fields here:

https://brainly.com/question/14621988

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

Answer:

in the direction of the applied force

Explanation:

Answer:

a) m=1, y₁ = 0.08 m , θ₁ = 4.57º ,  b) m=2,  y₂ = 0.16 m , θ₂ = 9.09º ,  c)  m=3,   y₃ = 0.24 m ,   θ₃ = 13.5º

Explanation:

After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.

They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.

The expression for constructive interference for two slits is

             d sin θ = m λ

where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference

Let's use trigonometry to find the distance from the central maximum

         tan θ = y / L

in all interference experiments the angle is small,

         tan θ = sin θ / cos θ = sin θ

         sint θ = y / L

let's replace

        d y / L = m λ

        y = m λ L / d

let's calculate

distance to the first maximum m = 1

          y₁ = 1  400 10⁻⁹ 1/5000 10⁻⁹

          y₁ = 0.08 m

distance to second maximum m = 2

          y₂ = 2  400 10⁻⁹ 1/5000 10⁻⁹

          y₂ = 0.16 m

distance to the third maximum m = 3

          y₃ = 3  400 10⁻⁹ 1/5000 10⁻⁹

          y₃ = 0.24 m

with these values ​​we can search for each angle

           tan θ = y / L

           θ = tan⁻¹ y / L

for m = 1

            θ₁ = tan⁻¹ (0.08 / 1)

            θ₁ = 4.57º

for m = 2

            θ₂ = tan⁻¹ (0.16 / 1)

            θ₂ = 9.09º

for m = 3

            θ₃ = tan⁻¹ (0.24 / 1)

            θ₃ = 13.5º

Explanation:

velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]

lets call (h) 1 m to make it simple.

= 3.614 m/s

[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:

[tex]4×V_d=\sqrt(4/3hg)[/tex]

[tex]V_h=\sqrt(hg)[/tex]

velocity of hoop=[tex]\sqrt(gh)[/tex]

lets call (h) 1m to make it simple again.

[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s

[tex]\sqrt(gh) = sqrt(hg)

so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]

The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball [tex]=0.7v^2= gh[/tex]

solid disc [tex]= 0.75v^2 = gh[/tex]

hoop [tex]=v^2=gh[/tex]

The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]

let (h) be 1m again to compare.

[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s

solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]

uniform hoop speed [tex]=\sqrt(gh)[/tex]

solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]

Answer:

1. c. Same in both

2. a. Case 1

Explanation:

1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.

2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

Answer:

the size are components relative to the whole.

Explanation:

they are particularly good at showing percentage or proportional data

Answer:

in CGS system G is denoted as gram

Answer:

The correct answer is option 3 .

Please check the answer once :)

Answer:

The magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

Answer:

8.9 seconds

Explanation:

The height of the object at time t is:

y = h + vt − 4.9t²

where h is the initial height, and v is the initial velocity.

Given h = 30 and v = 40:

y = 30 + 40t − 4.9t²

When y = 0:

0 = 30 + 40t − 4.9t²

4.9t² − 40t − 30 = 0

Solving with quadratic formula:

t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)

t = [ 40 ± √(1600 + 588) ] / 9.8

t = 8.9

It takes 8.9 seconds for the object to land.

Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.

Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

Given Information:  

Radius of wire = r = 0.405 mm = 0.405×10⁻³ m

Length of wire = L = 15 m

Voltage = V = 12 V

Resistivity =  ρ = 100×10⁻⁸ Ωm

Required Information:  

Current = I = ?

Answer:  

Current = I = 0.412 A

Explanation:  

The current flowing through the wire can be found using Ohm's law that is

V = IR

I = V/R

Where V is the voltage across the wire and R is the resistance of the wire.

The resistance of the wire is given by

R = ρL/A

Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(0.405×10⁻³)²

A = 0.515×10⁻⁶ m²

So the resistivity of the wire is

R = ρL/A

R = (100×10⁻⁸×15)/0.515×10⁻⁶

R = 29.126 Ω

Finally, the current flowing through the wire is

I = V/R

I = 12/29.126

I = 0.412 A

Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.

Answer:

the ray of light should approach normal

Explanation:

When light passes through two means of different refractive index, it fulfills the equation

              n₁ sin  θ₁ = n₂ sin θ₂

where index 1 and 2 refer to each medium

In this problem, they tell us that light passes to a medium with a higher index, which is why

               n₁ <n₂

let's look for the angle in the second half

            sinθ₂ = n₁ /n₂  sin θ₁

            θ₂ = sin⁻¹ (n₁ /n₂  sin θ₁)

let's examine the angle argument the quantity n₁ /n₂ <1   therefore the argument decreases, therefore the sine and the angle decreases

Consequently the ray of light should approach normal

Answer:

Explanation:

The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .

Let the charge be - q .

force on charge

= q x E where E is electric field

= q x 680

weight = 1.6 x 10⁻³ x 9.8

so

q x 680 = 1.6 x 10⁻³ x 9.8

q = 1.6 x 10⁻³ x 9.8 / 680

= 23 x 10⁻⁶ C

- 23 μ C .

