Answer:
The 3 main energy matter components are Coal, natural gas and hydro
Explanation:
The 3 most important energy components that are frequently being used for the generation of the energy i.e. coal which constitutes 41% it has the maximum contribution or can say mostly utilized after that comes natural gas which again constitutes 22% and followed by Hydro which is just 16%. with the rise of civilization people get aware of these energy components by coming across it in various phase of life.
An electron is accelerated from rest by a potential differ- ence of 350 V. It then enters a uniform magnetic field of magni- tude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
Answer:
a) the speed of the electron is 1.11 × 10⁷ m/s
b) the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m
Explanation:
a) Let's assume that we have an electron accelerated using a potential difference of V = 350, which gives the ion a speed of v. In order to find this speed we set the potential energy of the electron equal to its kinetic energy. Thus,
eV = 1/2 m v²
where
e is the charge of the electronm is the mass of the electronv is the speed of the electronThus,
v = √[2eV / m]
v = √[2(1.6 × 10⁻¹⁹ C)(350 V) / 9.11 × 10⁻³¹ kg]
v = 1.11 × 10⁷ m/s
Therefore, the speed of the electron is 1.11 × 10⁷ m/s
b) Then the electron enters a region of uniform magnetic field, it moves in a circular path with a radius of:
r = mv / eB
where
m is the mass of the electronv is the speed of the electrone is the charge of the electronB is the magnetic fieldThus,
r = (9.11 × 10⁻³¹ kg)(1.11 × 10⁷ m/s) / (1.6 × 10⁻¹⁹ C)(200 × 10⁻³ T)
r = 3.16 × 10⁻⁴ m
Therefore, the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m
A glass window is coated with a transparent film of refractive index n2 =1.25. This film will cause destructive interference in the reflected light when light of wavelength λ = 550nm is incident normally on the window. What is the minimum thickness of film required to produce this anti-reflection coating?
Answer:
0.00000011 m
Explanation:
n = 0 for minimum thickness
t = Thickness
[tex]n_2[/tex] = Refractive index of the film = 1.25
[tex]\lambda[/tex] = Wavelength = [tex]550\times 10^{-9}\ m[/tex]
We have the formula
[tex]2n_2t=(2n+1)\dfrac{\lambda}{2}[/tex]
[tex]\\\Rightarrow t=(2n+1)\dfrac{\lambda}{4n_2}\\\Rightarrow t=(2\times 0+1)\dfrac{550\times 10^{-9}}{4\times 1.25}\\\Rightarrow t=0.00000011\ m[/tex]
The minimum thickness of the film is 0.00000011 m
An 8.31-m, 267-N uniform ladder rests against a smooth (frictionless) wall. The coefficient of static friction between the ladder and the ground is 0.582, and the ladder makes a 52.6? angle with the ground. A 928 N person is standing on the ladder a distance d from the bottom end of the ladder.
(a) Draw a free-body diagram of the ladder. (Turn in part (a) on the sheet provided in your packet.)
(b) How far up the ladder (distance d) can the person climb before the ladder begins to slip? Hint: Among other things, you may want to look at the sum of the torques about the point where the ladder touches the wall. Be careful with angles! ___________ m ( ± 0.02 m)
To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force.
Explanation:To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The forces include the normal reaction force from the floor, the static friction force between the ladder and the floor, the weight of the ladder, and the normal reaction force from the wall. The torques are calculated by multiplying the force by the lever arm, which in this case is the distance between the center of mass of the ladder and the point of contact with the floor.
The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force. We can calculate the torque due to the person's weight by multiplying the person's weight by the distance between their position on the ladder and the point of contact with the floor. Next, we calculate the torque due to the static friction force by multiplying the static friction force by the distance between the point of contact with the floor and the point of contact with the wall.
Setting these two torques equal to each other and solving for the distance d yields the distance up the ladder that the person can climb before it begins to slip.
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During any process involving a heat transfer, the entropy of a system must always increase, it can never decrease.
True or False?
Answer:
True
Explanation:
Entropy is a thermodynamic variable that always increases or remains constant, but does not decrease.
This indicates that energy is always transferred from a higher temperature system to a lower temperature system and not the other way around.
