Answer:Colloid - well compacted together but not one
Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
A battery is two or more individual cells connected together. Some large trucks utilize large 24 volt lead acid batteries. How many lead acid cells would be required to construct a battery with this voltage
Answer:
#_pile = 12 celdas
Explanation:
Lead acid sulfur batteries generate each cell a potential of 2 volts. By colonato to reach the voltage of 24 volts
#_pile = 24/2
#_pile = 12 cledas
serially connected
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
Both of these questions are the same but their answers in the answer key are different. Why?
A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
A student wants to start a small business in school. Write down six items that
he/she can sell in school at a profit.
Answer:
packets of pen
packets of pencil
copies
books
bottles
mask
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
What is a profit?Profit is the difference between the revenue earned by a business or individual and the costs incurred to produce the goods or services sold.
It is an important measure of financial success for companies and is often used to determine the value of a business.
We have,
Here are six items that a student can sell in school at a profit:
Homemade baked goods - cupcakes, cookies, brownies, and other treats can be sold individually or as a pack.
School supplies - items such as pens, pencils, erasers, rulers, notebooks, and binders are always in demand.
Drinks - bottled water, juices, and sodas are popular beverages that students may purchase during the school day.
Healthy snacks - fresh fruit, granola bars, and trail mix are nutritious snacks that many students are interested in buying.
Personalized accessories - items like keychains, bracelets, and bookmarks with unique designs or student names can be popular among peers.
Stickers - fun and colorful stickers can be sold individually or in packs and are often a favorite of younger students.
Thus,
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
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A conductor is placed in a steady external electric field. Which of the following is FALSE?
a) All excess charge is distributed on the surface of the conductor
b) The electric field inside the conductor is the same as the external electric field
c) The electric field is zero inside the conductor
d) the electric field just outside the surface of the conductor is perpendicular to the surface
Answer:
a
Explanation:
because the electric field doesn't effect the conductor and its goes into storage for later
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
A parallal capacitor consists of two Squere plates each of Side 25cm, 3. Omm apart. If a potential difference of 2000volts is applied, calculate the change in the plate with
1.air
2. paper of relative permittity 2.5, fully the space between them E=8.9×10^-12
Answer:poop
Explanation:
poop
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
list at least types of motion
Answer:
These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.
(I got this off the web so credits to the rightful owner and I hope you have good day :)
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
Calculate the magnitude of a gravitational force between two object 400kg and 800kg separated by a distance of of 45m (take G =6.67 * 10^-11 Nm^2 kg^-2)
Answer:
Explanation:
The formula is
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:
[tex]F_g=\frac{(6.67*10^{11})(400)(800)}{(45^2)}[/tex] and multiply and divide all that out to get
[tex]F_g=1.1*10^{-8}[/tex] It should really only be 1 significant digit since 400 and 800 both have only 1 significant digit, but I used 2. It should be
[tex]F_g=[/tex] 1 × 10⁻⁸ N
2. What is the average speed of an athlete who runs 1500 m in 4 minutes?
Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing
Answer:
The correct answer is "4.443 sec".
Explanation:
Given:
Mass of child,
= 34 kg
Mass of swing,
= 18 kg
Length,
= 4.9 m
The time period of pendulum will be:
T = [tex]2 \pi \sqrt{4g}[/tex]
= [tex]2 \pi \sqrt{\frac{4.9}{9.8} }[/tex]
= [tex]4.443 \ sec[/tex]
Answer:
The time taken to back and forth is 4.4 s .
Explanation:
Length, L = 4.9 m
let the time period is T.
Acceleration due to gravity, g = 9.8 m/s^2
Use the formula of time period
[tex]T = 2 \pi\sqrt{L}{g}\\\\T = 2 \times 3.14\sqrt{4.9}{9.8}\\\\T = 4.4 s[/tex]
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3
Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
difference between wavefront and wavelets
Answer:
A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?
Answer:
opposite direction
Explanation:
An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.
Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.
Answer:
Explanation:
negatively charged particle is placed inside uniform electric field
The force on the charge due to the electric field is
F = q E
when the charge is negative so the force on the charge is opposite to the direction of electric field.
The electric field is opposite to the force.
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C.
Answer:
Electric field at A = 9.28 x 10¹² N/C
Explanation:
Given:
K = 8.99 x 10¹² N/C
Missing information:
Length = 11 cm = 11 x 10⁻² m
q = 12.5 C
Find:
Electric field at A
Computation:
Electric field = Kq / r²
Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²
Electric field at A = 9.28 x 10¹² N/C
A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
Learn more about astronaut.
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While an elevator of mass 984 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between t= 0 and t=
4.00 s, the elevator's displacement is 5.00 m downward. What is the elevator's speed at t= 4.00 s?
m/s
Answer:
v = 5.15 m/s
Explanation:
At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N
As the cable tension is less than this value, the car must be accelerating downward.
7730 = 984(9.81 - a)
a = 1.95 m/s²
kinematic equations s = ut + ½at² and v = u + at
-5.00 = u(4.00) + ½(-1.95)4.00²
u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s
v = 2.65 + (-1.95)(4.00)
v = -5.15 m/s
A transparent. dielectric coating is applied to glass (εr = 4.μr=1, σ= 0) to eliminate the reflection of red light (wavelength in air of 750 nm).
a. What is the required dielectric constant and minimum thickness of the coating?
b. If violet light (wavelength in air of 420 nm) is shone onto this glass coating (6-0). what percentage of the incident power will be reflected?
Answer:
a) Dielectric constant ( λ ) = 750 * 10^-9 m
minimum thickness of coating ( d ) = 187.5 nm
b) 3.6%
Explanation:
Given data:
wavelength of red light in air = 750 nm
εr = 4
μr = 1, σ = 0
a) Determine the required dielectric constant and min thickness of coating used
Refractive index of coating ( n ) = √εr * μr = √4*1 = 2
the refractive index of glass( ng) = 1.5 which is < 2
λ = 750 * 10^-9 m
Dielectric constant ( λ ) = 750 * 10^-9 m
To determine the minimum thickness we will apply the formula below
d = m λ/2n ; where m = 1
∴ d = 750 nm / 2 ( 2 )
= 187.5 nm
minimum thickness of coating ( d ) = 187.5 nm
b) Determine the percentage of the incident power that will be reflected
R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2
= [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2
= 0.03628 = 3.6%
