Answer:
2.1576 MeV
Explanation:
r = Distance = [tex]2\times 10^{-15}\ m[/tex]
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
The resulting energy of the system is
[tex]U=\dfrac{k3q^2}{r}\\\Rightarrow U=\dfrac{8.99\times 10^9\times 3(1.6\times 10^{-19})^2}{2\times 10^{-15}}\\\Rightarrow U=3.45216\times 10^{-13}\ J[/tex]
Converting to MeV
[tex]\dfrac{3.45216\times 10^{-13}}{1.6\times 10^{-13}}=2.1576\ MeV[/tex]
The work needed to assemble an atomic nucleus is 2.1576 MeV
All things in the universe are made up of elements. There are different types of elements in our surroundings. These elements are as follows:-
CarbonHydrogenOxygen and etcAccording to the question, the formula used in the question is [tex]U= \frac{k3q^{2} }{r}[/tex]
As the data is given in the question the solution is:-
[tex]U =\frac{8.99*10^{9} * )3(1.6 *10^{-19})^{2} }{2* 10^{-15} }[/tex]
After solving the question, the value of U we get is [tex]U =3.45216 \ X \ 10^{-13} J[/tex]
The joule must be converted into Mev.Hence, the solution is -[tex]\frac{3.45216 \ X \ 10^{-13} J}{1.6*10^{-13} } = 2.1576 Mev[/tex]
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You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body is 10% efficient at converting chemical energy in your body into mechanical work.
A.) What is your internal chemical energy change?
B.) How long must you bike to convert 3.8×105 J of chemical potential while staying at this speed? (This amount of energy equals the energy released by the body after eating three slices of bread.)
A) The change in internal chemical energy is [tex]1.15\cdot 10^7 J[/tex]
B) The time needed is 1 minute
Explanation:
First of all, we start by calculating the power output of you and the bike, given by:
[tex]P=Fv[/tex]
where
F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed
v = 8.0 m/s is the velocity
Substituting,
[tex]P=(80)(8.0)=640 W[/tex]
The energy output is related to the power by the equation
[tex]P=\frac{E}{t}[/tex]
where:
P = 640 W is the power output
E is the energy output
[tex]t = 30 min \cdot 60 = 1800 s[/tex] is the time elapsed
Solving for E,
[tex]E=Pt=(640)(1800)=1.15\cdot 10^6 J[/tex]
Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by
[tex]\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J[/tex]
B)
From the previous part, we found that in a time of
t = 30 min
the amount of internal chemical energy converted is
[tex]E=1.15\cdot 10^7 J[/tex]
Here we want to find the time t' needed to convert an amount of chemical energy of
[tex]E'=3.8\cdot 10^5 J[/tex]
So we can setup the following proportion:
[tex]\frac{t}{E}=\frac{t'}{E'}[/tex]
And solving for t',
[tex]t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min[/tex]
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Which observation provided Albert Einstein the clue that he needed to explain the photoelectric effect? a) Light is made up of extremely small particles. b) Particles that are emitted from a strip of metal are electrons. c) Shining light on a metal strip produces emissions. d) Energy of electrons depends on light’s frequency, not intensity.
Answer:
b) Particles that are emitted from a strip of metal are electrons.
Explanation:
Which observation provided Albert Einstein the clue that he needed to explain the photoelectric effect?
b) Particles that are emitted from a strip of metal are electrons.
the metal is injected by photon which leads to electron emission called photo-electron.
Answer:
D is the correct answer
Explanation:
1. Describe the conditions at altitude that could limit the ability to perform physical activity.
2. What types of exercise are detrimentally influenced by exposure to high altitude and why?
3. When someone ascends to an altitude of over 1,500 m, describe the physiological adjustments that occur within the first 24 h.
