These steps are followed when using the half-life of carbon-14 to determine
the age of an object that contains carbon. What is the correct order of these
steps?
A. Use the half-life of carbon-14 to determine the number of half-lives
that have passed.
B. Measure the ratio of parent nuclei to daughter nuclei.
C. Use the number of half-lives that have passed to determine the age
of the object.
A. A,B,C
B. A,C,B
0 0
C. B, A,C
D. C, A, B
Answer: a different one is a.b.c
Explanation: still for ape.x
The correct order to determine the age of the an object using carbon-14 is C, A, B. Thus, option D is correct.
What is half life?
The half-life time is defined as the time taken by the radioactive element to reduce one half of its initial value. It is denoted by t(1/2).
To measure the age of an object, a radioactive isotope called carbon-14 is used. The half-life of carbon-14 is 5,730 years. All the objects in the universe consumes carbon in their lifetime and hence, carbon-14 is used to measure the age of the objects.
The process of determining the age of objects using carbon-14 is called Radiocarbon dating. All living organisms consume carbon in means of food and from atmosphere and when the plant and animals dies, the radioactive carbon atoms start decaying.
When it starts decaying, by using Carbon-14 the age of an object is calculated. The age is estimated by measuring the amount of carbon-14 present in the sample and comparing this carbon with the reference Carbon-14 isotope.
The amount of carbon in preserved plants is identified by:
f(t) = 10e {₋ct}
t = time in years when the plant dies( t= 0)
c = the amount of carbon-14 remaining in preserved plants.
The steps include to find the age of an object is :
1. Use the number of half-lives that have passed to determine the age of the object.
2. Use the half-life of carbon-14 to determine the number of half-lives that have passed.
3.Measure the ratio of parent nuclei to daughter nuclei.
Hence, from these steps the age of an object is determined. Therefore the correct solution is D) C, A, B.
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use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
What is the answer to question 4
In order to make it all the way from the opening to the detector, a wave has to travel through air, glass, water, plastic, and vacuum.
-- The siren and the tuning fork are sounds.
-- Sound cannot travel through vacuum.
-- That knocks out choices B, C, and D.
The answer is A.
Note: Making a big exception here. We don't do test questions on Brainly. That would be cheating. Don't let me catch you doing it again.
NO LINKS; The graph shows the motion of a train first moving, then stopping, then traveling again at a slower speed. Calculate the average speed for the entire trip.
20 m/s
8.3 m/s
10 m/s
0 m/s
Answer: 0 m/s. that is your answer i hope this help sorry if i am wrong
Explanation:
What does the area under the curve on a velocity-versus-time graph represent? ... then slows down to travel the last 40 miles in three hours. ... 20. A bicyclist travels the first 700 m of a trip at an average speed of 8 m/s, travels the ... to complete the trip at an average speed, for the entire trip of 440 km/h. ... 125) v1= 0 m/s.
wegut.
c) A body weighs 1.2N on the moon and 120N on
the earth. Calculate the density of the moon,
taking acceleration of free fall as 10ms on the
earth surface and gravitational constant as
6.67 x 10Nm’kg. The radius of the moon is
2740 km.
We/Wm = ge/gm = 120N/1.2N
or
gm = ge/100 = 0.1 m/s^2
density = mass/volume = 3M/(4pir^3)
Re-arranging this equation, we get
M/r^2 = (4/3)×pi×(density)×r
From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is
gm = G(M/r^2) = G×(4/3)×pi×(density)×r
Solving for density, we get the expression
density = 3gm/(4×pi×G×r)
= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)
= 130.6 kg/m^3
The dwarf planet Ceres contains over 50% of the mass of the main asteroid belt.
True
False (if is why)
False
Explanation:
Called an asteroid for many years, Ceres is so much bigger and so different from its rocky neighbors that scientists classified it as a dwarf planet in 2006. Even though Ceres comprises 25 percent of the asteroid belt's total mass, tiny Pluto is still 14 times more massive.
18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
heno desde una altitud de 60.0 m. Si la paca de heno pesa 175 N, ¿cuál es el momentum
de la paca antes de que golpee el suelo?
