You measure the power delivered by a battery to be 4.26 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it

Answers

Answer 1

Answer:

The power delivered by the battery is 17.04 W

Explanation:

Power through a circuit is given as

P = IV    ....1

where P is the power

I is the current through the circuit

V is the voltage through the circuit

The voltage in a circuit is given as

V = IR    ....2

Let us take the value of each resistor as equal to R

when connected in series, the total resistance will be

[tex]R_{t}[/tex] = R + R = 2R

If we assume constant voltage through the circuit, then from equation 2, the current in this case is

I = V/2R

If the resistors are connected in parallel, then the total resistance will be

[tex]\frac{1}{R_{t} }[/tex] = [tex]\frac{1}{R}[/tex] +

[tex]R_{t}[/tex] = R/2

The current in this case will be increased since the resistance is reduced

I = 2V/R

comparing the two situations, we can see that the current increased when connected in parallel to a ratio of

[tex]\frac{2V}{R}[/tex] ÷  [tex]\frac{V}{2R}[/tex] =  

This means that the current increased 4 times

From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to

P = 4 x 4.26 = 17.04 W


Related Questions

If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ray in the air make with the normal?

Answers

Answer:

29°

Explanation:

because the refracted ray angle is small than angle of incidence

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]

a person Travels along a straight road for half the distance with velocity V1 and the remaining half the distance with velocity V2 the average velocity is given by​

Answers

Answer: (V1+V2)/2

Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2

Nuclear energy is over ___ times stronger than the chemical bonds between the atoms

Answers

Answer:

1 millions times stronger

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.

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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g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube

Answers

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

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

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

The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at time t = -0.0s, with a velocity of 30m/s. The acceleration due to gravity at the surface of the moon is 1.62m/s2 the time when the rock is ascending at a height of 180m is closest to:______.
a. 8s .
b. 12s.
c. 17s.
d. 23s.
e. 30s
For the previous situation, the height of the rock when it is descending with a velocity of 20m/s is closest to:_____.
A. 115m.
B. 125m.
C. 135m.
D. 145m
E. 155m.

Answers

Explanation:

Given that,

Initial speed of the rock, u = 30 m/s

The acceleration due to gravity at the surface of the moon is 1.62 m/s².

We need to find the time when the rock is ascending at a height of 180 m.

The rock is projected from the surface of the moon. The equation of motion in this case is given by :

[tex]h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2[/tex]

It is a quadratic equation, after solving whose solution is given by:

t = 7.53 s

or

t = 8 seconds

(e)If it is decending, v = -20 m/s

Now t' is the time of descending. So,

[tex]v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s[/tex]

Let h' is the height of the rock at this time. So,

[tex]h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m[/tex]

or

h' = 155 m

The valid digits in a measurement are called _____ digits. Question 10 options: insignificant significant uncertain non-zero

Answers

Answer:

Significant

Explanation:

Valid digits in measurements are called significant digits, or also called significant figures.

These significant digits allow data and measurements to be more accurate and exact.

Answer:

Significant digits

Explanation

Took the test got it right

A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?

Answers

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

The bar magnet is pushed toward the center of a wire loop. Looking down from the top view (would appear the magnet is coming up toward the observer); Which is true? A. There is no induced current in the loop B. There is a counterclockwise induced current in the loop C. There is not enough information to correctly answer the question D. There is a clockwisee induced current in the loop

Answers

Answer:

Explanation:

B. There is a counterclockwise induced current in the loop

Explanation:

This in line with the right hand grip rule,

The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

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

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 34.1 m/s2 with a beam of length 5.55 m , what rotation frequency is required?

Answers

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

f = 0.4 Hz

Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. SOLUTION

Answers

Answer:

Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone

answer : -6661.59 volts

Explanation:

The total electric potential can be calculated using this relation

V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]

q 1 = 1.62 uc

r1 = 4.00 m

q2 = -5.73 uc

r2 = 5.00 m  

k = 8.99 * 10^9 N.m^2/c^2

insert the given values into the above equation

V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex]  =  -6661.59 volts

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

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 jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in meters per second squared right before taking off? Neglect air resistance and friction.

Answers

Answer:

The acceleration is   [tex]a =2\ m/s^2[/tex]

Explanation:

From the question we are told that

       The  mass of the jumbo jet is  [tex]m_j = 100000\ kg[/tex]

        The thrust is  [tex]F_k = 50000 \ N[/tex]

Generally given that the jet has four engines the total thrust is  

        [tex]F_t = 4 * F_k[/tex]

substituting values

       [tex]F_t = 4 * 50000[/tex]

      [tex]F_t = 200000 \ N[/tex]

Generally the acceleration of the is mathematically represented as

         [tex]a = \frac{F_t}{m}[/tex]

substituting values

       [tex]a =2 \frac{N}{kg}[/tex]

Now  

        [tex]N = kg \cdot m/s^2[/tex]

Hence

         [tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]

        [tex]a =2\ m/s^2[/tex]

You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.

Answers

Answer:

It will take. the same distance up as before, but take a longer time

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

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]

For more information on this visit

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A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3

Answers

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

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

Suppose there are only two charged particles in a particular region. Particle 1 carries a charge of +q and is located on the positive x-axis a distance d from the origin. Particle 2 carries a charge of +2q and is located on the negative x-axis a distance d from the origin.

Required:
Where is it possible to have the net field caused by these two charges equal to zero?

1. At the origin.
2. Somewhere on the x-axis between the two charges, but not at the origin.
3. Somewhere on the x-axis to the right of q2.
4. Somewhere on the y-axis.
5. Somewhere on the x-axis to the left of q1.

Answers

Answer:

x₂ = 0.1715 d

1) false

2) True

3) True

4) false

5) True

Explanation:

The field electrifies a vector quantity, so we can add the creative field by these two charges

             E₂-E₁ = 0

             k q₂ / r₂² - k q₁ / r1₁²= 0

             q₂ / r₂² = q₁ / r₁²

suppose the sum of the fields is zero at a place x to the right of zero

          r₂ = d + x

          r₁ = d -x

we substitute

           q₂ / (d + x)² = q₁ / (d-x)²

we solve the equation

           q₂ / q₁ (d-x)² = (d + x) ²

           

let's replace the value of the charges

       q₂ / q₁ = + 2q / + q = 2

          2 (d²- 2xd + x²) = d² + 2xd + x²

          x² -6xd + d² = 0

we solve the quadratic equation

          x = [6d ± √ (36d² - 4 d²)] / 2

          x = [6d ± 5,657 d] / 2

          x₁ = 5.8285 d

          x₂ = 0.1715 d

      with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d

this value remains on the positive part of the x axis, that is, near charge 1

now let's examine the different proposed outcomes

1) false

2) True

3) True

4) false

5) True

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

An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.

Answers

Answer:

(A). 1620 watt.

(B).0.8885.

Explanation:

So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;

Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.

(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).

(B). the rms values to find the power efficiency Pout/Pin of the inverter.

P(in) = 195 × 9.35 = 1823.3 watt.

Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.

what is angular frequency​

Answers

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

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

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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