4. Describe the theoretical advantage of living high and training low.
5. What are the health risks associated with acute exposure to high altitude and how can they be minimized?
Answer:
1. Lower Barometric Pressure
2. Aerobic exercise
3. Basal Metabolic Rate Increase
4. improvements in some short duration sea-level performances
5. Vomiting, insomnia , dyspnea, nausea, and headache
Explanation:
1. Lower Barometric Pressure
-Hypobaric environment of the particular altitude
-Low atmospheric pressure or Hypobaria
Reduced atmospheric pressure
- Reduced PO2 air breathed in
Reduced PO2
-Caps pulmonary oxygen diffusion from the lungs
-Restricts O2 transport to the tissues
Low PO2 content in the air
-Also known as hypoxia
Hypoxia
-Low O2
Associated Low PO2 in the blood
-Hypoxemia
2. what types of exercise are detrimentally influenced by exposure to high altitude and why
aerobic exercise
The oxygen requirements of aerobic exercise leads to increased heart rate and , resulting tiredness within a short period of time
3. When someone ascends to an altitude of over 1500 m, describe the physiological adjustments that occur within the first 24 h
- Basal Metabolic Rate Increase
- Arterial PO2 Reduction
- Progressive Decrease In Blood plasma volume
- Blood pH increases
- Decreases in Pressure Gradient
- Drop in Alveolar PO2
- Increases in Ventilation
4. Describe the theoretical advantage of living high and training low
The logic in this system is that acclimatising the body to altitude by living there, and maintaining training intensity by training at sea level provides the harnessing of the gains of altitude exposure and avoiding some of the negative effects of altitude exposure. There are improvements in some short duration sea-level performances by keeping to the above technique.
5. Vomiting, insomnia , dyspnea, nausea, and headache
Consider the ascent rate, the altitude intended to be ascended, and the experience of the individual with the altitude to reduce the likelihood for symptoms to manifest.
At high altitudes, low oxygen levels can lead to acute mountain sickness and limit sustained aerobic exercise. Physiological adjustments such as increased breathing rate (hyperpnea) and red blood cell production occur within the first 24 hours, and 'living high, training low' offers competitive advantages by leveraging acclimatization benefits. Acute high altitude exposure risks can be mitigated through gradual ascent, hydration, and avoiding strenuous activity.
Explanation:Conditions at high altitudes can significantly impede physical activity due to the low partial pressure of oxygen, which leads to low blood and tissue levels of oxygen. Acute mountain sickness (AMS) is a potential consequence of such exposure, marked by symptoms like headaches, disorientation, and nausea. These can become more severe during physical exertion, underlining the importance of acclimatization.
Exercise types that involve sustained aerobic performance, like endurance running or cycling, are especially detrimentally influenced at high altitudes. The availability of oxygen is crucial for aerobic respiration, and with less oxygen available, performance can drop significantly.
Within the first 24 hours of ascending to over 1,500 m, physiological adjustments occur, including an increase in breathing rate (hyperpnea) and depth to compensate for lower oxygen levels, a phenomenon also known as hyperventilation. Over days to weeks, the body starts to produce more red blood cells, improving the oxygen-carrying capacity of the blood.
The theoretical advantage of living high and training low centers around the physiological benefits of acclimatization, such as increased red blood cell count, while still being able to train at peak capacity at lower altitudes with more oxygen available.
Health risks associated with acute exposure to high altitudes include AMS, cerebral edema, and pulmonary edema. Minimizing these risks involves a gradual ascent to allow for acclimatization, staying hydrated, and avoiding over-exertion.
Write the definition of a function print Grade, which takes one parameter containing a string value and returns nothing. The function prints "Grade: " followed by the string parameter.
Answer:
def print_Grade(grade):
print("Grade: ",grade)
Explanation:
The above code segment is a function that takes grade as a string parameter and returns nothing.
The line 1 starts with def keyword.
The keyword is used to start a function and the function will be uniquely identified with it.
The name of the declared function is print_Grade
Variable grade is then declared along with the function.
Line 2 prints "Grade: " without the strings along with the value of grade variable.
The code segment is written in Python
To define a function named printGrade that takes a string parameter and prints 'Grade: ' followed by the string, you'd create a function in a programming language like Python without a return statement, which outputs the concatenated string.
Explanation:In programming, when you are asked to write the definition of a function like printGrade, which takes a string parameter and prints it with some additional text, you are essentially writing a small part of a program. For example, in Python, a simple function that meets the requirements might look like this:
def printGrade(grade):This function definition includes the name printGrade, a single parameter grade, and a body that executes the print statement, which is returning nothing. When you call printGrade with a string argument, it will output the text "Grade: " concatenated with the string provided as an argument to the function.
What type of material makes up the functional aspect of the transducer that creates the high-frequency sound?
a. Tungsten
b. Iron oxide
c. Silver/chromium alloy
d. Crystals
The type of material makes up the functional aspect of the transducer that creates the high-frequency sound is Crystals which is a ceramic.
Explanation:
High frequency of sound can be produced in the transducer with the help of materials like ceramic materials.These are non-metallic compounds. The ceramic materials are those that are brittle in nature. Crystals, glasses are some of the examples of ceramic materials.
Ceramic materials like crystals are used in transducers that makes them functionally fit to produce high frequency sounds. The materials that are given in the options tungsten, iron oxide and silver/chromium alloy are not fit for these type of materials manufacturing. These belongs to metals and metal oxides and alloy types of metals.