Answer:
Definimos momento como el producto entre la masa y la velocidad
P = m*v
(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)
Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.
Peso = m*9.8m/s^2 = 175N
m = (175N)/(9.8m/s^2) = 17.9 kg
Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.
Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:
Vx = 36m/s
Mientras que para la velocidad vertical, usamos la conservación de la energía:
E = U + K
Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)
Entonces al principio solo hay energía potencial:
U = m*g*h
donde:
m = masa
g = aceleración gravitatoria
h = altura
Sabemos que la altura inicial es 60m, entonces la energía potencial es:
U = 175N*60m = 10,500 N
Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:
10,500N = (m/2)*v^2
De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.
√(10,500N*(2/ 17.9 kg)) = 34.25 m/s
La velocidad vertical es 34.25 m/s
Entonces el vector velocidad se podrá escribir como:
V = (36 m/s, -34.25 m/s)
Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.
Reemplazando esto en la ecuación del momento obtenemos:
P = 17.9kg*(36 m/s, -34.25 m/s)
P = (644.4 N, -613.075 N)
Help please ( with out links ).
You are asked to design a spring that will give a 1070 kg satellite a speed of 3.75 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible.
(a) What must the force constant of the spring be?
(b) What distance must the spring be compressed?
Answer:
[tex]380697.33\ \text{N/m}[/tex]
[tex]0.138\ \text{m}[/tex]
Explanation:
m = Mass rocket = 1070 kg
v = Velocity of rocket = 3.75 m/s
a = Acceleration of rocket = 5g
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow kx=\dfrac{mv^2}{x}\\\Rightarrow kx=\dfrac{1070\times 3.75^2}{x}\\\Rightarrow kx=\dfrac{7250}{x}[/tex]
The force balance of the system is given by
[tex]ma=kx\\\Rightarrow m5g=\dfrac{7250}{x}\\\Rightarrow x=\dfrac{7250}{1070\times 5\times 9.81}\\\Rightarrow x=0.138\ \text{m}[/tex]
The distance the spring must be compressed is [tex]0.138\ \text{m}[/tex]
[tex]k=\dfrac{7250}{x^2}\\\Rightarrow k=\dfrac{7250}{0.138^2}\\\Rightarrow k=380697.33\ \text{N/m}[/tex]
The force constant of the spring is [tex]380697.33\ \text{N/m}[/tex].
Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the maximum value of the electric field associated with this radiation? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
The right answer is "1010 V/m".
Explanation:
The given values are:
Intensity,
[tex]I=1350 \ W/m^2[/tex]
[tex]c = 3.00\times 108 \ m/s[/tex]
[tex]\mu_0= 4\pi\times 10^{-7} \ T.m/A[/tex]
Now,
The electric field's maximum value will be:
= [tex]\sqrt{2\times u\times c\times I}[/tex]
On substituting the values in the above formula, we get
= [tex]\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}[/tex]
= [tex]\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}[/tex]
= [tex]1010 \ V/m[/tex]
The maximum value of the electric field associated with this radiation is :
1010 V/mGiven data :
Intensity ( I ) = 1350 W/m²
C = 3.00 * 10⁸ m/s
The maximum value of the electric field can be calculated using the equation below
[tex]E_{max} = \sqrt{2*u*c*I}[/tex] ----- ( 1 )
where : I = 1350 W/m², μ = 4π * 10⁻⁷, c = 3.00 * 10⁸
Insert values into equation ( 1 )
∴ [tex]E_{max}[/tex] = 1010 V/m.
Hence we can conclude that the maximum value of the electric field is 1010 V/m .
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Why does one side of the Moon’s surface always remain much colder than the other side?
A) The colder side of the Moon has a cloud cover over it.
B) The colder side of the Moon is entirely covered with ice.
C) The colder side of the Moon always faces away from the Sun.
D) The colder side of the Moon has many high-altitude mountains.
Answer:
it's C
Explanation:
The colder side of the Moon always faces away from the Sun.
A bicycle possesses 1000 units of momentum. what would be the bicycle's momentum if,
A.its velocity is doubled
B. its mass is tripled
How does the intensity of a sound wave change if the distance from the
source is increased by a factor of 3?