Entropy can also be thought of as what drives a system into equilibrium, and systems reach equilibrium through heat transmission.
The fact that the entropy of a closed system never decreases is a statement of the second law of thermodynamics.
How does using a ramp make work easier when loading a piano onto a truck three feet above the ground?A. It allows a larger force to be applied over a shorter distance.B. It allows a larger force to be applied over a longer distance.C. It allows a smaller force to be applied over a shorter distance.D. It allows a smaller force to be applied over a longer distance.
Answer:
D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano
Explanation:
In this exercise the work to raise the piano is given by
W = F d cos θ
The distance and the displacement have an angle of zero degrees
W = F d
If we use energy conservation
W = ΔEm
W = [tex]Em_{f}[/tex] - Em₀
Suppose the piano starts from rest and reaches the top also at rest
W = mh y₂ - mg y₁
W = mg h
h= 3 ft
We see that the work to raise the body is always the same, regardless of the path, so, in the work relationship or is the product of force for distance if one goes up the other must decrease so that the product is the same.
Let's examine the answers
A) False. The distance on a ramp is greater
B) False. Change the value of work
C) False. Change the value of work
D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano
A steel, rigid container is filled with an ideal gas. The gas inside is heated such that its temperature, as measured in kelvin, doubles. Choose the following:A. The pressure inside will rise by a factor of 4.A. The pressure inside will rise by a factor of 2. B. The pressure inside will drop by a factor of 4.C. The pressure inside will drop by a factor of 2.D. Nothing will happen to the pressure because this is an isobaric process.
Answer:
If the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2.
Explanation:
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T} =k[/tex]
where V = volume, T = temperature, K = Constant
This law indicates that the ratio between pressure and temperature is constant.
This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature. This must be fulfilled since the relationship must remain constant
In short, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.
It is desired to study two different states, an initial state and an final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. When the temperature varies to a new T2 value, then the pressure will change to P2, and the following will be true:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
Given the above, it is possible to say that if the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2 (doubles too)
In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:
a. LinkedIn.
b. Google.
c. Yahoo.
d. Microsoft Windows Live.
e. All of these sites include discussion boards.
Answer:
A. Linkedln
Explanation:
In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:
A. Linkedln
All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.
Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.
The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex].
How to apply moment formulas to determine the tension in the left ropeA representation of the system is presented in the image attached below. Let suppose that the entire system represents a rigid body, that is, a body whose geometry cannot be neglected and does not experiment deformation due to the application of external loads.
The tension in the left rope ([tex]T_{L}[/tex]) can be found by applying Newton's laws and D'Alembert's Principle applied on the rightmost point of the bar:
[tex]\Sigma F = W\cdot (L-x)-T_{L}\cdot L = 0[/tex]
[tex]T_{L} = \frac{W\cdot (L-x)}{L}[/tex]
[tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex] (1)
The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex]. [tex]\blacksquare[/tex]
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Considering a situation where there is no horizontal acceleration, the tension in the left rope (TL) equals the weight of the person when x equals zero (x=0). Since the only forces on the static system are weight (W) and the tensions in the two ropes, and they are in equilibrium, the tension in either rope will be the same as the weight.
Explanation:The question requires us to find the tension in the left rope (TL), given that x equals zero. Using the equilibrium equation under the condition of no horizontal acceleration and no external forces except the weight (W) and the two tensions, TL and TR, we can conclude the magnitudes of the tensions TL and TR must be equal.
Given the scenario, the only forces acting on the system are the weight of the walker (W) and the tensions in the ropes. If the system is static, the net external force is zero, which means the sum of the forces should also equal zero according to Newton's second law. Hence, expressing the tension TL in terms of W becomes simple given that the system is in equilibrium. Under these conditions, the tension in the rope should equal the weight of the supported mass, making TL equal to W.
Therefore, since x is zero and we are neglecting other variables like rope weight and the angle of the ropes, the tension TL in the left rope becomes equivalent to the weight of the person. Hence the tension (TL) equals W.
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When push-starting your car, what is the 3rd Law partner to the friction force you exert on the ground?
A) The horizontal force you exert on the car.
B) The normal force you exert on the ground.
C) The normal force the ground exerts on you.
D) The friction force the ground exerts on you.
