A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?

Answers

Answer 1

Answer:

Around the center of the mirror


Related Questions

May someone help...please. Pretty please...

If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height? Assume that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same.
f(short)/f(avg)=?

Answers

We have that  the ratio of his walking pace  to the walking pace of a person of average height is

[tex]\frac{V_2}{V_1}=1.10[/tex]

given the assumption and the calculation given below

From the question we are told that:

Consider a person 18\% shorter than average

Let average height of a person be [tex]10m[/tex]

Therefore

The height of an [tex]18\%[/tex] shorter man is mathematically given as

H=10*0.18

H=8.2m

Generally, the equation for velocity is mathematically given by

[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]

Where we have the  Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same

Therefore

[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]

[tex]\frac{V_1}{V_2}={82}{100}[/tex]

[tex]\frac{V_2}{V_1}=1.10[/tex]

In conclusion

The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is

[tex]\frac{V_2}{V_1}=1.10[/tex]

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What is the smallest value of n for which the wavelength of a Balmer series line is less than 400 nm

Answers

Answer:

The smallest value is n= 2

Explanation:

The balmer equation is given below

1/λ = R(1/4 - 1/n₂²).

R= 1.0973731568508 × 10^7 m^-1

λ= 400*10^-9 m

(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)

(400*10^-9)/1.0973731568508 × 10^7

= 1/4 - 1/n²

364.51 *10^-16= 1/4 - 1/n²

1/n²= 1/4 -364.51 *10^-16

1/n² = 0.25-3.6451*10^-14

1/0.25= n²

4= n²

√4= n

2= n

The smallest value is N= 2

HELP!!! 35 point question. answer at least 3 correctly. please include equations and how you did it

Answers

Answer:

9. (B) ¼ Mv²

10. (A) √(3gL)

11. 20 N

12. 5 m/s²

Explanation:

9. The rotational kinetic energy is:

RE = ½ Iω²

RE = ½ (½ MR²) (v/R)²

RE = ¼ Mv²

10. Energy is conserved.

Initial potential energy = rotational energy

mgh = ½ Iω²

Mg(L/2) = ½ (⅓ ML²) ω²

g(L/2) = ½ (⅓ L²) ω²

gL = ⅓ L² ω²

g = ⅓ L ω²

ω² = 3g / L

ω = √(3g / L)

The velocity of the top end is:

v = ωL

v = √(3gL)

11. Sum of torques about the hinge:

∑τ = Iα

-(Mg) (L/2) + (T) (r) = 0

T = MgL / (2r)

T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)

T = 20 N

12. Sum of forces on the block in the -y direction:

∑F = ma

mg − T = ma

Sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a / R)

T = ½ Ma

Substitute:

mg − ½ Ma = ma

mg = (m + ½ M) a

a = mg / (m + ½ M)

Plug in values:

a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))

a = 5 m/s²

a reagent is added to the blue solution to identify the copper 2 ions name the blue solution​

Answers

Answer:

buriret i belive

Explanation:

.

Answer:

The blue solution is named copper sulfate

HELP ME PLZ FAST
There is more than 1 answer,
The picture is down

Answers

Answer:

test her prototype and collect data about its flight

help... Please help!!!!!!!!!!!

Answers

Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

The timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500 micro F capacitor and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?

Answers

Answer:

check 2 photos for answer

check 2 photos for answer

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm

Answers

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

A ball is thrown upward from a height of 432 feet above the​ ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. ​a) Find​ s(t), the function giving the height of the ball at time t. ​b) How long will the ball take to reach the​ ground? ​c) How high will the ball​ go?

Answers

Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet

By applying the integrations,

(a) [tex]S = 96t-16t^2+432[/tex]

(b) Time will be "t = 9".

(c) Height will be "576"

Given:

Height,

423 feet

Initial velocity,

96 feet/sec

According to the question,

(a)

Integrate v:

[tex]S = 96t-16t^2+C[/tex]

Initial Condition,

→ [tex]S = 96t-16t^2+432[/tex]

(b)

Hits the ground when,

S = 0

→ [tex]0=96t-16t^2+432[/tex]

→ [tex]t =9[/tex]

(c)

Maximum height when,

v = 0

→ [tex]0 = 96-32 t[/tex]

→ [tex]t = 3[/tex]

Now,

→ [tex]S = 96\times 3-16\times 3^2+432[/tex]

      [tex]= 576[/tex]

Thus the answer above is correct.