O A. The intensity decreases by a factor of 3.
O B. The intensity decreases by a factor of 9.
O c. The intensity increases by a factor of 9.
O D. The intensity increases by a factor of 3.
Answer: C- The intensity increases by a factor 9
Explanation: The intensity of a sound wave follows an inverse square law, that means that it is inversely proportional to the square of the distance: so the new distance is the intensity will increase by a factor 9.
My sentence- I hope that helped!
Intensity of a sound wave decreases by a factor of 9, if the distance from the source is increased by a factor of 3. Hence option B is correct.
Intensity of Sound is inversely proportional to the square of the distance from the sound source. Since sound waves carry its energy though a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing as second power of distance form the source.
Mathematically,
Intensity I ∝ 1/D²
If the distance from the source is increased by a factor of 3, Then
I ∝ 1/3² ∝ 1/9
Intensity ∝ 1/9
Hence option B is correct.
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What is the acceleration of a bicycle that goes from 3 m/s to 1 m/s in 2 seconds?
0.5 m/s2
1.0 m/s2
1.5 m/s2
-1.0 m/s2
Answer:
a=vf - vi/t
vf=final velocity
vi= initial velocity
t=time period
Now The bicycle went from 3ms to 1ms...
1.. It decelerated
2... Since it went from 3 to 1... 1 is the final velocity while 3 is the initial Velocity
Applying the Formula
a= 1-3/2
= -2/2
a= -1ms-²
Option D
Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows.
Liquid 1: ????1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: ????2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: ????3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]
P = 9622.9 Pa = 9.62 KPa
50 POINTS!!!!!!!
Two charged objects are positioned 5 cm away from each other. Describe the change in the force between these two objects when:
(a) the charge on one of the objects is increased
(b) the distance between the objects is increased to 10 cm
(c) the charge on both of the objects is decreased
Please answer all 3
Answer:
A. The shorter the distance the greater the force, so now the force would have increased because they haven't moved away from each other.
B. The force would decrease because they are now further apart.
C. the force wouldn't be as great as before but the force would still be high because they are not far apart from each other.
I hope this helps you( ◜‿◝ )♡
The plank you are going to walk has a length of 8.76 m and a mass of 16.35 kg. The plank is not bolted down but is to be placed so that 3.17 m of the plank are actually in contact with the boat. The rest of the board overhangs the jelly fish infested waters. You have a mass of 60.3 kg. What is the minimum mass you must put on the end of the board that is still on the ship so that you can successfully walk to the end of the plank without have the board rotate you into the water
Answer:
Explanation:
The center of gravity will act at the middle point that is at 4.38 m from the end of the plank that is on boat . 16.35 x 9.8 N will act at this point .
The plank will turn around a point at 3.17 m from the end of plank that is on boat .
Suppose the weight required be W . It will be placed at the end of the plank that is on boat . The position of man is on the extreme end of the plank that is over water . So its distance from turning point will be 8.76 - 3.17 = 5.59 m .
Taking torque about turning point of all the forces like , weight of plank , weight of the person walking and force due to W
W x 3.17 = 16.35 x ( 4.38 - 3.17 ) + 60.3 x ( 8.76 - 3.17 )
W x 3.17 = 19.78 + 337.07 = 356.85
W = 112.57 kg .
A wave is a disturbance that carries
A. water from one place to another.
B. sound from one place to another.
C. matter from one place to another.
D. energy from one place to another.
Answer:
(D) energy from one place to another
11. A candle is placed in front of a plane mirror. The calculated value of m,
the lateral magnification, is positive. What does the positive sign indicate
about the image?
O The image is enlarged.
The image is on the same side of the mirror as the object.
The image distance is greater than the object distance.
The image is upright.
Answer: If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.
(I think, I was also stuck on this question for a bit)
The positive sign indicate about the image that
b) The image is on the same side of the mirror as the object.
d) The image is upright.
What is magnification ?
Magnification is a quantification of comparing the size of the image with respect to the size of the object. It gives us information about the image in terms of how large or small is the image formed.
magnification = height of image / height of object
since , height of object is always positive as it is always kept upright hence , in order to make m positive , height of image need to be positive
also magnification = -(v/u )
v = image distance
u = object distance
u is always positive hence , in order to make m positive , v needs to be negative , which implies on the same side of the mirror as the object
correct option
b) The image is on the same side of the mirror as the object.
d) The image is upright.