Answer:
C
Explanation:
Newton third law states that to every action force there is an equal and opposite reaction force. In the case of the car, the person pushes against the ground in other to get the car to move and consequently the ground exert a normal force opposite to the push on the person.
Two cars travel westward along a straight highway, one at a constant velocity of 97 km/h, and the other at a constant velocity of 113 km/h. Assuming that both cars start at the same point, how much sooner does the faster car arrive at a destination 17 km away? Answer in units of h.
Answer:
0.025hr
Explanation:
The full solution is on the image below. The two cars cover the same distance at different time intervals. Since the distance is constant, the velocity is inversely proportional to the time taken to cover the constant distance
Answer:
0.025 h
Explanation:
Let's assume for the first car, the destination is [tex]x_{1}[/tex], the time is [tex]t_{1}[/tex], the velocity is [tex]v_{1}[/tex] and for the second car the destination is [tex]x_{2}[/tex], the time is [tex]t_{2}[/tex], the velocity is [tex]v_{2}[/tex].
We are given:
[tex]v_{1}[/tex] = 97 km/h
[tex]v_{2}[/tex] = 113 km/h
If we are asked the time, the destinations must be equal which are also given:
[tex]x_{1}[/tex] = [tex]x_{2}[/tex] = 17
For constant velocity, the equation is x = v * t
Hence,
[tex]x_{1}[/tex] = [tex]v_{1}[/tex] * [tex]t_{1}[/tex] = [tex]x_{1}[/tex] = 97 * [tex]t_{1}[/tex] = 17
⇒ [tex]t_{1}[/tex] = 17/97 = 0.175 h
[tex]x_{2}[/tex] = [tex]v_{2}[/tex] * [tex]t_{2}[/tex] = [tex]x_{2}[/tex] = 113 * [tex]t_{2}[/tex] = 17
⇒ [tex]t_{2}[/tex] = 17/113 = 0.150 h
So,
[tex]t_{1}[/tex] - [tex]t_{2}[/tex] = 0.175 - 0.150 = 0.025 h
The second car arrives 0.025 h sooner.
A 2.5-kg book slides horizontally and falls from a shelf 3.0 m above the floor. How much work does the force of gravity do on the book as it falls 3.0 m?
Answer:
73.5 N
Explanation:
W = PE = mgh = 2.5 * 9.8 * 3.0 = 73.5 N
Final answer:
The work done by gravity on a 2.5-kg book falling from a 3.0 m height is calculated to be 73.5 joules, using the formula for work in the presence of gravitational force.
Explanation:
To calculate the work done by gravity on the 2.5-kg book as it falls 3.0 m, you can use the formula work (W) = force (F) × distance (d). Since we're dealing with gravity, the force is the weight of the book, which is its mass multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, the force is 2.5 kg × 9.8 m/s². The distance the book falls is given as 3.0 m.
The formula simplifies to W = m × g × d, which gives us W = 2.5 kg × 9.8 m/s² × 3.0 m, resulting in 73.5 J. The work done by gravity on the book as it falls to the ground is 73.5 joules.
Keaton is asked to solve the following physics problem:
What determines the speed at which a ball will drop from a five-story building?
He is given six balls and a tall box. Demonstrating his ability to think within the formal operational stage, Keaton will likely __________.
a. begin by climbing onto the box and dropping all of the balls
b. lower the height at which the balls are dropped
c. think about the problem first, systematically consider all factors, and form a hypothesis
d. take a vivid guess that the larger the ball, the faster the speed
Answer:
The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"
Explanation:
In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;
1) v=u+at
2)v^2=u^2+2as
3)s=ut+(1/2)(at^2)
the variables are explained below;
u= initial velocity of the body
a=acceleration/Speed of the body
t= time taken by the body while travelling
s= displacement of the body.
Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)
To test the resiliency of its bumper during low-speed collisions, a 3 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 6.00 106 N/m and compresses 3.26 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall? m/s
Answer:
1.45549 m/s
Explanation:
m = Mass of car = 3010 kg
v = Velocity of car
k = Spring constant = [tex]6\times 10^6\ N/m[/tex]
x = Displacement of spring = 3.26 cm
As the energy of the system is conserved
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow v=\sqrt{\dfrac{6\times 10^6\times 0.0326^2}{3010}}\\\Rightarrow v=1.45549\ m/s[/tex]
The speed of the car before impact is 1.45549 m/s
energy expenditure due to physical activity generally accounts for __________ of total energy expenditure.