The scientific theory for the origin of our solar system, involving the condensation and collapse of interstellar material in a spiral arm of the Milky Way galaxy, is the ____________________ theory.
Answer:
The Solar Nebula theory
Explanation:
The solar nebula theory describes the origin and formation of the solar system. It explains how the solar system originated from a gaseous cloud, that was comprised of interstellar dust particles and gases. From this giant cloud, the sun formed at the center, and the rest of the planets were formed by the process of condensation. These planets gradually started to rotate and revolve around the sun and were controlled within an elliptical or circular path due to the presence of a strong gravitational force of attraction.
This solar system was constructed in a spiral arm of the huge milky way galaxy.
This solar nebula theory was the ancient accepted theory for the formation of the solar system.
The scientific theory for the origin of our solar system is the nebular hypothesis, which states that the Sun and the planets formed from the collapse of a giant cloud of gas and dust called a nebula.
Explanation:The scientific theory for the origin of our solar system, involving the condensation and collapse of interstellar material in a spiral arm of the Milky Way galaxy, is the nebular hypothesis. According to this hypothesis, the Sun and the planets of our solar system formed about 4.6 billion years ago from the collapse of a giant cloud of gas and dust, called a nebula. The material in the nebula gradually came together due to gravity, forming a spinning disk. Within this disk, smaller clumps of matter called planetesimals started to form, which eventually coalesced to form the planets and moons.
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the electric field at twice the distance from the point charge?
A) At twice the distance, the strength of the field is E/2.B) At twice the distance, the strength of the field is 2E.C) At twice the distance, the strength of the field is 4E.D) At twice the distance, the strength of the field remains equal to E.E) At twice the distance, the strength of the field is E/4.
Answer:
E
Explanation:
Using Coulomb's law equation
Force of the charge = k qQ /d²
and E = F/ q
substitute for F
E = ( K Qq/ d² ) / q
q cancel q
E = KQ / d²
so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²
What is the net force that accelerates an object on an inclined plane due to the effect of gravity, normal force and friction of the inclined plane on the object?
The net force that accelerates an object on an inclined plane is the parallel component of the weight minus the force of friction.
Explanation:When an object rests on an inclined plane that makes an angle with the horizontal surface, the weight of the object can be resolved into components that act perpendicular and parallel to the surface of the plane.
The net force that accelerates an object on an inclined plane is the component of the weight that acts parallel to the plane, minus the force of friction. The force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane (normal force) and a force acting parallel to the plane. The parallel component of the weight causes the object to accelerate down the incline.
If there is friction present on the inclined plane, the force of friction opposes the motion of the object and must be subtracted from the parallel component of the weight. The net force can then be calculated by subtracting the force of friction from the parallel component of the weight.
A rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0 degrees above the horizontal. How far from the base of the cliff will the rock hit?
Answer:
4.8063m
Explanation:
Horzontal range is given by the formula;
R=(u²sin2θ)/g
u=7m/s, θ=53°, g=9.8m/s
[tex]R=\frac{7^{2}sin2*53 }{9.8}[/tex]
[tex]R=\frac{7^{2}sin106 }{9.8}[/tex]
[tex]R=\frac{49*sin2*53 }{9.8}[/tex]
[tex]R=\frac{47.102 }{9.8}[/tex]
R=4.8063m
A rock thrown at an angle of 53 degrees with an initial speed of 7.0 m/s from a 50-m-height cliff will hit the ground approximately 8.3 meters away from the base of the cliff.
Explanation:The problem regards the range of a projectile which is given by the formula R = (v²/g) * sin(2*Theta), where v is the initial velocity, g is the acceleration due to gravity, and Theta is the launch angle. Given the initial velocity v = 7.0 m/s, launch angle Theta = 53 degrees, and g = 9.8 m/s², it's a matter of substituting these values into the range formula: R = ((7.0 m/s)² / 9.8 m/s²) * sin(2 * 53 degrees).
After performing the calculations, you obtain that the rock will hit ground approximately 8.3 meters away from the base of the cliff.
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If the cold temperature reservoir of a Carnot engine is held at a constant 331 K , what temperature should the hot reservoir be kept at so that the engine efficiency is 0.55 ?
Answer:
735.56 K
Explanation:
Carnot Engine: A Carnot engine is a an engine that operates on Carnot cycle.
A Carnot cycle: is defined as an ideal reversible closed thermodynamic operation, in which there are four successive and reversible cycles. These operations include isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
The efficiency of a Carnot engine is given below,
η=(Th-Tc)/Th.................................................... Equation 1
η = 1-(Tc/Th)........................................ Equation 2
Where Th = temperature of the hot reservoir, Tc = temperature of the cold reservoir, η = efficiency of the Carnot engine.