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Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five seconds.
Answer:
E = 75 J
Explanation:
First, we will calculate the total power consumed by the five lamps:
[tex]Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W[/tex]
Now, the energy supply can be calculated as follows:
[tex]E = Pt[/tex]
where,
E = Energy = ?
t = time = 5 s
Therefore,
E = (15 W)(5 s)
E = 75 J
two 0.5 kg carts, one red and one green, sit about half a meter apart on a low friction track, you push on the red one with the constant force of 4N for 0.17m and then remove your hand. the cart moves 0.33 m on the track and then strikes the green cart. what is the work done by you on the two cart system?
Answer:
The work done by you on the two cart system is 2 N-m
Explanation:
Work done is the product of force and displacement.
W = F * D
Substituting the given values we get -
W =
[tex]4 * (0.17+0.33)\\= 2[/tex]
The work done by you on the two cart system is 2 N-m
Please help!!!!!! Motion and Forces
Please help I’ll mark you brainliest
Answer:
Percentage:
Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)
rr = also 50%, because it's also 2/4.
Phenotype:
Rr = heterozygous
rr = "hozygous" recessive
In addition, RR is "hozygous" dominant
Explanation:
They said the hozygous is a swearword LOL.
The school is 3, 000m meters from the mall. What is the distance in kilometers?
Answer:
3,000 meters = 3 kilometers
Explanation:
A rock is thrown by applying a force of 2.50N while doing 6.00J of work. Over what distance was the force applied?
O 0.417m
8.50m
2.40m
15.0m
Answer:
[tex]\boxed {\boxed {\sf 2.40 \ meters}}[/tex]
Explanation:
Work is the product of force and distance, so the formula is:
[tex]W= F \times d[/tex]
The work is 6.00 Joules, but
1 Joule (J) is equal to 1 Newton meter (N*m). Therefore, the work of 6.00 J is equal to 6.00 N*mThe force is 2.50 Newtons.
The known values are:
W= 6.00 N*m F= 2.50 NSubstitute the values into the formula.
[tex]6.00 \ N*m= 2.50 \ N *d[/tex]
We are solving for distance, so we must isolate the variable, d. It is being multiplied by 2.50 Newtons and the inverse of multiplication is division. Divide both sides by 2.50 N.
[tex]\frac {6.00 \ N*m}{2.50 \ N}=\frac{2.50 \ N * d}{2.50 \ N}[/tex]
[tex]\frac {6.00 \ N*m}{2.50 \ N}=d[/tex]
The units of Newtons cancel out.
[tex]\frac {6.00 m}{2.50 }=d[/tex]
[tex]2.40 \ m =d[/tex]
The force was applied to the rock over a distance of 2.40 meters and choice C is correct.
The chart lists the masses of four planets.
Planetary Masses
Planet
Mass
Neptune
1.02 x 1026
Uranus
8.68 x 1025
Mars
6.42 x 1023
Venus
4.87 X 1024
According to evidence that supports Einstein's general
theory of relativity, which list shows the planets that
would cause curvature in space-time from the least
amount of curvature to the greatest?
O Mars, Venus, Uranus, Neptune
O Neptune, Uranus, Venus, Mars
O Neptune, Uranus, Mars, Venus
O Venus, Mars, Uranus, Neptune
Answer:
I think it's A
Explanation:
It's definitely not B on edge
The correct option for the given question about Einstein's general
theory of relativity is Option A) Mars, Venus, Uranus, Neptune.
What is the Einstein's general theory of relativity?Albert Einstein established that the rules of physics apply to all non-accelerating observers in his theory of special relativity.He also demonstrated that the speed of light in a vacuum remains constant regardless of the velocity of an observer.The theory may be used to anticipate everything and describes how objects behave in space and time. For instance: that if there are black holes or not, if Gravity can causes light to bend, The way Mercury behaves when it is in orbit and many more interesting things.As a Conclusion, we can state that the planets who would cause curvature in space time from the least amount of curvature to the greatest will be in order Mars, Venus, Uranus, Neptune.