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When the adjustable mirror on the Michelson interferometer is moved 20 wavelengths, how many fringe pattern shifts would be counted

Answers

Answer:

The number of  fringe pattern shift is   m  = 40

Explanation:

 From the question we are told that

      The  Michelson interferometer is moved 20 wavelengths i.e  [tex]20 \lambda[/tex]

Generally the distance which the  Michelson interferometer is moved is mathematically represented as

         [tex]d = \frac{m * \lambda}{2}[/tex]

Here [tex]m[/tex] is the number of  fringe pattern shift

     So

         [tex]20 \lambda = \frac{m * \lambda}{2}[/tex]

         [tex]40 \lambda = m * \lambda[/tex]

        m  = 40

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?

Answers

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit

Answers

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

A sinusoidal voltage Δv = (100 V) sin (170t) is applied to a series RLC circuit with L = 40 mH, C = 130 μF, and R = 50 Ω.

Required:
a. What is the impedance of the circuit?
b. What is the maximum current in the circuit?

Answers

Answer:

See attached file

Explanation:

please help !!!!!!!!!!!!!!!!!! give the answer to the question i. which lighthouse will be warmer during the day time and why ? ii. which lighthouse will be warmer during the night time and why ? please help

Answers

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

Explanation:

Because the land conducts heat better than water the light house farthest away from the water will get hotter during as the ground will heat up faster than the water. But this also means that the ground will lose heat faster at night where the water won't making the light house closest to the water hotter at night.

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resultant magnetic field at the point y

Answers

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

what is angular frequency​

Answers

measures angular displacement per unit time.
Angular frequency =2pie/time period

Which one of the following frequencies of a wave in the air can be heard as an audible sound by human ear

Answers

Answer:

1,000 Hz hope this helps.

Explanation:

The sound said to be audible if it comes in the range of audible sound range. The audible sound is the specific frequency range of sound, which can be heard by human ears. The audible sound frequency range is 20Hz−20,000H

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?

Answers

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]

Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?

Answers

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainland.

Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?

Answers

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

   [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Explanation:

From the question we are told that

     The  area of  Manhattan is  [tex]a_k = 87.46 *10^{6} \ m^2[/tex]

      The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]

        The  thickness is  [tex]t = 500 \ m \\[/tex]

       

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

         [tex]V = 500 * 3.498*10^{8}[/tex]

         [tex]V = 1.75 *10^{11}\ m^3[/tex]

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

Here [tex]\rho_i[/tex] is the density of ice the value is  [tex]\rho _i = 916.7 \ kg/m^3[/tex]

=>   [tex]m_i = 916.7 * 1.75*10^{11}[/tex]

=>    [tex]m_i = 1.60 *10^{14} \ kg[/tex]

Generally the energy needed for the ice to melt is mathematically represented as

        [tex]Q = m _i * H_f[/tex]

Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is  [tex]H_f = 3.33*10^{5} \ J/kg[/tex]

=>    [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]

=>    [tex]Q = 5.34 *10^{19} \ J[/tex]

Considering part b

  We are told that the annual energy consumption is  [tex]G = 1.2*10^{20 } \ J / year[/tex]

So  the time taken to melt the ice is

      [tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]

        [tex]T = 0.445 \ years[/tex]

converting to days

      [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y

Answers

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

What is perfect machine ​

Answers

Answer:

Explanation:

A machine is which no part of the work done on the machine is wasted, is called an ideal or perfect machine

A perfect machine is when nothing Of the components break over time and last over long periods of use and over long periods of years used!

Difference between matter and energy

Answers

Answer:

Energy is the strength and vitality required for sustained physical or mental activity.

Matter occupies space and possesses rest mass, especially as distinct from energy.