Given: Th = 331 K, η = 0.55
Substitute into equation 2
0.55 = 1-(331/Th)
0.55-1 = -331/Th
-0.45 = -331/Th
331/Th = 0.45
Th = 331/0.45
Th = 735.56 K.
Thus the temperature of the hot reservoir is 735.56 K.
To determine the hot reservoir temperature required for a Carnot engine to operate with a given efficiency, you can use the equation ηc = 1 - Tc/Th, where ηc is the engine's efficiency, Tc is the cold reservoir temperature, and Th is the hot reservoir temperature. Given an engine efficiency of 0.55 and a cold reservoir temperature of 331K, you can rearrange the equation to solve for the hot reservoir temperature.
Explanation:The efficiency of a Carnot engine is given by the equation ηc = 1 - Tc/Th, where ηc is the engine's efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir. Given that the engine's efficiency (ηc) is 0.55 and the cold reservoir temperature (Tc) is 331K, you can rearrange the equation to solve for the hot reservoir temperature (Th):
Th = Tc / (1 - ηc)
Substituting the given values into this equation yields the hot reservoir temperature required for the engine to operate with an efficiency of 0.55.
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From the edge of a cliff, a 0.49 kg projectile is launched with an initial kinetic energy of 1670 J. The projectile's maximum upward displacement from the launch point is 160 m. What are the (a) horizontal and (b) vertical components of its launch velocity?
Answer:
(a) Horizontal component of the launch velocity = 60.7m/s
(b) Vertical component of the launch velocity = 56.0m/s
Explanation:
Initial Kinetic Energy = (IKE)= 1670J, mass (m) = 0.49kg, hmax = 160m
IKE = 1/2mu^2
1670×2/0.48 = u^2
u^2 = 6816.3
u = √6816.3 = 82.6m/s
hmax = u^2sin^2A/2g
160×2×9.8/6816.3 = sin^2A
Sin^2A = 0.46
SinA = √0.46 = 0.6782
A = inverse (sin 0.6782)
A = 42.7° (angle of inclination to the horizontal)
(a) Horizontal component (Ux) = ucosA = 82.6×cos42.7° = 82.6×0.7349 = 60.7m/s
(b) Vertical component (Uy) = usinA = 82.6×sin42.7° = 82.6×0.6782 = 56.0m/s
To find the horizontal and vertical components of the projectile's launch velocity, use the principle of conservation of energy.
Explanation:To find the horizontal and vertical components of the projectile's launch velocity, we'll use the principle of conservation of energy. The projectile's initial kinetic energy is equal to its potential energy at its maximum displacement.
The projectile's initial kinetic energy is given as 1670 J. At the maximum displacement of 160 m, the potential energy is given as m*g*h, where m is the mass of the projectile (0.49 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum displacement (160 m).
Using these values, we can solve for the vertical component of the launch velocity using the equation for potential energy: m*g*h = 0.49 kg * 9.8 m/s^2 * 160 m. The horizontal component of the launch velocity remains unchanged throughout the projectile's motion.
A mountain is 1100 feet above sea level. A trench is 420 feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
Answer:1520feet
Explanation:
The difference in elevation between the mountain top and the bottom of the trench will be addition of the distance between the sea and mountain top and the distance between the sea level and bottom trench i.e 1100 + 420
= 1520feet
Planet X has a diameter that is 3 times the diameter of Earth and a mass that is 30 times the mass of Earth. In SI units, what is the gravitational acceleration on planet X?
Final answer:
The gravitational acceleration on planet X is approximately 34.4 m/s².
Explanation:
The gravitational acceleration on planet X can be determined using the equation:
g = (G * M) / r^2
Where:
g is the gravitational acceleration
G is the gravitational constant (approximately 6.674 × 10^-11 N m^2/kg^2)
M is the mass of the planet
r is the radius of the planet
Since the diameter of planet X is 3 times the diameter of Earth, its radius would be 1.5 times that of Earth. The mass of planet X is 30 times the mass of Earth.
Let's assume the radius of Earth (rE) is 6371 km and the mass of Earth (ME) is 5.972 x 10^24 kg.
Using these values, the radius of planet X (rX) would be 1.5 * rE = 9556.5 km, and the mass of planet X (MX) would be 30 * ME = 1.7916 x 10^26 kg.
Now, we can plug these values into the equation to calculate the gravitational acceleration on planet X:
gX = (G * MX) / rX^2
gX = (6.674 x 10^-11 N m^2/kg^2 * 1.7916 x 10^26 kg) / (9556.5 km)^2
Converting km to meters and solving for gX gives us approximately 34.4 m/s².
Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Earth is approximately 385,000 km, and data travels over the link at the speed of light (3 x 10⁸ meters per sec.).
Calculate the minimum RTT (two-way propagation time) for the link.(In other words, how long would it take to send a single bit to the moon and immediately bounce it right back to Earth?)
Answer:
The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s
Explanation:
Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m
Speed of data travel = speed of light ≈ 3 x 10⁸m/s
As, v=d/t
t=d/v
[tex]t=\frac{3.85*10^{8} }{3*10^{8}}[/tex]
t=1.28s
RTT = Double of single way time taken = 2x1.28
RTT=2.56s
Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by a 102.0 kg player traveling at 5.2 m/s toward her. If the 102.0 kg player is knocked backwards at 2.9 m/s, what is the resulting velocity of the 87.2 kg player?
The problem is about the conservation of linear momentum in a collision between two basketball players. The combined momentum before the collision equals the combined momentum after the collision. By setting these two equal and solving for the unknown, we find the velocity of the 87.2 kg player after the collision.
Explanation:In this problem, we are dealing with the conservation of linear momentum. The combined momentum of the two basketball players before the collision should equal the combined momentum after the collision given that no external forces are acting on the system.
Before the collision, the momentum is calculated as the mass of the 1st player times her velocity (87.2 kg x 7.0 m/s in the positive direction) plus the mass of the 2nd player times his velocity (102.0 kg x 5.2 m/s in the negative direction, because he is moving in the opposite direction).
After the collision, the momentum equals the mass of the 2nd player times his velocity (102.0 kg x -2.9 m/s) plus the mass of the 1st player times her final velocity (87.2 kg x v, where v is the velocity we're trying to find).
Setting these two expressions for momentum equal to one another and solving for v, we find the velocity of the 87.2 kg player after the collision.
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The resulting velocity of the 87.2 kg basketball player after the collision is calculated to be approximately 4.3 m/s in the positive direction.
To solve this problem, we use the principle of conservation of momentum. The total momentum before and after the collision must be the same.
The initial velocities and the masses of the players are:
Player 1: mass (m₁) = 87.2 kg, running at velocity (u₁) = 7.0 m/s in the positive direction
Player 2: mass (m₂) = 102.0 kg, running at velocity (u₂) = -5.2 m/s (toward the first player)
Therefore, the initial momentum of the system will be:
m₁ u₁ + m₂ u₂ = 87.2 kg × 7.0 m/s + 102.0 kg × (-5.2 m/s) = 80 kgm/s -(i)
After the collision:
Player 2 ends up moving with velocity (v₂) = -2.9 m/s (knocked backward)
We have to find the velocity of the Player (1) given as v₁.
Therefore, the final momentum of the system will be:
m₁ v₁ + m₂ v₂ = 87.2 kg × v₁ + 102.0 kg × (-2.9 m/s) = (87.2 v₁ - 295.8) kgm/s -(ii)
According to the principle of conservation of momentum, the initial momentum of the system before collision and the final momentum of the system after collision should be the same. So, equations (i) and (ii) must be equal:
80 kgm/s = (87.2 v₁ - 295.8) kgm/s
375.8 kgm/s = 87.2 v₁ kgm/s
or, v₁ = 375.8 kgm/s ÷ 87.2 kg = 4.3 m/s
The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 40-m-diameter loop-the-loop. What is the car's speed at the top?
Answer:
Explanation:
Given
Diameter [tex]d=40\ m[/tex]
radius [tex]r=20\ m[/tex]
From diagram, at top point
If Normal force is equal to Gravitational force
[tex]N=mg[/tex]
where N=normal reaction
m=mass of car
Normal reaction will provide centripetal force
[tex]N=\frac{mv^2}{r}[/tex]
thus
[tex]\frac{mv^2}{r}=mg[/tex]
[tex]v=\sqrt{2gr}[/tex]
[tex]v=\sqrt{2\times 9.8\times 20}[/tex]
[tex]v=19.79\ m/s[/tex]
The speed of a roller coaster car at the top of a 40-m-diameter loop, when the normal force equals the gravity, is 20 m/s.
Explanation:The question asks for the speed of a roller coaster car at the top of the loop. In this scenario, it is said that the normal force equals the magnitude of the gravitational force. The normal force is the force exerted by a surface to support the weight of an object resting on it. When the roller coaster car is at the top of the loop, the gravitational force (Fg) and normal force (Fn) are acting in the same direction and their sum provides the necessary centripetal force (Fc) for the car to continue its circular motion.
Therefore, we have Fn + Fg = Fc, but as it's mentioned Fn = Fg, so it will be 2Fn = mv²/r. Here, Fn=mg so we can replace it to the equation which is 2mg = mv²/r. Because 'm' (mass) is on both sides of the equation, we can eliminate it, which leaves 2g = v²/r. Rearranging the equation to solve for 'v' (velocity) gives v = square root of (2gr). Given that g (gravity) = 9.8 m/s² and r (radius) = 20 m (half of the 40 m diameter), we solve the equation: v = square root of (2 * 9.8 m/s² * 20 m) = 20 m/s.
Therefore, the roller coaster's speed at the top of a 40-m-diameter loop is 20 m/s.
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A tennis pro charges $15 per hour for tennis lessons for children and $30 per hour for tennis lessons for adults. The tennis pro is practicing A) first-degree price discrimination. B) second-degree price discrimination. C) third-degree price discrimination. D) fourth-degree price discrimination. E) fifth-degree price discrimination.
Answer:
Third-degree price discrimination
Explanation:
Third degree price discrimination according to Investopedia.com occurs when companies price products and services differently based on the unique demographics of subsets of its consumer base, such as students, military personnel, or seniors.
One liter (1000cm3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil slick. Assume the oil molecules have a diameter of 2 × 10-10 m.
Answer:
he diameter of the oil slick is 2523 m
Explanation:
given information?
V = 1 L = 1000 cm³ = 0.001 m³
h = 2 x 10⁻¹⁰ m
first we have to find the radius using the following equation
V = πr²h
r = √V/(πh)
= √(0.001)/(π x 2 x 10⁻¹⁰ )
= 1261.56 m
now, we can calculate the diameter of the oil slick
d = 2r
= 2 (1261.56)
= 2523 m
To estimate the diameter of oil slick, we can calculate the volume of one oil molecule and then divide the total volume of the spilled oil by the volume of one molecule. The diameter of the oil slick would be the diameter of one molecule multiplied by the square root of the number of oil molecules.
To estimate the diameter of the oil slick, we can first calculate the volume of one oil molecule. The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius. Given the diameter of the oil molecule is 2 × 10-10 m, the radius would be half of that, which is 10-10 m. Plugging in the values, we get V = (4/3)π(10-10)^3 = 4.19 × 10-29 m3.
The total volume of the spilled oil is given as 1000 cm3, which is equal to 1000 × 10-6 m3. To find the number of oil molecules in the spilled oil, we can divide the total volume of oil by the volume of one molecule: 1000 × 10-6 m3 / 4.19 × 10-29 m3 = 2.39 × 1022.
Since the oil slick is one molecule thick, the diameter of the slick would be the diameter of one oil molecule multiplied by the square root of the number of oil molecules: 2 × 10-10 m × √ (2.39 × 1022) ≈ 4.89 × 106 m.
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A driver driving along a highway at a steady 41 mph (60 ft/sec) sees an accident ahead and slams on the brakes. What constant deceleration is required to stop the car in 200 ft? To find out, carry out the following steps.
Answer:
constant deceleration required is 9 m/s²
Explanation:
Data provided in the question:
Initial Speed of the driver = 41 mph = 60 ft/s
Stopping distance = 200 ft
Now,
Since the car stops after 200 ft therefore final speed, u = 0 ft/s
from the Newton's equation of motion
we have
v² - u² = 2as
where,
v is the final speed
u is the initial speed
a is the acceleration
s is the distance
thus,
0² - 60² = 2a(200)
or
-3600 = 400a
or
a = - 9 m/s²
here, negative sign means deceleration
Hence,
The constant deceleration required is 9 m/s²
The difference in arrival times of P and S waves at a seismic observatory can be used to calculate the distance to a quake but not the ____ from the station.
Answer:
Direction
Explanation:
The arrival of seismic waves namely P and S wave are essential as it helps in the determination of the distance from the recording seismic station. In order to cover all the possibilities that are related to an earthquake, seismologists draw a circle around the station but this does not help in obtaining the proper information. By the use of two seismic stations, it can draw two circles that intersect at two points, which again does not help in determining the exact epicenter location. So, at least three seismic stations are needed, by which the triangulation method can be implemented, and as the three circles drawn from the three seismic stations intersect at one common point, it represents the exact location of the earthquake epicenter, and the exact direction also is obtained from this.
Two force vectors are perpendicular, that is, the angle between their directions is ninety degrees. If their magnitudes are 4.52 newtons and 7.33 newtons, then what is the magnitude of their sum?