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A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Answer:
n = 1.56 x 10¹⁷ electrons
Explanation:
First of all, we will calculate the current passing through wire:
[tex]I = \frac{q}{t}[/tex]
where,
I = current = ?
q = charge = 9 mC = 0.009 C
t = time = 3.6 s
Therefore,
[tex]I = \frac{0.009\ C}{3.6\ s}\\\\I = 0.0025\ A = 2.5\ mA[/tex]
Now, for the same current in 10 s time the charge will be:
q = It = (0.0025 A)(10 s)
q = 0.025 C
Now, the number of electrons can be given as:
[tex]q = ne\\\\n = \frac{q}{e}\\\\[/tex]
where,
n = no. of electrons = ?
q = charge = 0.025 C
e = charge on single electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]n = \frac{0.025\ C}{1.6\ x\ 10^{-19}\ C}[/tex]
n = 1.56 x 10¹⁷ electrons
!! How much voltage is needed to generate a current of 20 Amps if a line has a resistance of 10 ohms ? How much power does the appliance from question number one give off ? If the appliance runs for 3 minutes , how much energy is used ? Please help me
Answer:
Power = 4000watts
Energy = 22.22Joules
Explanation:
Power = I²R
I is the current
R is the resistance
t is the time
Given the following
I = 20Amps
R = 10ohms
t = 3miuntes = 180secs
Substitute
P = 20²*10
P = 400*10
P = 4000Watts
Hence the amount of power used is 4000Watts
Energy used = Power/time
Energy used= 4000/180
Energy used = 22.22Joules
A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round
Answer:
The new angular speed of the merry-go-round = [tex]1.6rad/sec[/tex]
Explanation:
From angular momentum conservation
[tex]Iw_1 = (I + mr^2)w_2\\\\3*100 = (100 + 22*2^2)w_2\\\\300 = (100 + 88)w_2\\\\w_2 = \frac{300}{188}\\\\w_2 = 1.6rad/sec[/tex]
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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.10 m2 and whose thickness is 8 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
the percentage of heat lost by the window is 93.18%
Explanation:
Given the data in the question;
Area of glass [tex]A_{glass[/tex] = 0.10 m²
Thickness of glass [tex]t_{glass[/tex] = 8 mm = 0.008 m
Area of Styrofoam [tex]A_{styrofoam[/tex] = 11 m²
Thickness of Styrofoam [tex]t_{styrofoam[/tex] = 0.15 m
we know that;
Thermal conductivity of glass [tex]k_{glass[/tex] = 0.80 J/smC°
Thermal conductivity of Styrofoam [tex]k_{styrofoam[/tex] = 0.010 J/smC°
Now, temperature difference between outside and inside the walls and window is ΔT
So, In time t, heat lost due to conduction in the window will be;
[tex]Q_{glass[/tex] = [[tex]k_{glass[/tex] × [tex]A_{glass[/tex] × ΔTt] / [tex]t_{glass[/tex]
we substitute
[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008
[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008
[tex]Q_{glass[/tex] = 10(ΔT)t J
Also, the heat lost due to conduction in the wall be;
[tex]Q_{styrofoam[/tex] = [[tex]k_{styrofoam[/tex] × [tex]A_{styrofoam[/tex] × ΔTt] / [tex]t_{styrofoam[/tex]
we substitute
[tex]Q_{styrofoam[/tex] = [ 0.010 × 11 × ΔTt] / 0.15
[tex]Q_{styrofoam[/tex] = 0.7333(ΔT)t J
Now, Net heat lost in the wall and window is;
Q = [tex]Q_{glass[/tex] + [tex]Q_{styrofoam[/tex]
Q = 10(ΔT)t J + 0.7333(ΔT)t J
Q = 10.7333(ΔT)t J
So, the percentage of heat lost by the windows will be;
% of heat lost = [tex]Q_{glass[/tex] / Q
= 10(ΔT)t J / 10.7333(ΔT)t J
= 0.93167
= ( 0.93167 × 100 )%
= 93.18%
Therefore, the percentage of heat lost by the window is 93.18%