Answer:
Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.
Explanation:
The total amount of energy your body uses daily is usually divided as follows:
1. Your Basal Metabolic Rate at rest (50-70%)
Basal Metabolic Rate is the number of calories required to keep your body functioning at rest
2. Consumption by your daily physical activities (20%)
3. Energy to digest food. (10-20%)
This energy has to be deducted from the overall energy content of the food itself.
Then, the answer is:
Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.
Final answer:
The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure. Physical activity guidelines recommend adults to engage in aerobic and muscle-strengthening activities for substantial health benefits.
Explanation:
The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure.
According to the 2018 Physical Activity Guidelines for Americans issued by the Department of Health and Human Services, adults should do at least 150 minutes to 300 minutes per week of moderate-intensity aerobic activity, or 75 minutes to 150 minutes per week of vigorous-intensity aerobic physical activity, or an equivalent combination of both. Engaging in physical activity beyond 300 minutes per week can result in additional health benefits.
Therefore, the energy expenditure due to physical activity is a crucial component of total energy expenditure and plays a significant role in maintaining overall health.
What force is described as the attraction between a sample of matter and all other matter in the universe?
Answer:
Gravitational Force
Explanation:
Gravitational force also called gravity or gravitation is an attractive force that keeps two objects in space. Gravitational force is an attractive force that tends to pull matters together. Every objects in the universe experience gravitational pull. Planets, stars, galaxies, are held together by gravity. It is a weak force. The weight of an object is the product of gravitational force acting on its mass.
Newton's Law of Universal Gravitation states the force of attraction between two masses m₁ and m₂ in the universe is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.
[tex]F = G\frac{m_{1}m_{2}}{r^2}[/tex]
Where;
F is the gravitational force,
G is the gravitational constant = 6.67 × 10¹¹ m³/kg/s,
m1 and m2 are the masses of the objects,
r is the distance between the centers of the masses
A large storage tank open at the top and filled with water develops a small hole in its side at a point 16.0 m below the water level. The rate of flow from the leak is found to be 2.50 × 10^(-3) m^3/min.
Determine; (a) the speed (m/s) at which the water leaves the hole and (b) the diameter of the hole (mm).
Final answer:
The speed at which the water leaves the hole is 17.7 m/s, and the diameter of the hole is approximately 4.8 mm using Torricelli's theorem and the flow rate equation.
Explanation:
The scenario described involves principles of fluid dynamics within the field of physics, specifically Bernoulli's equation and the equation of continuity. To calculate the speed at which the water leaves the hole (a) and the diameter of the hole (b), we employ the Torricelli's theorem and the flow rate equation.
Using Torricelli's theorem, the speed (v) of water exiting the hole can be determined by the formula v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the water column above the hole (16.0 m).
v = √(2 * 9.8 m/s² * 16.0 m) = √(313.6 m²/s²) = 17.7 m/s
To find the diameter of the hole, we use the flow rate Q = A * v, where Q is the flow rate (2.50 × 10³ m³/min), A is the area of the hole, and v is the speed of water leaving the hole. First, we convert the flow rate to m³/s by dividing by 60 (since there are 60 seconds in a minute):
Q (in m³/s) = (2.50 × 10³ m³/min) / 60 = 4.17 × 10µ m³/s
To find the area: A = Q / v, then to find the diameter (d), we use the area of a circle A = π * (d/2)². Rearranging and solving for d gives us:
d = 2 * √(Q / (π * v))
d = 2 * √((4.17 × 10µ m³/s) / (π * 17.7 m/s)) = 4.8 mm
(a) The speed at which the water leaves the hole is approximately 17.7m/s.
(b) The diameter of the hole is approximately 1.74mm.