Answer:

θ = 33.8

a = 3.42 m/s²

Explanation:

given data

mass m = 2 kg  

coefficient of static friction μs = 0.67

coefficient of kinetic friction μk = 0.25

solution

when block start slide

N = mg cosθ    .............1

fs = mg sinθ   ...............2

now we divide equation 2 by equation 1 we get

[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]

[tex]\frac{\mu s N }{N}[/tex]  = tanθ

put here value we get

tan θ = 0.67

θ = 33.8

and

when block will slide  then we apply newton 2nd law

mg sinθ - fk = ma    ...............3

here fk = μk N = μk mg cosθ

so from equation 3 we get

mg sinθ -  μk mg cosθ = ma

so a will be

a = (sinθ - μk cosθ)g

put here value and we get

a = (sin33.8 - 0.25 cos33.8) 9.8

a = 3.42 m/s²

Answer:

f = 357.29Hz

Explanation:

In order to calculate the fundamental frequency in the closed tube, you use the following formula:

[tex]f_n=\frac{nv}{4L}[/tex]       (1)

n: order of the mode = 1

v: speed of sound = 343m/s

L: length of the tube = 24cm = 0.24m

You replace the values of the parameters in the equation (1):

[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]

The fundamental frequency of in the tube is 357.29Hz

Answer:

2.95m

Explanation:

Using h= 2.5+ v²/2g

Where v= 3m/s

g= 9.8m/s²

h= 2.95m

Answer:

2P

Explanation:

See attached file

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

[tex]\omega=\omega_{0}+\alpha t[/tex]

Put the value in the equation

[tex]\omega=0.17+1.3\times1.7[/tex]

[tex]\omega=2.38(k)\ m/s[/tex]

We need to calculate the angular displacement

Using angular equation of motion

[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]

Put the value in the equation

[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]

[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]

[tex]\theta= 124.18^{\circ}[/tex]

We need to calculate the velocity at point A

Using equation of motion

[tex]v_{A}=v_{0}+\omega\times r[/tex]

Put the value into the formula

[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]

[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]

[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]

We need to calculate the acceleration at point A

Using equation of motion

[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]

Put the value in the equation

[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]

[tex]a_{A}=0.791j-0.851i[/tex]

[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]

Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]

(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]

Answer:

D. All of the above

Explanation:

The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.

James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure,  moratorium, and achievement.

Answer:

The kinetic energy is  [tex]KE = 5.67*10^{-18} \ J[/tex]

Explanation:

From the question we are told that

    The potential difference is  [tex]\Delta V = 36 \ volts[/tex]

The potential energy of the end  is mathematically represented as

        [tex]PEs = - q * \Delta V[/tex]

q  is the charge on an electron with a constant value of [tex]q = 1.60 *10^{-19} \ C[/tex]

       substituting values

      [tex]PE = - 1.60*10^{-19} * 36[/tex]

      [tex]PE = - 5.67*10^{- 18} \ J[/tex]

Now from the law of energy conservation

     The [tex]PE_e = KEe[/tex]

Where  [tex]KE _e[/tex] is the potential  energy at the end

 So  

        [tex]KE = 5.67*10^{-18} \ J[/tex]

The  negative sign is not includes because kinetic energy can not be negative

Answer:

Majorly the electric field is reduced among other effect listed in the explanation

Explanation:

In capacitors the presence of di-electric materials

1. decreases the electric fields

2. increases the capacitance of the capacitors.

3. decreases the voltage hence limiting the flow of electric current.

 The di-electric material serves as an insulator between the metal plates of the capacitors

0.17T

When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e

[tex]F_{C}[/tex] = [tex]F_{M}[/tex]     --------------(i)

But;

[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]   [m = mass of the particle, r = radius of the path, v = velocity of the charge]

[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]

Substitute these into equation (i) as follows;

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

Make B subject of the formula;

B = [tex]\frac{mV}{qr}[/tex]            ---------------(ii)

Known constants

m = 1.67 x 10⁻²⁷kg

q = 1.6 x 10⁻¹⁹C

From the question;

v = 1.4 x 10⁷m/s

r = 0.85m

Substitute these values into equation(ii) as follows;

B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]

B = 0.17T

Therefore, the magnetic field strength is 0.17T

Answer:

The ratio  is  [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              [tex]I = \frac{2}{3} * m r^2[/tex]

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       [tex]RE = \frac{1}{2}* I * w^2[/tex]

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

             [tex]w = \frac{v}{r}[/tex]

=>           [tex]w^2 = [\frac{v}{r}] ^2[/tex]

So

             [tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]

            [tex]RE = \frac{1}{3} * mv^2[/tex]

Generally the transnational  kinetic energy of this motion is  mathematically represented as

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

So  

      [tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]

       [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Answer:

The answer is A

Explanation:

Its A because he created a telescope to be able to observe stars.

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Answer:

The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.

Explanation:

Answer:

Key points that can be found in the realist philosophical position​ are as follows: