Answer:
B or 2
Explanation:
Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems
B
digestive, circulatory, and excretory systems
C
circulatory, immune, and respiratory systems
D
digestive, respiratory, and circulatory systems
Answer:
The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.
Explanation:
so the answer is D
TRUE OR FALSE
2 QUESTIONS
NEED HELP ASAP
THX :)
LOTS OF POINTS :>
Answer: Both False
Explanation:
Our Milky Way Galaxy is a spiral galaxy. Some spiral galaxies are what we call "barred spirals" because the central bulge looks elongated
Irregualuar glaxyices are all over the place
Use the drop-down menus to complete each statement about tornado safety.
Before a tornado warning occurs, you should
✔ have a disaster plan ready.
If you’re indoors during a tornado, you should
✔ go to the basement.
If you’re outdoors during a tornado, you should
✔ lie in a ditch or low-lying area.
Answer:
correct tysmm
Explanation:
Two students Tim and Alane travel to South Dakota. Tim stands on Earth’s surface and enjoys some sunshine. At the same time, Alane descends into a gold mine where neutrinos are detected, Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
Explanation:
Neutrinos are otherwise called leptons. They are principal particles. A lepton is a rudimentary half-spin molecule that doesn't go through solid reactions. Neutrinos are not usually charged and exceptionally light weighted so they once in a while interface with other matter. Neutrinos are light weighted. Their mass is around 10⁻⁷ kg. A neutrino possesses a small radius, too little to ever be estimated. A little span and very less mass make them imperceptible. Since neutrinos have next to no mass. they travel at almost the speed of light and thus they arrive at the outside of the Sun in only 2 seconds, dissimilar to photons which take convoluted ways to arrive at the Sun's surface in a huge number of years.
The photon and neutrino, both were made in the Sun's center yet on various occasions. The neutrino is only a couple of minutes old though the photon is around 1,000,000 years of age. At the point when the photon was made in the Sun's center. it needed to venture out to the outside of the Sun. in any case, rather because of its hefty mass and cooperation with other matter, it headed out a crisscross way to the surface. Ordinarily, it was repulsed and it was sent back to the middle where it needed to begin once more. It required a large number of years for a photon to arrive at the outside of the Sun.
Nonetheless, when it arrived at the Sun's surface, it required just 8.8 minutes for the photon to arrive at Earth. The neutrino was anyway made only a couple of minutes prior in the Sun's center. Since it has an entirely irrelevant mass, little size, and no charge, it didn't interface with its environmental factors. So it just required 2 seconds for the neutrino to arrive at the Sun's surface. When it arrived at the Sun's surface, it arrived at the earth in about 8.8 minutes. with the photon. So both, photon and neutrino have various histories as the two of them were made at a hole of around 1,000,000 years.
What happens in the crushing can experiment?
Explanation:
When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped
What is local business?explain with exampls
what is the definition of mutual flux?
Answer:
Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another
in winter soap does not dissolve properly in winter why
Answer:
In winter, the temperature of the water is very low which makes the soap dissolve in a very small amount. ... Since the water molecules move faster, they come in contact with soap more often causing it to dissolve faster
Hope it helps...
Have a great day : )
Mars orbits the Sun in 1.87 Earth years. How far is Mars from the Sun?
Answer:
151.12 million miles i actually just did this question on a test
Explanation:
Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.
What is meant by one revolution?
The term "revolution" describes how an item moves in its orbit around another object. For instance, the 24-hour day is created as the Earth rotates on its own axis.
The 365-day year is made possible by the Sun's rotation of the Earth. A planet spins around a satellite.
While orbiting the Sun, Mars travels at an average speed of 53,979 miles per hour, which equates to 86,871 km per hour.
Thus, Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.
To learn more about One revolution, refer to the below link:
https://brainly.com/question/26289977
# SPJ2
You can hear sounds transmitted through alr, water, or steel but not through the empty vacuum of space. What type of wave carries sound?
gravity
light
mechanical
electromagnetic
Lil' Ricky is trying on his Halloween costume in front of a flat (plane) mirror. Lil Ricky stands 1.5 m from the mirror.
Which statement correctly describes the image formed in the mirror?
A)
It is upright and 1.5 m behind the mirror.
B)
It is upright and 1.0 m behind the mirror.
C)
It is inverted and 1.5 m behind the mirror.
D)
It is inverted and 1.5 m in front of the mirror.
Answer:
The correct answer is A) It is upright and 1.5m behind the mirror
Explanation:
Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.
A physics professor wants to demonstrate the large size of the henry unit. On the outside of a 16-cm-diameter plastic hollow tube, she wants to wind an air-filled solenoid with self-inductance of 1.0 H using copper wire with a 0.79-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).