Given:
- The depth of the hole below the water level, [tex]\( h = 16.0 \)[/tex] m
- The rate of flow, [tex]\( Q = 2.50 \times 10^{-3} \) m\(^3\)/[/tex]min
Part (a): Speed of Water Leaving the Hole
We can use Torricelli's Law to determine the speed at which the water leaves the hole:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity[tex](\( 9.8 \) m/s\(^2\))[/tex]
- [tex]\( h \)[/tex] is the height of the water above the hole
Plugging in the values:
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{313.6} \][/tex]
[tex]\[ v \approx 17.7 \text{ m/s} \][/tex]
Part (b): Diameter of the Hole
First, convert the flow rate to cubic meters per second:
[tex]\[ Q = 2.50 \times 10^{-3} \text{ m}^3/\text{min} \][/tex]
[tex]\[ Q = \frac{2.50 \times 10^{-3}}{60} \text{ m}^3/\text{s} \][/tex]
[tex]\[ Q = 4.17 \times 10^{-5} \text{ m}^3/\text{s} \][/tex]
The flow rate ( Q ) can also be expressed as:
[tex]\[ Q = A \cdot v \][/tex]
where:
- ( A ) is the cross-sectional area of the hole
- ( v ) is the speed of the water leaving the hole
Rearranging to solve for ( A ):
[tex]\[ A = \frac{Q}{v} \][/tex]
[tex]\[ A = \frac{4.17 \times 10^{-5}}{17.7} \][/tex]
[tex]\[ A \approx 2.36 \times 10^{-6} \text{ m}^2 \][/tex]
The area ( A) of a circular hole is given by:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Solving for the diameter[tex]\( d \)[/tex]:
[tex]\[ d = 2 \sqrt{\frac{A}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{\frac{2.36 \times 10^{-6}}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{7.51 \times 10^{-7}} \][/tex]
[tex]\[ d \approx 1.74 \times 10^{-3} \text{ m} \][/tex]
[tex]\[ d \approx 1.74 \text{ mm} \][/tex]
A military base a few miles from a medium- sized city is being decommissioned. The base is a large one, sited along a coastline, and includes some small hills, pristine woodlands and wetlands, streams, a river, and a small lake. There is a railroad spur that connects the base with the nearby city, as well as a four- lane highway. The base contains housing, schools, a hospital, shops, and recreational areas for a population of 10,000. The city's options for using, selling, or leasing the land are numerous, and many groups come forward with proposals. Members of several local environmental groups propose that nearly all of the buildings be removed and the area turned into a regional hiking and camping area, with very limited vehicle access. This will preserve the natural areas while providing recreation for the city dwellers. One issue that is raised by opponents is ________.
Answer:
The lack of jobs and lack of transportation to jobs
Explanation:
As it is seen, the establishment of the camping and hiking areas will dispatch buildings which were hospitals, shops and schools.
These buildings and facilities previously employed a number of workers and their removal hence, will reduce the number of available jobs.
Also, the new constructions will prove difficult in switching from one job to another.
While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (b) When she tightens the string slightly, she hears 3.00 beats/s. What is the frequency of the string now? (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?
Answer
given,
frequency = 523 Hz
piano tuner hears = 2.00 beats/s
a) possible frequency = 523 ± 2
525 Hz and 521 Hz
b) On tightening piano wire, tension will be increased which will cause an increase of frequency of piano wire. Since no. of beats heard is also increased, the frequency of piano wire must be greater than the frequency of oscillator.
f p' = 5+ f₀
f p' = 3 + 523 Hz = 526 Hz.
c) f = k x √T
Let the tension at 523 Hz be T₀ and at 526 Hz it be T₁
523 = k x √T₀
T₀ = 523²/ k²
T₁ = 526²/ k²
% change in tension
= [tex]\dfrac{T_0- T_1}{T_1}[/tex]
= [tex]\dfrac{523^2- 526^2}{526^2}[/tex]
= 1.13 %
Two train whistles have identical frequencies of 1.54 102 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 8.00 beats/s when the whistles operate together. What are the two possible speeds that the moving train can have? moving away from the station m/s moving towards the station m/s
Final answer:
The two possible speeds that the moving train can have are 343.292 m/s when moving away from the station and 355.182 m/s when moving towards the station.
Explanation:
To determine the two possible speeds that the moving train can have, we can use the formula for the beat frequency:
beat frequency = |v - v_s| / v * f_w
v is the speed of sound,
v_s is the speed of the moving train,
f_w is the frequency of the train whistle,
|v - v_s| is the absolute value of the difference between the speed of sound and the speed of the moving train.