F=__________ newtons
Explanation:
Resultant, R of two vectors P and Q is given by,
[tex]R=\sqrt{P^2+Q^2+2PQcos\theta }[/tex]
Here P = 4.52 N
Q = 7.33 N
θ = 90°
Substituting
[tex]R=\sqrt{4.52^2+7.33^2+2\times 4.52\times 7.33\times cos90 }\\\\R=8.61N[/tex]
Magnitude of their sum, F = 8.61 N
Releasing the clutch pedal reapplies the clamping force and reconnects the engine and transmission by firmly clamping the clutch disc between the ______ and the flywheel, allowing them to rotate as a unit.
Answer:
Preassure plate
Explanation:
The flywheel, clutch disc, pressure plate, throw-out bearing clutch fork, and the pilot bearing are the components of a clutch. It works when the flywheel bolts onto the crankshaft and therefore the pressure plate bolts back into the flywheel.
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If a liquid twice as dense as mercury were used in a barometer, approximately how high would the column of that liquid be under normal sea-level conditions?
A fluid twice as dense as mercury would rise to approximately half the height of mercury in a barometer under normal sea-level conditions. This is due to the relation between hydrostatic pressure, density, and height of the fluid column.
Explanation:Under normal sea-level conditions, atmospheric pressure supports a column of mercury about 760 mm high. This occurs due to the hydrostatic pressure, which is essentially the pressure exerted by a fluid due to gravity. In the case of a liquid twice as dense as mercury, the height of the liquid column would be half that of mercury under the same conditions. This is because the hydrostatic pressure is directly proportional to the density and height of the fluid column, which implies that for a given pressure, if we increase the density, the height decreases correspondingly. So, a liquid twice as dense as mercury would rise to approximately 380 mm in the barometer.
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A orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the altitude of the orbit (called a "synchronous orbit")?
The altitude of a geosynchronous orbit is [tex]3.59\cdot 10^7 m[/tex]
Explanation:
A geostationary (or geosynchronous) orbit is the orbit of a satellite that stays over the same spot on the equator of the rotating Earth.
This means that the period of a geostationary satellite is equal to the period of rotation of the Earth, which is 24 hours:
[tex]T=24 h \cdot 3600 s/h = 86400 s[/tex]
We can find the altitude of the orbit in the following way. First, we notice that the orbital speed of the satellite is given by
[tex]v=\frac{2\pi r}{T}[/tex]
where r is the radius of the orbit.
Then we also notice that the gravitational force between the satellite and the Earth is equal to the centripetal force, so we can write:
[tex]\frac{GMm}{r^2}=m\frac{v^2}{r}[/tex]
where
G is the gravitational constant
M is the mass of the Earth
m is the mass of the satellite
Re-arranging the equation,
[tex]\frac{GM}{r}=v^2[/tex]
And substituting the expression for the velocity,
[tex]\frac{GM}{r}=(\frac{2\pi r}{T})^2=\frac{4\pi^2 r^2}{T^2}[/tex]
Solving for r,
[tex]r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}[/tex]
And substituting:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}\\M=5.98\cdot 10^{24} kg\\T=86400 s[/tex]
we find:
[tex]r=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(86400)^2}{4\pi^2}}=4.225\cdot 10^7 m[/tex]
And since the radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
The altitude of the satellite is
[tex]h=r-R=4.225\cdot 10^7 - 6.37\cdot 10^6 = 3.59\cdot 10^7 m[/tex]
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A satellite in a synchronous orbit stays over a certain spot on the equator as the Earth rotates beneath it. The altitude of a synchronous orbit can be calculated using the formula: altitude = (radius of the Earth) + (height of the geostationary orbit), which gives an altitude of approximately 42,157 kilometers.
Explanation:
In order for a satellite to stay over a certain spot on the equator of Earth, it needs to be in a synchronous orbit. A synchronous orbit is an orbit in which the satellite's orbital period matches the rotation period of the Earth. This means that the satellite stays above the same spot on the equator as the Earth rotates beneath it.
The altitude of a synchronous orbit can be determined using the formula:
altitude = (radius of the Earth) + (height of the geostationary orbit)
The radius of the Earth is approximately 6,371 kilometers, and the height of the geostationary orbit is approximately 35,786 kilometers. So the altitude of a synchronous orbit is:
altitude = 6,371 km + 35,786 km = 42,157 km
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After the process of survey, excavation, recording, mapping, and post-excavation analysis, all these efforts are largely wasted unless?
Answer: The results are Documented and published.
Explanation: The process of surveying, recording,mapping,and post-excavation analysis are activities that has to do with Real estates like Lands. All these activities must be documented and published according to the enabling laws guiding it,in all the States of the Federation their are Archeological guidelines which clearly states how all this activities are to be conducted.
