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
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
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)
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.
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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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
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
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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True or False - Water changing between states of matter is an example of a physical change.
true. true is the correct answer
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?
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.
THe Wilson cloud chamber is used to study the appearance of individual atoms?
Answer:
The Wilson cloud chamber is used to study the direction, speed, and distance of charged particles. Explanation; ... The Wilson cloud chamber works by producing a super-saturated vapor, as explained by florin.
Explanation:
I hope it's help u
If mass is a measure of inertia of a body at rest, then momentum is a measure of inertia of a body _____
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?
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
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]
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
Answer:
3M/S
Explanation:
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
Help please ASAP will give brainliest
Answer:
42.64
Explanation:
2x=2
7x=5
12x=1.4
17.5x=21.5
3
Explain the working principles
of 아
has/ Constant volume
thermometer
a gas
Answer:
A constant volume gas thermometer works on a thermometric property of " change in pressure with temperature at a constant volume"
a girl stands on a boat and feels the boat rise and fall 7 times In 1 minute. what is the period of the water wave? give your answer to 3 significant figures
Explanation:
hope it's helpful
have a nice day!!!! ;)
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?
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
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
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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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass. e.42 cm
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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?
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]
The key insight that Bohr introduced to his model of the atom was that the angular momentum of the electron orbiting the nucleus was quantized. He introduced the postulate that the angular momentum could only come in quantities of nh/(2π), where h is Planck's constant and n is a nonnegative integer (0,1,2,3,…). Given this postulate, what are the allowable values for the velocity v of the electron in the Bohr atom? Recall that, in circular motion, angular momentum is given by the formula L= mvr.
Answer:
v = [tex]n \frac{\hbar }{m r}[/tex]
the sppedof the electron is also quantized
Explanation:
The angular momentum of a rotating body is
L = m v r
in Bohr's atomic model the quantization postulate is that the angular momentum is equal to
L = n [tex]\hbar[/tex]
we substitute
n [tex]\hbar[/tex] = m v r
v = [tex]n \frac{\hbar }{m r}[/tex]
where n is an integer.
Therefore, the sppedof the electron is also quantized, that is, sol has some discrete values.
A ball is thrown downward from the top of a building with an initial speed of 25 m/s.
It strikes the ground after 2.0 s. How high is the building?
20 m
30 m
50 m
70 m
Answer:
h = 69.6 m
Explanation:
Data:
Vo = 25 m/st = 2.0 sg = 9.8 m/s²h = ?Formula:
[tex]\boxed{\bold{h=V_{0}*t+\frac{g*(t)^{2}}{2}}}[/tex]Replace and solve:
[tex]\boxed{\bold{h=25\frac{m}{s}*2.0\ s+\frac{9.8\frac{m}{s^{2}}*(2.0\ s)^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\frac{m}{s^{2}}+\frac{9.8\frac{m}{s^{2}}*4\ s^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+\frac{39.2\ m}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+19.6\ m}}[/tex][tex]\boxed{\boxed{\bold{h=69.6\ m}}}[/tex]The building has a height of 69.6 meters.
Greetings.
Hi!!! Please help me, 25 points!! :D
Which material refracts the light rays the least: air, water, or glass?
Answer:
air
Explanation:
water is a liquid and can be refracted.
glass is a solid which can also be refracted.
however,
air you can't see just see mainly.
air is the answer.
hope this helps :)
Which of the following statements correctly describes the index of refraction of a material? Select all that apply.
The index of refraction is the ratio of the speed of light in a material to the speed of light in a vacuum.
The index of refraction of a material must be greater than 1.
The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material.
The index of refraction of a material must be less than 1.
The index of refraction of water is less than the index of refraction of air.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.
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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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
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 = 0.5106Hope this helps
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
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].
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: 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
Plz help w answer 1:/ confused ash
Answer:
I would say d I had the same question yesterday and I got it correct so hope that helps
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
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]
Match these items.
1. effect observed when light from an object in cool air passes through warm air
2. line perpendicular to a surface
3. splitting of light into its component colors angle of incidence
4. alignment of light into a single vibrational direction
5. larger angle as light passes from air to water
6. cause of the sky's color
7. bouncing of light rays reflection
8. bending of light between media
9. ratio of speeds of light
10. inability of light to escape a low-velocity medium due to a large angle of approach refraction
a. scattering
b.dispersion
c.angle of incidence
d.index of refraction
e.mirage
f.reflection
g.normal
h.total inertial reflection
i.polarization
j.refraction
Answer:
1. effect observed when light from an object in cool air passes through warm air
→ h. total internal reflection
2. line perpendicular to a surface
→ g. normal.
3. splitting of light into its component colors angle of incidence
→ b. dispersion.
4. alignment of light into a single vibrational direction
→ i. polarization
5. larger angle as light passes from air to water
→ c. angle of incidence
6. cause of the sky's color
→ f. reflection
7. bouncing of light rays reflection
→ f. reflection
8. bending of light between media
→ j. refraction.
9. ratio of speeds of light
→ d. index of refraction.
10. inability of light to escape a low-velocity medium due to a large angle of approach refraction
→ scattering.