Let's solve this equation for two possible speeds:
Case 1 - Moving away from the station:
8.00 beats/s = |331 m/s - v_s| / 331 m/s * 1.54 *[tex]10^2 Hz[/tex]
Solving for v_s, we find v_s = 343.292 m/s (rounded to three decimal places).
Case 2 - Moving towards the station:
8.00 beats/s = |331 m/s + v_s| / 331 m/s * 1.54 * [tex]10^2 Hz[/tex]
Solving for v_s, we find v_s = -355.182 m/s (rounded to three decimal places). Since the speed cannot be negative, we take the absolute value and find v_s = 355.182 m/s (rounded to three decimal places).
Therefore, the two possible speeds that the moving train can have are:
Moving away from the station: 343.292 m/s (rounded to three decimal places)
Moving towards the station: 355.182 m/s (rounded to three decimal places)
A person with which of the following BMI values has the highest increased risk for premature mortality?
a.29 kg/m2
b.35 kg/m2
c.25 kg/m2
d.18.5 kg/m2
Answer:
b.
Explanation:
BMI is body mass index, it is calculated by dividing the weight of a person in kilogram by square of height in meters.
A high BMI is an indicator of obesity. Obese people are prone to death causing diseases like High Blood pressure, cardiac arrest and diabatese. So, a high value of BMI values has the highest increased risk for premature mortality.
So the correct answer is 35 kg/m^2.
For all turning vehicles, the rear wheels follow a ______ than the front wheels.
a. slower pathb. faster pathc. longer pathd. shorter path
Answer:
Shorter path
Explanation:
For all turning vehicles, the rear wheels follow Shorter path than the front wheels.
Any turning vehicle, the rear(the back part of something, especially a vehicle.) wheels follow a shorter path than the front wheels. The longer the vehicle is, the greater the difference will be in path. Trucks initially swing out before making a turn
What is the farthest distance at which a typical "nearsighted" frog can see clearly in air?
A. 12m
B. 80 m
C. 6.0m
D.17m
Answer: the correct option is D (17m).
Explanation: The farthest distance at which a typical "nearsighted" frog can see clearly in air is 17m.
A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 10 cm below yi.
(a) What is the frequency of the oscillation?
(b) What is the speed of the object when it is 8.0 cm below the initial position?
(c) An object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object?
(d) How far below yi is the new equilibrium (rest) position with both objects attached to the spring?
While certain parts of your question cannot be answered without specific data such as the spring constant, we can deduce from the half frequency that the mass of the initial object must also be 300g. More mass on the spring would pull the equilibrium further down but we can't quantify without more details.
Explanation:This question involves simple harmonic motion and Hooke's Law. Hooke's Law is F = -kx where F is the force, k is the spring constant, and x is the displacement. Simple Harmonic Motion deals with oscillations like that of a spring. However, without values for the spring constant (k) or initial speed of the object, we cannot calculate the frequency of the oscillation or speed of the object at any given point.
As for part (c), the frequency of the oscillation halves when a second mass is introduced. This implies that the system's mass has doubled as frequency is inversely proportional to the square root of the mass. Therefore, the initial object's mass would be 300g as well.
For part (d), without any specific figures such as the spring constant, the new equilibrium position below yi cannot be determined. Commonly, the force produced by the weight of the objects will be counteracted by the spring force when the system is in equilibrium. More mass would pull the spring further down until this equilibrium is achieved.
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The frequency of oscillation can be determined using the formula f = 1 / T. The speed of the object when it is 8 cm below the initial position can be calculated using the formula v = A * ω. The mass of the first object can be found by setting up two equations and solving for M. The new equilibrium position is slightly lower than the initial position yi.
Explanation:(a) The frequency of an oscillation can be determined using the formula:
frequency (f) = 1 / period (T)
Since the object oscillates up and down, it completes one full cycle (2π radians) every time it goes from the highest point to the lowest point and back. The period (T) is therefore the time it takes for one complete cycle. The object reaches its lowest position when it is 10 cm below yi, so the total distance traveled is 10 cm. Since the object moves up and down symmetrically, we can calculate the period by dividing the total distance traveled by 2 times the amplitude (A):
T = 2π * sqrt(m / k)
where m is the mass of the object and k is the spring constant. Given that the spring is massless, the mass of the object does not affect the period. Therefore, we can calculate the period as:
T = 2π * sqrt(k)
From the period, we can compute the frequency using the formula provided above.
