TAKE 100 POINTS!!!please help look the picture​!!!!​

TAKE 100 POINTS!!!please Help Look The Picture!!!!

Answers

Answer 1

Answer:

A major challenge in the drug delivery field is to enhance transport of therapeutics across biological barriers such as the blood brain barrier (BBB), the small intestine, nasal, skin and the mouth mucosa.

Answer 2

Answer:

An aqueous stagnant layer that overlies the apical membrane and the subepithelial blood flow are potential barriers to the absorption of drugs that readily penetrate the absorbing cell of the epithelium. The apical, basal, and basement membranes are potential barriers to the absorption of less permeable drugs


Related Questions

During a football game player A (mass 120kg velocity 5.5m/s [L]) tackles and grabs player B (mass 105kg velocity 7.9m/s [R]). What is the final speed of the two players immediately after the collision?

Answers

Answer:

6.62m/s

Explanation:

Given data

M1= 120kg

U1= 5.5m/s

M2= 105kg

U2= 7.9m/s

The system experiences an inelastic collision, the expression for inelastic collision is

M1U1+ M2U1= (M1+M2)V

Subsitute

120*5.5+ 105*7.9= (120+105)*V

660+ 829.5= (225)*V

1489.5= (225)*V

Divide both sides by 225

V= 1489.5/225

V= 6.62m/s

Hence the common velocity is 6.62m/s

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around the inner core. The spool is free to roll without slipping on a horizontal surface. If the string unwinds horizontally from the bottom of the core with a constant speed of 25.0 cm/s, what is the speed of the spool

Answers

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

        v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

         w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

         w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

        [tex]\frac{v_o}{r_o} = \frac{v_1}{r_1}[/tex]

        v1 = [tex]\frac{r_1}{r_o} \ \ v_o[/tex]

let's calculate

       v₁ = [tex]\frac{1.50}{1.00} \ \ 25.0[/tex]

       v₁ = 37.5 cm / s

Pedro is planning to model how changes in weather affect evaporation from lakes for his first experiment he wants to test how humidity affects the evaporation rate. he places one beaker with 300 mL of water in a dry area. and places another beaker with 300 mL of water near a humidifier which of the following variables does Pedro need to control during his experiment
A. humidity only
B. humidity and evaporation rate
C. volume of water and tempature
D. volume of water only

Answers

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

What is meant by humidity ?

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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Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:


atoms
compound
element
heterogeneous
homogeneous
mixture
molecule
particles
solution
suspension

1. An _______________ is made up of just one kind of atom.

2. Matter is made up of tiny _______________.

3. In a mixture called a ________________ the particles are larger and can settle out.

4. A _______________ of water is made up of two hydrogen atoms and one oxygen atom.

5. Two or more elements chemically combined form a _______________.

6. _______________ are the building blocks of matter.

7. Seawater is an example of a _______________ mixture because it is the same throughout.

8. Vegetable soup is an example of a _______________.

9. In a _______________ one substance is dissolved in another.

10. A mixture that is not the same throughout is called a _______________ mixture.

Answers

Answer:

1).atoms (3). mixture. (5). Element

2). particles (4). molecules (6). suspension

Explanation:

(7). Homogeneous (8). Heterogeneous

(9). compound (10). solutions

While scuba diving, you have injured yourself and must signal the boat on the surface for help. As you always do when diving, you have five laser pointers with you, each emitting light of a different color: red, orange, yellow, green, and blue. When you shine the yellow light up toward the boat, the beam makes such a small angle with the surface of the water that all the light is reflected back into the water.

Required:
a. If you are not able to move closer to the boat, which color or colors of light might you try?
b. Are these other lights likely to be successful in penetrating the surface of the water?

Answers

Answer:

a. Red and Orange

b. Yes, they are

Explanation:

a. This is because, since the yellow light is reflected back into the water, it undergoes total internal reflection and its wavelength in water is not long enough to allow it penetrate the water surface. Since the wavelength decreases from left to right, both the green and blue light have shorter wavelength than the yellow light, so they get reflected back. The red and orange lights are more likely to penetrate since they have longer wavelengths than the yellow light.

b. This is because the red and orange light have longer wavelengths and are thus more likely to penetrate the water surface and not get reflected back.

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?​

Answers

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

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density ρ. ( R2 is measured from the center of the planet, not its surface.) Use G for the universal gravitational constant.

A) Find the kinetic energy of this satellite, K
Express the satellite's kinetic energy in terms of G, m, π, R1, R2, and ρ.



B) Find U, the gravitational potential energy of the satellite. Take the gravitational potential energy to be zero for an object infinitely far away from the planet.
Express the satellite's gravitational potential energy in terms of G, m, π, R1, R2, and ρ.