Hope this helps! (づ ̄3 ̄)づ╭❤~

3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.​

Answers

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

How many heartbeats in a typical human lifetime? Enter your answer as a number (NOT as a power of ten) and in one significant figure.​

Answers

Answer:

20,000,000,000

Explanation:

As we've seen, humans have on average a heart rate of around 60 to 70 beats per minute, give or take. We live roughly 70 or so years, giving us just over 2 billion beats all up.Apr

When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass

Answers

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

A sports car accelerates uniformly from rest to 24 m/s in 6 seconds. Calculate the acceleration of the car

Answers

Answer:

a = 4m/s^2

Explanation:

Velocity(V) = uniform = 24m/s

time(t) = 6sec

Acceleration(a) = V/t

= 24/6

= 4m/s^2

When a sports car accelerates uniformly from rest to 24 m/s in 6 seconds,then acceleration of the car would be 4 m/s²

What are the three equations of motion?

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Note that these equations are only valid for a uniform acceleration.


As given problem sport car accelerates uniformly from rest to 24 m/s in 6 seconds then the acceleration of the car can be calculated by using the first equation of motion

v = u + at

As given the initial velocity u= 0

The final velocity v = 24 m/s

The time taken is t= 6 seconds

By substituting the respective values of velocity and  time

24 = 0+ a*6

a = 24/6

a = 4 m/s²

Thus, when a sports car accelerates uniformly from rest to 24 m/s in 6 seconds,then acceleration of the car comes out to be 4 m/s²

Learn more about  equations of motion from here

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Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave

Ez Physics question will mark brainliest.

Answers

Answer:

The answer is B. longitude wave

A force of pounds makes an angle of with a second force. The resultant of the two forces makes an angle of to the first force. Find the magnitudes of the second force and of the resultant.

Answers

This question is incomplete, the complete question is;

A force of 193 pounds makes an angle of  79°14' with a second force. The resultant of the two forces makes an angle of 27°0' to the first force. Find the magnitudes of the second force and of the resultant.

Answer:

magnitudes of the second force (vector) is 110.84 lb

the resultant force force has a magnitude of 239.85 lb

Explanation:

Given that;

Magnitude of resultant vector R = ?

Direction of resultant vector α = 27°0'

Magnitude of vector p = 193

Magnitude of vector Q = ?

Angle between two vectors ∅ = 79°14'

Using the formula

tan∝ = [ Qsin∅ / P + Qcos∅]

tan27°0' = [ Qsin79°14' / 193 + Qcos79°14' )

we cross multiply

(193tan27°0') + (Qcos79°14'tan27°0' ) = Qsin79°14'

Q = 193tan27°0' / (sin79°14' - cos79°14'tan27°0')

Q = 110.84 lb

Therefore magnitudes of the second force (vector) is 110.84 lb

Now

R = √( p² + Q² + 2PQcos∅ )

R = √( 193² + 110.84² + ( 2 × 193 × 110.84 × cos79°14'))

R = 239.85 lb

Therefore the resultant force force has a magnitude of 239.85 lb

In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a slab of glass with a thickness of 0.800 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.What is the wavelength of the light in the glass? Use 3.00×108 m/s for the speed of light in a vacuum. Express your answer using two significant figures.

Answers

Answer:

196 nm

Explanation:

Given that

Value of wavelength, = 490 nm

Time spent in air, t(a) = 17.5 ns

Thickness of glass, th = 0.8 m

Time spent in glass, t(g) = 21.5 ns

Speed of light in a vacuum, c = 3*10^8 m/s

To start with, we find the difference between the two time spent

Time spent on glass - Time spent in air

21.5 - 17.5 = 4 ns

0.8/(c/n) - 0.8/c = 4 ns

Note, light travels with c/n speed in media that has index of refraction

(n - 1) * 0.8/c = 4 ns

n - 1 = (4 ns * c) / 0.8

n - 1 = (4*10^-9 * 3*10^8) / 0.8

n - 1 = 1.2/0.8

n - 1 = 1.5

n = 1.5 + 1

n = 2.5

As a result, the wavelength of light in a medium with index of refraction would then be

490 / 2.5 = 196 nm

Therefore, our answer is 196 nm

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