You are working in the finance department of Innotech Ltd (INT). The Company has spent $5 million
in research and development over the past 12 months developing cutting-edge battery technology
which will be incorporated into the electric vehicle market. INT now need to choose between the
following three options for bringing the product to market. These options are:
Option 1: Manufacturing the product “in-house” and selling directly to the market
Option 2: Licensing another company to manufacture and sell the product in return for a royalty
Option 3: Sell the patent rights outright to the company mentioned in option 2
Your task
Your manager, INT’s CFO, Mr Barry Smith, has asked you to evaluate the three different options and
draft a memo to the Board of Directors providing recommendations on the alternatives, along with
supporting analyses.
Mr Smith has outlined the following three (3) areas you need to cover in your memo:
a) Analyse base case figures for the three options and using NPV as the investment decision
rule;
b) Provide recommendations based on the base-case analyses;
c) Provide recommendations on further analyses and discuss factors that should be considered
prior to making a final decision on the three options (Note. You do NOT have to undertake
any further financial analyses).
Further details for the various options are as follows:
Option 1: Manufacturing the product “in-house” and selling directly to the market
Three months ago, INT paid an external consultant $1.5 million for a production plan and demand
analysis. The consultant recommended producing and selling the product for five years only as
technological innovation will likely render the market too competitive to be profitable enough after
that time. Sales of the product are estimated as follows:
In the first year, it is estimated that the product will be sold for $45,000 per unit. However, the price
will drop in the following three years to $40,000 per unit and fall again to $36,000 per unit in the
final year of the project, reflecting the effects of anticipated competition and improving technology
Year Estimated sales volume
(units)
1 5,200
2 4,600
3 4,200
4 3,800
5 3,600
In the first year, it is estimated that the product will be sold for $45,000 per unit. However, the price
will drop in the following three years to $40,000 per unit and fall again to $36,000 per unit in the
final year of the project, reflecting the effects of anticipated competition and improving technology
in the market. Variable production costs are estimated to be $29,000 per unit for the entire life of the
project.
Fixed production costs (excluding depreciation) are predicted to be $3 million per year and marketing
costs will be $1.6 million per year.
Production will take place in factory space the company owns and currently rents to another business for $2.5 million per year. Equipment costing $87 million will have to be purchased. This
equipment will be depreciated for tax purposes using the prime cost method at a rate of 10% per
annum. At the end of the project, the company expects to be able to sell the equipment for $37
million.
Investment in net working capital will also be required. It is estimated that accounts receivable will
be 30% of sales, while inventory and accounts payable will each be 25% of variable and fixed
production costs (excluding depreciation). This investment is required from the beginning of the
project because credit sales, inventory stocks and purchases on trade credit will begin building up
immediately. All accounts receivable will be collected, suppliers paid and inventories sold by the end
of the project, thus the investment in net working capital will be returned at that point. (Refer to
spreadsheet example provided in Assessment Details).
Option 2: Licensing another company to manufacture and sell the product in return for a royalty
Lion Batteries Ltd (LIB), a multinational corporation, has expressed an interest in manufacturing and
marketing the pro
​

Answer:

Im sorry I do not know how to do this, I hope you will be able to figure it out

Explanation:

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: [tex]f=\frac{v}{\lambda}[/tex]

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

[tex]f=\frac{340}{0.22} \\\\f=1545.454...[/tex]

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

Answer:

L = 0.1379 m = 13.79 cm

Explanation:

The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction

            a sin θ = m λ

the first zero occurs when m = 1

let's use trigonometry to find the angle

          tan θ = y / L

as in these experiments the angles are very small

          tan θ = sin θ /cos θ = sin θ

          sin θ = y / L

we substitute

          a y /L = λ

In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced

           a y / L = 1.22 λ

           L = a y / 1.22 λ

We search the magnitudes to the SI system

           a = 1.04 cm = 1.04 10⁻² m

           y = 9.09 10⁻⁶ m

           λ = 562 10⁻⁹ m

let's calculate

           L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]

           L = 1.379 10⁻¹ m

           L = 0.1379 m = 13.79 cm

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)

Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.

The amount of water vapor in the air is known as humidity. The humidity will be high if there is a lot of water vapour in the atmosphere.

Water can evaporate even at very low temperatures, but as the temperature rises, the rate of evaporation increases.