(b) The speed of the object when it is 8.0 cm below the initial position can be determined using the formula for the speed of an object in simple harmonic motion:
speed (v) = amplitude (A) * angular frequency (ω)
The angular frequency is related to the period by the formula:
angular frequency(ω) = 2π / period (T)
Using the given values, we can calculate the angular frequency and then substitute it into the equation for speed.
(c) When an object of mass 300 g is attached to the first object, the system oscillates with half the original frequency. Let's denote the mass of the first object as M and the spring constant as K. We can set up two equations to solve for the mass of the first object:
Equation 1: f = (1 / (2π)) * sqrt(K / M) (original frequency)
Equation 2: (f/2) = (1 / (2π)) * sqrt(K / (M + 0.3)) (new frequency)
By solving these equations simultaneously, we can find the mass of the first object.
(d) The new equilibrium position, with both objects attached to the spring, is determined by the combined mass of the two objects. Since the first object is now heavier due to the addition of the second object, the equilibrium position will be slightly lower than the initial position yi. The exact displacement can be calculated using the formulas for equilibrium position in simple harmonic motion.
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Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t ≥ 0 . After her parachute opens, her velocity satisfies the differential equation dvdt=−2v−32 , with initial condition v(0) = −50 . It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
Answer:
t = 1.07 seg
Explanation:
First we are going to solve the differential equation for the velocity:
[tex]\frac{dv}{dt} = -2v-32[/tex]
This is a differential equation of separable variables
[tex]\frac{dv}{-2v-32} = dt[/tex]
Multiplying by -1 to both sides of the equation
[tex]\frac{dv}{2v+32} = -dt[/tex]
We integrate the left side with respect to the velocity and the right side with respect to time
[tex]\frac{ln(2v+32)}{2} = -t +k[/tex]
where k is a integration constant
ln(2v+32) = -2t + k
[tex]2v +32 = e^{-2t+k}[/tex]
[tex]2v + 32 = ce^{-2t}[/tex]
[tex]v = ce^{-2t} -16[/tex]
We determine the constant c with the initial condition v(0) = -50
[tex]-50 = ce^{-2(0)} -16[/tex]
-50 + 16 = c
c = -34
Then
[tex]v(t) = -34e^{-27}-16[/tex]
When the velocity is -20 ft/s the time is:
[tex]-20 = -34e^{-2t}-16[/tex]
[tex]\frac{-4}{-34} = e^{-2t}[/tex]
[tex]ln(\frac{4}{34} ) = -2t[/tex]
t = 1.07 seg
To determine the time at which the skydiver's velocity reaches 20 feet per second, we integrate the given differential equation dv/dt = -2v - 32 with initial conditions and solve for the time t when the velocity function v(t) equals -20.
Explanation:The student's question involves solving a first-order linear differential equation to find the time at which a skydiver's velocity reaches a certain value after her parachute opens. The differential equation dv/dt = -2v - 32 represents the velocity of the skydiver, and we must use initial conditions to solve it. Given that the skydiver's initial velocity is -50 feet per second and the safe landing speed is 20 feet per second, we need to integrate the differential equation to find the function v(t) and then solve for t when v(t) equals -20 feet per second (since velocity is negative when descending).
The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.076 V exists across the membrane. The thickness of the cell membrane is 7.30 10-9 m.
What is the magnitude of the electric field in the membrane?
Answer:[tex]E=1.041\times 10^7 V/m[/tex]
Explanation:
Given
Potential Difference [tex]\Delta V=0.076 V[/tex]
thickness of cell membrane [tex]d=7.30\times 10^{-9} m[/tex]
Electric Field for this Potential is given by
[tex]E=\frac{\Delta V}{d}[/tex]
[tex]E=\frac{0.076}{7.30\times 10^{-9}}[/tex]
[tex]E=0.01041\times 10^9 V/m[/tex]
[tex]E=1.041\times 10^7 V/m[/tex]
600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.