When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.
1. What is the nature of the force between balls A and B?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive.
2. What is the nature of the force between balls A and C?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive .
Answer:
1. (a)strongly attractive
2 (c) weakly attractive.
Explanation:
Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces shown in the figure when brought very near to the individual balls. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D.
should be the concluding part to this question as presented above
1. What is the nature of the force between balls A and B?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive.
since there is an attractive force between ball A and the test charge, the charge on ball A must be negative
A = -Q
since ball B is repulsive to the test charged , then B must be positively charged
B = +Q
since A is negative and B is positive , then they will experienced a strong attraction
option A
2. What is the nature of the force between balls A and C?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive .
Since C is weakly attracted to the test charge, we can say that Ball C will be weakly attracted to A, because it possess some weka charges
option C
Answer:its c trust me
Explanation:
If the surface of Earth were to suddenly turn white, the temperature of the planet would ________ because ________ insolation would be absorbed.
Answer: decrease; less
Explanation:
Insolation, refers to the amount of solar radiation that is received by earth. If the surface of earth was white, the radiation, will be reflected back into space. This reflective activity of a white surface will in turn cause the temperature of the planet to drop as there would be less energy to be converted to heat.
Answer:
A. Earth would cool down because it would reflect more solar radiation.
Explanation:
was right on edge :)
Learning Goal: To practice Problem-Solving Strategy 7.1 Rotational dynamics problems. Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical?
Final answer:
To find the angular acceleration of a falling pencil, use the equation angular acceleration = gravitational acceleration * sin(theta) / length. Plug in the provided values and solve for angular acceleration. With the given angle and assuming a pencil length, the angular acceleration is 1.07 rad/s^2.
Explanation:
To find the angular acceleration of a falling pencil, we can use the equation:
[angular acceleration = gravitational acceleration * sin(theta) / length].
Given that the angle is 10.0 degrees and the length of the pencil is not provided, we can assume a typical length of a pencil (15 cm). The gravitational acceleration can be taken as 9.8 m/s^2. Plugging in these values into the formula, we get:
[angular acceleration = 9.8 m/s^2 * sin(10.0 degrees) / 0.15 m].
Simplifying the equation, we get angular acceleration = 1.07 rad/s^2.
The scientific method is a process by which scientists: prove theories to be known facts gain confidence in theories by failing to prove them wrong show all theories to be wrong test the ideas of Aristotle?
Answer:
The scientific method is the process by which scientists gain confidence in theories by failing to prove them wrong.
Explanation:
Scientific method is a step by step process of creating and carrying out experiments. These processes to be performed are derived from logical and rational application of knowledge concerning a particular subject matter. Through these processes, scientists reach a conclusion about the world, these makes them confident in their findings.
The steps involved in scientific method include:
Observation: it is the first step. It enables you organise in your mind how you want the experiment to.
HYPOTHESIS is the result you imagine you'll find.
PREDICTION is how you think about the scientific idea- It solely relies on the hypothesis made.
EXPERIMENT is used to test your hypothesis. It is the tool designed to check if your idea is right or wrong.
CONCLUSION is the final step. It is the result gotten from the experiment. It can approve the hypothesis or reject the hypothesis.
A theory is formed when hypothesis has been tested to be true by many scientists. Hence scientists gain confidence in theories when the scientific method is approved.
A scientific theory is a well-approved explanation of an aspect of the natural world. The approval is based on thoroughly confirmed hypothesis and experiments carried out by many scientists.
A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above the ground, how long will it take him to reach a net that is 8 ft above the ground
This physics problem can be solved using the principles of motion under gravity, we plug our known values into the equation of motion and solve the resulting quadratic equation for time.
Explanation:This is an example of a physics problem under the theme of motion, specifically vertical motion under gravity. To answer this, you must apply the principles of physics to determine the total time the acrobat will spend in the air from the moment he is shot out of the cannon until the moment he touches the net.
In this situation,l we'll use the equation of motion, Y = Yo + Vot + 0.5gt² , where
Y is the final height (which is the height of the net, 8 ft), Yo is the initial height (which is the height the acrobat is shot from, 4 ft), Vo is the initial vertical speed (which is 34 ft/s), g is the acceleration due to gravity (which on earth is -32 ft/s² because it is downward), andt is the time we are trying to solve for.Plugging the known values into the equation, we obtain a quadratic equation in terms of t. Solving this quadratic equation will give us two solutions for t, one for the upward journey and one for the downward journey. The acrobat reaches the height of the net along his upward journey from the cannon, so we pick the smaller t as the correct solution (as it occurs first).
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