C) What is the ratio of the kinetic energy of this satellite to its potential energy?
Express K/U in terms of parameters given in the introduction.

Answers

Answer:

a)  

get mass of planet:  

ρ = M / V  

V = 4/3 * R_1^3  

M = ρ * V  

M = ρ * 4/3 * R_1^3  

equate force equations:

F = (GMm) / r^2     // r = R_2  

F = ma  

a = v^2/R_2  

F = m * (v^2/R_2)    

m * (v^2/R_2) = (GMm) / R_2^2  

plug in and solve v^2:  

m * (v^2/R_2) = (G * (ρ * 4/3 * R_1^3) *m) / R_2^2  

v^2 = (G * ρ * (4/3) * π * R_1^3) / R_2

 

put into kinetic energy equation:  

K = 1/2 * m * v^2  

K = 1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2

B)

givens:  

U = -(GmM) / R_2

 

plug in mass of planet:  

U = -(G * m * ρ * 4/3 * R_1^3) / R_2

C)

use previous equations and do some algebra:

K/U = (1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2) * -(R_2 / (G * m * ρ * 4/3 * R_1^3))  

K/U = -1/2

Which of the following best explains the greater difficulty in stopping a 1000-kg car
moving at 174 km/h compared to an identical 1000-kg car moving at 100 km/h?

Answers

Stopping a 1000 kg car moving at a speed of 174 km/h will be more difficult than stopping an identical car moving at 100 km/h.

This can be explained by using Newton's second law of motion which says that the rate of change of momentum of a body is equal to the force applied.

We know that momentum of a body is the product of its mass and velocity.The momentum of the car moving at 174 km/h = [tex]1000\, kg \times 174\, km/h = 174000 \, kg.km/h[/tex].The momentum of the car moving at 100 km/h = [tex]1000\, kg \times 100\, km/h=100000 kg.km/h[/tex].

Therefore, the car moving at 174 km/h has higher momentum and from Newton's second law of motion, a higher force would be required to stop this car.

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A bat emits a 40 kHz chirp to locate flying insects. If the speed of sound is 340 m/s and a bat hears the echo from the moth after 0.6 seconds, then how far away is the moth?

Answers

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

Please help me ASAP!!

Answers

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

g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops

Answers

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]

Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on the moon fM to its frequency on the earth fE. Suppose that gE is the free-fall acceleration on the earth and gM is the free-fall acceleration on the moon.
Express your answer in terms of some or all of the variables l, m, gE, gM.
fM/fE = ?

Answers

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

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

What frequency change would occur to the pendulum if it was taken to the moon?

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:

the density of brick is 1,600 kg/m3. what is the mass of a brick with a volume of 0.0006 m3? WILL MARK BRAINLIEST

Answers

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.

HOW TO CALCULATE MASS?

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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What is the answer to question 4

Answers

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.

If mass is a measure of inertia of a body at rest, then momentum is a measure of inertia of a body _____​

Answers

Answer:

Momentum is a measure of inertia in motion. Momentum is equal to mass multiplied by velocity. A 2250 kg pickup has a velocity of 25 m/s east.

Explanation:

Brainliest please?

Which diagram shows magnets that will attract each other? 2 bar magnets side by side with their long axes vertical, both red S on top and blue N on bottom. 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 blue N on top and red S on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with blue N on top and red S on bottom and the bottom magnet with red S on top and blue N on bottom. 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.

Answers

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.

What is Magnet ?

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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For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles

Answers

Answer:

A/A₀ = 0.5106

Explanation:

To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:

ω = 2π/T   (1)

But angular frequency (ω) can be also be calculated using:

ω = √(k/m)   (2)

Using (1) and (2), we can solve for the period T:

2π/T = √(k/m)    (mass in kg)

2π/T = √(130/0.360)

2π/T = √361.11

2π/T = 19.003

T = 2π/19.003 = 0.331 s

Now, the amplitude A at a time t, is:

A = x exp(-bt/2m)    (3)

At time 0, A = x. so A₀ = x

The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:

A = x exp(-b(17t/2m))     using b as kg/s = 0.086 kg/s

Replacing the data we have:

A = x exp(-0.086(17*0.331)/2*0.36)

A = x exp(-0.086 * 7.815)

A = x exp(-0.6721)

A = 0.5106x    (4)

Now, doing the ratio with the innitial we have:

A / A₀ = 0.5106x / x

The ratio is:

A/A₀ = 0.5106

Hope this helps

A ball is kicked at 10.4 m/s at an angle of 32 degrees to the horizontal

how long (time) is the ball in the air?

find the horizontal displacement (range) of the ball

Answers

Answer:

3M/S

Explanation:

Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution.
In rigid-body rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is I = cmr^2, where c is a dimensionless constant, m is the mass of the body, and r is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass m of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass?