More surface molecules per unit of volume may be able to escape from a substance with a larger surface area, so it will evaporate more quickly.

The control variables in an experiment are the variables that the experimenter intends to keep constant always so as to limit their effect on the measurements of the relationship between the dependent and the independent variable.

Therefore, in order to have a proper measurement of the effect of humidity on evaporation rate, other variables such as temperature, and the volume of the water in the experiment investigations which affect evaporation rate by the provision of heat, (temperature) and their heat capacity, the volume, etc. should be controlled.

Hence,

Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.

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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

Answer:

the ball lose kentic energy and gains potential energy rolling upward

Answer:

c

Explanation:

when a ball is rolling down it loses potential and gains kinetic, but in this case since the ball is going upwards it is losing kinetic energy and gaining potentail like a roller coaser

Answer:

The depth of the diver is 28.01 m

Explanation:

Given;

density of the seawater, ρ = 1,015 kg/m³

standard sea level pressure, P₀ = 1.0 atm = 101,325 Pa

the final reading of her pressure, P₁ = 3.75 atm = 379968.75 Pa

acceleration due to gravity, g = 9.8 m/s²

Let the depth she was diving at the final pressure = h

This depth is calculated as;

P₁ = P₀  +  ρgh

P₁ - P₀ =  ρgh

[tex]h = \frac{ P_1 \ - \ P_o}{\rho g} = \frac{379968.75 \ - \ 101325}{1015 \ \times \ 9.8} = 28.01 \ m[/tex]

Therefore, the depth of the diver is 28.01 m

Answer:

I would say d I had the same question yesterday and I got it correct so hope that helps

Answer:

6.8 N.m

Explanation:

The computation of the magnitude of the magnetic torque on the coil is given below:

Given that

n = 400

d =  6.0 cm

Current  is I = 7.0 A

Angle is [tex]\theta[/tex] = 30 degree

Now

We know that

the magnitude of the magnetic torque is

= nIABsin[tex]\theta[/tex]

= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)

As

[tex]\theta[/tex] = (90° - Ф)

=  (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°

= 6.8 N.m

For the  pendulum taken to the moon, The frequency change that would occur is mathematically given as

[tex]\frac{Fmoon}{Fearth}=0.408[/tex]

Generally, the equation for the Time period  is mathematically given as

[tex]T=2\pi\sqrt{L/g}[/tex]

Therefore

[tex]\frac{Fmoon}{Fearth}=\frac{\sqrt{g/6L}}{\sqrt{g/6L}}\\\\\frac{Fmoon}{Fearth}=\sqrt{1/6}[/tex]

[tex]\frac{Fmoon}{Fearth}=0.408[/tex]

In conclusion, The frequency change

[tex]\frac{Fmoon}{Fearth}=0.408[/tex]

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Answer:

.408

Explanation:

Answer: [tex]7.96\ \mu C[/tex]

Explanation:

Given

The dimension of the plate is [tex]1\ cm\times 6\ cm[/tex]

The gap between the plate is [tex]0.001\ cm[/tex]

Voltage applied [tex]V=15,000\ V[/tex]

The capacitance of the capacitor is

[tex]C=\dfrac{\epsilon_o A}{d}\\\\C=\dfrac{8.85\times 10^{-12}\times 1\times 6\times 10^{-4}}{10^{-5}}\\\\C=53.1\times 10^{-11}\ F[/tex]

Charge acquired by the capacitor

[tex]\Rightarrow Q=CV\\\Rightarrow Q=53.1\times 10^{-11}\times 15,000\\\Rightarrow Q=796.5\times 10^{-8}\\\Rightarrow Q=7.96\times 10^{-6}\ C[/tex]

Answer: 5

Explanation: B is for sure 60°, c* cosB = 10*1/2 =5

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of [tex]131_{I}[/tex],

[tex]T_{\frac{1}{2} }= 8 \ days[/tex]

To find t when R will be "1%" of [tex]R_o[/tex], then

⇒ [tex]R=\frac{1}{100}R_o[/tex]

As we know,

⇒ [tex]R=R_o e^{-\lambda t}[/tex]

or,

∴ [tex]e^{\lambda t}=\frac{R_o}{R}[/tex]

By putting the values, we get

        [tex]=\frac{R_o}{\frac{R}{100} }[/tex]

        [tex]=100[/tex]

We know that,

Decay constant, [tex]\lambda = \frac{ln2}{T_{\frac{1}{2} }}[/tex]

hence,

⇒ [tex]\lambda t=ln100[/tex]

     [tex]t=\frac{ln100}{\lambda}[/tex]

        [tex]=\frac{ln100}{\frac{ln2}{8} }[/tex]

        [tex]=53.15 \ days[/tex]  

Answer:

0.5 m/s

Explanation:

acceleration= force times mass

Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².