How much heat energy is transferred to or from the system as heat?
Write positive value if the energy is transferred to the system and negative if the energy is transferred from the system.
Answer:
900 J
Explanation:
[tex]\Delta U[/tex] = Change in entropy = -300 J (Decrease)
[tex]W[/tex] = Work done = 600 J
[tex]Q[/tex] = Heat transferred
Change in internal energy is given by
[tex]\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J[/tex]
The heat transferred to or from the system as heat is -900 J.
900 J is transferred. As heat is transferred from the system the sign is negative.
When 600 J of work is done on a system and the thermal energy of the system decreases by 300 J, the heat energy transferred to or from the system can be calculated using the First Law of Thermodynamics. In this case, -900 J of heat is transferred from the system.
Explanation:When 600 J of work is done on a system and the thermal energy of the system decreases by 300 J, the change in internal energy of the system can be calculated using the First Law of Thermodynamics.
The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to or removed from the system plus the work done on or by the system. Mathematically, it can be expressed as:
ΔU = Q + W
Where ΔU represents the change in internal energy, Q represents the heat added to or removed from the system, and W represents the work done on or by the system.
In this case, we know that the work done on the system is 600 J, and the change in thermal energy is -300 J (since it decreases).Substituting these values into the equation, we get:ΔU = Q + 600 J
-300 J = Q + 600 J
Now we can solve for Q, which represents the heat transferred to or from the system.Rearranging the equation:Q = -300 J - 600 J
Q = -900 J
The negative sign indicates that the heat is transferred from the system.
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You and a friend are riding on a ferris wheel, which is moving with constant velocity. You, as a seeker of deep understanding, are sitting on a bathroom scale, wondering what will happen. Your friend observes 3 bathroom scale readings, one at the top, Rt , one at the middle, Rm, and one at the bottom, Rb .
Which of the following is the correct relationship between Rb , Rt , Rm? top middle bottom
1. Rt = Rb < Rm
2. Rt > Rm > Rb
3. Rt = Rb = Rm
4. Not enough information
The correct relationship between the bathroom scale readings at the top, middle, and bottom of a ferris wheel ride is that they are all the same.
Explanation:When you and your friend are riding on a ferris wheel moving with constant velocity, the bathroom scale readings at the top, middle, and bottom of the ride can be compared. Since the ferris wheel is moving at a constant velocity, there is no acceleration and thus no net force acting on you. This means that your apparent weight will be the same at all three points on the ride. Therefore, the correct relationship is option 3: Rt = Rb = Rm.
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Which of the following is an example of Newton's second law of motion?
a. Catie's skateboard hits a curb and she falls forward onto the grass. b. A tissue and a rock are dropped from a ladder.
c. The rock exerts a greater force when it hits the ground.
d. A person stands on a chair and pushes down, and the chair pushes upwards on the person.
e. Tires from a bike push against the road and the road push back.
Answer:
B
Explanation:
Newton’s Second Law of Motion
Newton’s Second Law of Motion states that ‘when an object is acted on by an outside force, the mass of the object equals the strength of the force times the resulting acceleration’.
This can be demonstrated dropping a rock or and tissue at the same time from a ladder. They fall at an equal rate—their acceleration is constant due to the force of gravity acting on them.
The rock's impact will be a much greater force when it hits the ground, because of its greater mass. If you drop the two objects into a dish of water, you can see how different the force of impact for each object was, based on the splash made in the water by each one.
Answer: B
Explanation:
Strange as it may seem, it is as hard to accelerate a car on a level surface on the moon as it is here on the Earth. This is because?1.the weight of the car is independent of gravity.2.the mass of the car is independent of gravity.3.Nonsense! A car is much more easily accelerated on the moon than on the Earth.
Answer:
2.the mass of the car is independent of gravity.
Explanation:
When mass and acceleration is multiplied we get the force.
[tex]F=ma\\\Rightarrow a=\dfrac{F}{m}[/tex]
Mass of an object is always constant anywhere in the universe. Acceleration of the car would be the ratio of the force and mass. As the mass is constant force the same force applied the acceleration is the same on Earth or Moon.
Hence, the mass of the car is independent of gravity.
Experimental Inquiry: What factors influence the loss of nutrients from a forest ecosystem?