Answers

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

The figure below shows regions of the electromagnetic
spectrum
Gamma
Radio Microwaves Infrared Visible Ultraviolet X-rays Rays
Which of the following waves has the highest frequency?
{GAMMA RAYS}

Answers

Answer:

The electromagnetic spectrum in order of increasing frequency is - radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

The frequency of the gamma rays is >3×10

19

m.

Hence, the gamma rays has the maximum frequency in the electromagnetic spectrum.

While diving in cancun Mexico where the seawater has a density of 1,015 kg/m3 Nana observed that her pressure meter device reading was 3.75 atm. The reading at sea level is standard 1.0 atm. At what depth is she diving when the meter read 3.75 atm g

Answers

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

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?

Answers

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)

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

Answers

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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Plz help w answer 1:/ confused ash

Answers

Answer:

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

Yeah that’s the same question I was asking

Help pls... 100points
1 Answer:
The pitch of a sound depends on
a. frequency
b. intensity
c. amplitude
d. quality
2 Answer:
The loudness of a sound depends on
a. frequency
b. intensity
c. overtones
d. timbre
3 Answer:
The number of sound waves per unit time is called
a. amplitude
b. wavelength
c. speed
d. frequency
4 Answer:
The blending of fundamental tone and overtones produces a sound's
a. intensity
b. pitch
c. timbre
d. loudness
5 Answer:
Sonar is used
a. to locate objects in water
b. to determine underwater distances
c. for both a and b
d. for neither a nor b
6 Answer:
Sounds of frequencies below 20 Hz are called
a. fundamentals
b. overtones
c. ultrasonic sounds
d. infrasonic sounds
7 Answer:
A sound is music if it has
a. a definite pitch
b. a pleasing quality
c. rhythm
d. all of these characteristics
8 Answer:
The stirrup is a part of the
a. outer ear
b. middle ear
c. inner ear
d. cochlea
9 Answer:
An example of a percussion instrument is a (an)
a. trumpet
b. harp
c. drum
d. flute
10 Answer:
The speed of sound depends on
a. temperature
b. density
c. elasticity
d. all of these factors
11 Answer:
Sounds with frequencies above 20,000 Hz are called _____.
12 Answer:
The science of sound is called _____.
13 Answer:
High-frequency waves are used in a sound navigation and ranging
system called _____.
14 Answer:
Sounds produced when sections of a string vibrate separately are
called _____.
15 Answer:
Sound without pleasing quality, identifiable pitch, and repeating
patterns is called _____.
16 Answer:
The speed of sound decreases with increasing temperature.
a. true
b. false
17 Answer:
Sound waves are transverse.
a. true
b. false
18 Answer:
The decibel is a unit of pitch.
a. true
b. false
19 Answer:
A change in the pitch of a sound due to motion of source or
receiver (observer) is called the Doppler effect.
a. true
b. false
20 Answer:
Guitars and cellos are examples of percussion instruments.
a. true
b. false

Answers

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

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

Answers

Answer: The Company has spent $5 million in research and development over the past 12 months developing cutting-edge battery technology which will be incorporated ...

Explanation: uhmmmmmm i dont know this one but it is pretty ez

A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1.50. At what angle with the plane of the surface must the incident ray strike the glass in order that the polarization of the reflected ray is the greatest

Answers

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]

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.

Answers

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.

What is the energy in a compressed spring?

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

Ek = 1/2 Kx^2

where

K is spring constant x is displacement of the spring

What is the kinetic energy of a body?

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:

KE = 1/2mv^2

How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground?

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.

Learn more about conservation of energy at: https://brainly.com/question/381281

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m/s. Then the vehicle travels for 39.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the self-driving car in motion (in s)?
(b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s

Answers

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].

The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reactor core spread over a wide area. The primary radiation exposure to people in western Europe was due to the short-lived (half-life 8.0 days) isotope 131I
131 I, which fell across the landscape and was ingested by grazing cows that concentrated the isotope in their milk. Farmers couldn't sell the contaminated milk, so many opted to use the milk to make cheese, aging it until the radiactivity decayed to acceptable levels. How much time must elapse for the activity of a block of cheese containing 131I 131 I to drop to 1.0% of its initial value?

Answers

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]

Let the initial activity be "[tex]R_o[/tex]".and, activity to time t be "R".

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]  

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