A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.

From Newton's Second Law, force is expressed as;

F = m × a

Where is mass of object and a is the acceleration

Given the data in the question;

F = m × a

10kgm/s² = 20kg × a

a = 10kgm/s² ÷ 20kg

a = 0.5m/s²

Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².

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Answer:

.0000004

Explanation:

The mass of a brick with a volume of 0.0006 m³ and a density of 1600 kg/m³ is 0.96kg.

The mass of a substance can be calculated by multiplying the density of the substance by its volume. That is;

Mass = density × volume

According to this question, the density of brick is 1,600 kg/m3 and it has a volume of 0.0006m³. The mass is calculated as follows:

Mass = 0.0006 × 1600

Mass = 0.96kg

Therefore, the mass of a brick with a volume of 0.0006m³ and a density of 1600 kg/m³ is 0.96kg.

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Answer:

The tension in string will be "3.62 N".

Explanation:

The given values are:

Length of string:

l = 3 ft

or,

 = 0.9144 m

frequency,

f = 60 Hz

Weight,

= 0.096 lb

or,

= 0.0435 kgm/s²

Now,

The mass will be:

= [tex]\frac{0.0435}{9.8}[/tex]

= [tex]0.0044 \ kg[/tex]

As we know,

⇒  [tex]\lambda=\frac{2L}{n}[/tex]

On substituting the values, we get

⇒     [tex]=\frac{2\times 0.9144}{4}[/tex]

⇒     [tex]=0.4572 \ m[/tex]

or,

⇒  [tex]v=f \lambda[/tex]

⇒      [tex]=0.4572\times 60[/tex]

⇒      [tex]=27.432 \ m/s[/tex]

Now,

⇒  [tex]v=\sqrt{\frac{T}{\mu} }[/tex]

or,

⇒  [tex]T=\frac{m}{l}\times v^2[/tex]

On putting the above given values, we get

⇒      [tex]=\frac{0.0044}{0.9144}\times (27.432)^2[/tex]

⇒      [tex]=\frac{752.51\times 0.0044}{0.9144}[/tex]

⇒      [tex]=3.62 \ N[/tex]

Answer:

[tex]56.5\ \text{s}[/tex]

[tex]21.13\ \text{m/s}[/tex]

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}[/tex]

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is [tex]12.5+39+5=56.5\ \text{s}[/tex]

Distance traveled

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}[/tex]

[tex]s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}[/tex]

[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}[/tex]

The total displacement of the car is [tex]156.25+975+62.5=1193.75\ \text{m}[/tex]

Average velocity is given by

[tex]\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}[/tex]

The average velocity of the car is [tex]21.13\ \text{m/s}[/tex].

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.

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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102 m

Explanation:

The time 0.6 sec is the time it took for the sound to travel from the bat to the moth and back. So it took 0.3 sec for the sound to reach the moth. From the definition of speed, the distance of the moth d to the bat is given by

v = d/t ---> d = vt = (340 m/s)(0.3 sec) = 102 m

Answer:

Increases

Explanation:

The light is emerging from a denser medium to rarer medium so the average light speed increases.

When a light beam emerges from water into air, the average light speed increases.

      The bending of a ray passes at an angle from one medium to another in which the speed will be different, as when light passes from air into water and also from water into air can be known as refraction.

      When light travels from water to air, the angle from water to air the light  gets speeds up and then its changes direction. The light turns away from the normal line. Because the light moves from a denser medium to the rarer medium. But from air to water the speed will be decreased because in that case the light travels from the rarer medium to the denser medium.

Hence, the light beam when passes, the average speed gets increased.

So, Option A is the correct answer.

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Answer:

 m₂ = 4 kg

Explanation:

The moment of inertia is defined by

         I = ∫ r² dm

for bodies with high symmetry it is tabulated, for a spherical shell

        I = 2/3 m r²

in this case the first sphere has a radius of r₁ = 2m and a mass of m₁ = 1 kg, the second sphere has a radius r₂ = 1m.