Required:
a. How long will the plastic tube need to be?
b. How many kilometers of copper wire will be required?
c. What will be the resistance of this solenoid?
Answer:
a) the plastic tube need to be 24.7 m long
b) the kilometer of copper wire required is 15.7
c) the resistance of this solenoid is 5538.8 2 ohms
Explanation:
Given the data in the question;
we determine the length of the plastic tube. assuming the solenoid is long.
the self inductance of a long solenoid is;
L = μ₀n²πr²l
μ₀ = 4π × 10⁻⁷ T-m/A
where
n = number of turns per unit length
r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)
l = length of the solenoid or the length of the plastic tube
we find n = number of turns per unit length
given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter
number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter
So from our previous formula, we find l
L = μ₀n²πr²l
we substitute
1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)
1 = 0.04048 × l
l = 1 / 0.04048
l = 24.7 m
Therefore, the plastic tube need to be 24.7 m long
b)
n = number of turns per unit length = 1265.8 turns/metre
so, the length of the plastic tube over which the copper wire is to be wound,
number of turns of copper wire required = n × l
= 1265.8 turns/meter × 24.7 m
= 31,265.26 turns
Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.
so for each turn length of copper wire required = 2π × r
= 2π × 0.08 m
= 0.5026548 m
So copper wire required for 31,265.26 turns will be;
⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km
Therefore, the kilometer of copper wire required is 15.7
c)
p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m
Resistance = pl/a
where l is length of copper wire, a is cross sectional area;
diameter of copper wire is 0.79-mm
radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m
area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²
Resistance = pl/a
we substitute
Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]
Resistance = 5538.8 2 ohms
Therefore, the resistance of this solenoid is 5538.8 2 ohms
5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.
Answer: 0.8 g/cm
Explanation:
p= m/V
= 4 kg/ 5 liter
= 0.8
Answer:
.8
Explanation:
give them brainliest
An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
If you push with a power of 20 Watts
on a 150 Newton object, how long would
it take to push it over the 4.3 m?
Answer:
32.25 s
Explanation:
From the question,
P = W/t.............. Equation 1
Where P = Power, W = work done, t = time.
But
W = F×d................. Equation 2
Where F = force and d = distance
Substitute equation 2 into equation 1
P = F×d/t............... Equation 3
make t the subject of euqation 3
t = (F×d)/P............. Equation 4
Givn: F = 150 N, d = 4.3 m, P = 20 watts.
Substitute these values into equation 4
t = (150×4.3)/20
t = 32.25 s
A child of mass 28 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 7 m/s. At this instant the cord is 3.30 m long. (Take the +x direction to be horizontal and to the right, the +y direction upward and +z direction to be out of the page.)
a. At this instant, what is the magnitude of the rate of change of the child's momentum?
b. At this instant, what is the (vector) net force acting on the child?
Answer:
Explanation:
(A)
Force is said to be the rate of change of momentum.
At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).
[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]
[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]
(B)
The net force acting on the child is:
[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]
[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]
A block is sliding along a horizontal surface with 100 J of kinetic energy. The kinetic frictional force stops the block in 5 seconds.
What is the rate (in J/s) at which the force of kinetic friction does work?
Answer:
20 J/s
Explanation:
Given data
Kinetic energy=100 J
Time= 5 seconds
Hence the rate at which the kinetic friction work is
=100/5
=20 J/s
Therefore the answer is 20 J/s
A 1500-kg car traveling due north with a speed of 25 m/s collides head-on with a 4500-kg truck traveling due south with a speed of 15 m/s. The two vehicles stick together after the collision. What is the total momentum of the system prior to the collision
Answer:
OMG IM ON THE SAME QUESTION
Explanation:
A box with a mass of 0.5 slugs lies on an inclined plane making an angle of 30o with the horizontal. If the coefficient of kinetic friction between the box and the plane is 0.4, what is the magnitude of the force that must be applied parallel to the incline to keep the box moving down the incline at a constant speed
Answer: 10.98 N
Explanation:
Given
mass of box is [tex]m=0.5\ slugs\approx 7.3\ kg[/tex]
The coefficient of kinetic friction is [tex]\mu= 0.4[/tex]
It is given that on the application of force box started moving with constant speed i.e. there is no net external force
[tex]\Rightarrow F+mg\sin 30^{\circ}=\mu mg\cos 30^{\circ}\\\Rightarrow F=mg(\mu \cos 30^{\circ}-\sin 30^{\circ})\\\Rightarrow F=7.3\times 9.8(0.866\times0 .4-0.5)=-10.98\ N\\[/tex]
The negative sign indicates direction of force is opposite i.e. it must be applied upwards
A banana peel has lots of friction.