They ask what is the masses of the second spherical shell so that the moment of inertia of the two is the same.

        I₁ = ⅔ m₁ r₁²

        I₂ = ⅔ m₂ r₂²

They ask that the two moments have been equal

        I₁ = I₂

        ⅔ m₁ r₁² = ⅔ m₂ r₂²

         m₂ = (r₁ / r₂) ² m₁

let's calculate

         m₂ = (2/1) ² 1

         m₂ = 4 kg

Answer:

[tex]\mu=41.5\textdegree[/tex]

Explanation:

From the question we are told that:

Water index of refraction [tex]i_w=1.33[/tex]

Glass index of refraction [tex]i_g=1.50[/tex]

Generally the equation for Brewster's law is mathematically given by

 [tex]\theta=tan^{-1}(\frac{i_g}{i_w})[/tex]

 [tex]\theta=tan^{-1}(\frac{1.50}{1.33})[/tex]

 [tex]\theta=48.44 \textdegree[/tex]

Therefore Angle of incident to plane  \mu (normal at 90 degree to the surface)

 [tex]\mu=90\textdegree-\theta[/tex]

 [tex]\mu=90\textdegree-48.44\textdegree[/tex]

 [tex]\mu=41.5\textdegree[/tex]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

         Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

         Em_f = K = ½ m v²

since friction is zero, energy is conserved

          Em₀ = Em_f

          1 / 2k x² = ½ m v²

          v = [tex]\sqrt{ \frac{k}{m} }[/tex]     x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

The energy in a compressed spring is elastic potential energy given by the formula:

where

The kinetic energy of a body is the energy the body the has due to it's motion.

Kinetic energy, KE, is givenby the formula below:

From the law of conservation of energy, the total energy in a closed system is conserved.

Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.

Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

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2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom. this diagram shows magnets that will attract each other. Hence option D is correct.

A permanent magnet is an item constructed of magnetised material that generates its own persistent magnetic field. A refrigerator magnet, for example, is commonly used to hold notes on a refrigerator door. Ferromagnetic (or ferrimagnetic) materials are those that can be magnetised and are strongly attracted to a magnet. These include the elements iron, nickel, and cobalt, as well as their alloys, some rare-earth metal alloys, and naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones that are strongly attracted to a magnet and are widely thought to be magnetic, all other substances respond weakly to a magnetic field via one of many different forms of magnetism.

Hence option D is correct.

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Answer:

ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]

Explanation:

In this experiment the system can be approximated as a tube with one end open and the other closed.

The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m

a node is formed at the closed end and a belly at the open end, so the resonance has the form

             λ = 4L                    1st harmonic

             λ = 4/3 L                third harmonic

             λ = 4/5 L                fifth harmonic

             λ = 4L / (2n + 1)     n = 0, 1, 2, (2n + 1)

This wave is a standing wave therefore energy density remains in place

                P = 1/2 ρ v (w S₀)²

angular velocity is related to frequency

               w = 2π f

we substitute

                E = P = 2π² ρ v f² S₀²

if this energy per unit area is transformed into heat

                E = m c_e DT

let's use the concept of density

                ρ = m / V

                m = ρ V

if there are no losses in the system

       ½ π² ρ v f² S₀² = ρ V c_e ΔT

       ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]

When analyzing this expression the temperature increase is

* quadratic at the frequency and maximum amplitude of the wave

* proportional to the speed of the wave in the tissue

* inversely proportional to tissue volume

we can approximate the volume of the tissue to the volume of a cylinder tube

            V = π r² L

           ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]

we calculate

            ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)

            ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]

Answer:

Material that allow the electrons to move freely in order to produce a current

Please mark as brainliest if answer is right

Have a great day, be safe and healthy  

Thank u  

XD  

Answer:

SiZe

Explanation:

bad bunny is the best!!!

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.

Answer:

Explanation:

1. a as pitch depends on frequency

2. b as loudness depends on intensity

3. d as that is definition of frequency

4. c as that is definition of timbre

5. c as sonar can do both

6. d as that is definition of infrasound

7. d

8. b as that is one of three bones in the middle ear

9. c as you hit the drum to make sound

10. d

11. ultrasound

12. acoustics

13. ultrasound

14.  harmonics? or resonance?

15. noise

16. b

17. b

18. b

19. a

20. b