True or False
Answer:
False
Explanation:
I learned it the hard way trust me T^T
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent
Answer:
they are right it is a new moon
Explanation:
took the test
how many cubic feet are in 6,117 ounces
6.388 or 6.388468 cubic feet..
not sure
The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. The water pressure in the line at the street is 130 psig. If I flow 10 gpm through the pipe, what pressure would I expect when I get to my house. My house is 10 ft higher in elevation than the water line at the street
Answer:
The right solution is "126 Psi".
Explanation:
The given values are:
P₁ = 130 psig
i.e.,
= [tex]130\times 6.894[/tex]
= [tex]896.22 \ Kpa[/tex]
or,
= [tex]896.22\times 10^3 \ Pa[/tex]
Z₂ = 10ft
= 3.05 m
[tex]\delta[/tex] = 1000 kg/m³
According to the question,
Z₁ = 0
V₁ = V₂
As we know,
⇒ [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]
On substituting the values, we get
⇒ [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]
⇒ [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]
⇒ [tex]P_2=866330 \ P_a[/tex]
i.e.,
⇒ [tex]=866330\times 0.000145[/tex]
⇒ [tex]=126 \ Psi[/tex]
Linearly polarized light whose Jones vector is [0 1] (horizontally polarized) is sent through a train of two linear polarizers. The first is oriented with its transmission axis at 45 degrees and the second has its transmission axis vertical. Show that the emerging light is linearly polarized in the vertical direction; that is, the plane of polarization has been rotated by 90 degrees.
Answer:
Following are the solution to the given question:
Explanation:
The input linear polarisation was shown at an angle of [tex]2 \mu[/tex]. It's a very popular use of a half-wave plate. In particular, consider the case [tex]\mu = 45 \pm[/tex], at which the angle of rotation is [tex]90\pm[/tex]. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.
A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8
Answer:
Heat loss per unit length = 642.358 W/m
The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m
Explanation:
From the information given:
Diameter D [tex]= 100 mm = 0.1 m[/tex]
Surface emissivity ε = 0.8
Temperature of steam [tex]T_s[/tex] = 150° C = 423K
Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]
Velocity of wind V = 8 m/s
To calculate average film temperature:
[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]
[tex]T_f = \dfrac{423+293}{2}[/tex]
[tex]T_f = \dfrac{716}{2}[/tex]
[tex]T_f = 358 \ K[/tex]
To calculate volume expansion coefficient
[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]
From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;
Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s
Thermal conductivity k = 30.608 × 10⁻³ W/m.K
Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s
Prandtl no. Pr = 0.698
Rayleigh No. for the steam line is determined as follows:
[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]
[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]
[tex]Ra_{D} = 5.224 \times 10^6[/tex]
The average Nusselt number is:
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]
[tex]Nu_D = 23.29[/tex]
However, for the heat transfer coefficient; we have:
[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]
[tex]h_D = 7.129 \ Wm^2 .K[/tex]
Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]
Now;
To determine the heat loss using the formula:
[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]
[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]
Now; here we need to determine the Reynold no and the average Nusselt number:
[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]
However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;
[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]
SO, the heat transfer coefficient for forced convection is determined as follows afterward:
[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]
Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
what is the distance a train can travel if its speed is 20mph over a time of 5.6 hours (show all 3 steps)
Answer:
distance = 112 miles
Explanation:
its 12 miles every 0.6 in a hour
35.15 .. Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen
Answer:
[tex]1.199\ \mu\text{m}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength = 600 nm
D = Distance of the light source from screen = 3 m
y = Distance of first order bright fringe from center = 4.84 mm
d = Distance between slits
m = Order = 1
We have the relation
[tex]y=\dfrac{D\lambda}{d}\\\Rightarrow d=\dfrac{D\lambda}{y}\\\Rightarrow d=\dfrac{3\times 600\times 10^{-9}}{4.84\times 10^{-3}}\\\Rightarrow d=0.0003719\ \text{m}[/tex]
From the question we have
[tex]y=\dfrac{\dfrac{1}{2}3\lambda}{d}\\\Rightarrow \lambda=\dfrac{2}{3}yd\\\Rightarrow \lambda=\dfrac{2}{3}\times 4.84\times 10^{-3}\times 0.0003719\\\Rightarrow \lambda=0.000001199\ \text{m}=1.199\ \mu\text{m}[/tex]
The required wavelength of light is [tex]1.199\ \mu\text{m}[/tex].