Answer:
λ = 157 μm
Explanation:
given,
[tex]\vec{E}(y,t)=[ 3.9\times 10^5\ V/m cos(ky-(1.20\times 10^{13}\ rad/s )t)]\vec{k}[/tex]
now,
ω = 1.2 x 10¹³ rad/s
[tex]f =\dfrac{\omega}{2\pi}[/tex]
[tex]f =\dfrac{1.2\times 10^{13}}{2\pi}[/tex]
f = 1.91 x 10¹² Hz
determine the wavelength of the E M wave
[tex]\lambda = \dfrac{c}{f}[/tex]
[tex]\lambda = \dfrac{3\times 10^8}{1.91\times 10^{12}}[/tex]
λ = 1.57 x 10⁻⁴ m
λ = 157 x 10⁻⁶ m
λ = 157 μm
hence, the wavelength of the wave is equal to λ = 157 μm
Given that,
[tex]\vec{E}(y,t) = [3.9\times 10^5 \ V/mcos] (ky-(1.20\times 10^{13} \ rad/s)t)] \vec{k}[/tex]here,
[tex]\omega = 1.2\times 10^{13} \ rad/s[/tex]As we know,
→ [tex]f = \frac{\omega}{2 \pi}[/tex]
By substituting the values, we get
[tex]= \frac{1.2\times 10 ^{13}}{2 \pi}[/tex]
[tex]= 1.91\times 10^{12} \ Hz[/tex]
hence,
The wavelength will be:
→ [tex]\lambda = \frac{c}{f}[/tex]
[tex]= \frac{3\times 10^8}{1.91\times 10^{12}}[/tex]
[tex]=157\times 10^{-6} \ m[/tex]
[tex]= 157 \ \mu m[/tex]
Thus the above response is correct.
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Elements that cycle in the environment and that also have a gaseous phase at some point in their cycle include what?
Answer:
Carbon, nitrogen and sulphur.
Explanation:
In carbon cycle, carbon dioxide is in the gaseous form in atmosphere. This gaseous carbon dioxide is emitted in the atmosphere through combustion of fossils, respiration, decomposition. In nitrogen cycle, atmospheric nitrogen is in the gaseous form which is emitted through denitrification. In sulphur cycle, sulphur is in the form of gaseous sulphur dioxide in the atmosphere. This sulphur is emitted in the atmosphere by the volatilization of hydrogen sulphide.
Elements such as carbon, nitrogen, and oxygen go through a gaseous phase in their respective nutrient cycles (the Carbon Cycle and Nitrogen Cycle for instance), moving through the environment in a regular pattern.
Explanation:Certain elements move through our planet's systems, such as the water, atmosphere, and soils, in cycles known as nutrient cycles or biogeochemical cycles. These elements often go through a gaseous phase. Carbon, nitrogen, and oxygen are examples of elements that have a gaseous phase in their cycle. For instance, Nitrogen Cycle involves nitrogen gas in the atmosphere being converted into usable forms by bacteria in a process known as nitrogen fixation. Similarly, in the Carbon Cycle, carbon dioxide gas is absorbed by plants and converted into organic carbon through photosynthesis.
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A compressed spring in a toy is used to launch a 5.00-gram ball.
If the ball leaves the toy with an initial horizontal speed of 5.00 meters per second, the minimum amount of potential energy stored in the compressed spring was:
a) 0.0125 J
b) 0.0250 J
c) 0.0625 J
d) 0.125 J
Answer:
c) 0.0625 J
Explanation:
How the mechanical energy is conserved, then ball’s kinetic energy is equal to stored energy in compressed spring.
Then:
[tex]K = U_{e}[/tex]
Where K: kinetic energy
[tex]U_{e}[/tex]: elastic potential energy
[tex]K = \frac{mv^{2} }{2}[/tex]
[tex]K = \frac{(0.005)(5)^{2} }{2}[/tex]
K = 0.0625 J
and
[tex]U_{e}[/tex] = 0.0625
A box is 25 kg with a kinetic force of .35 is pushed horizontally. What force must you continue continue to exert if you want to accelerate the box at a rate of 2.5 m/s squared
The force required is equal to 148.33 N.
Why?
To answer your question, I'll assume that the value of 0.35 is referring to the coefficient of kinetic friction. So, we can solve the problem using the following equations:
[tex]F-F_{f}=m*a\\\\FrictionalForce=\mu *m*g\\\\F_{f}=0.35*25kg*9.81m\frac{m}{s^{2}}=85.83N[/tex]
Now, substituting, we have:
[tex]F-F_{f}=m*a\\\\F=25kg*2.5\frac{m}{s^{2}}+85.83N\\\\F=62.5N+85.83N=148.33N[/tex]
Hence, we have that the force required to accelerate the box at a rate of 2.5 m/s2 is 148.33N.
Have a nice day!
TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of a 610-m-high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius?
Answer:
1.894 × 10^-4 m
Explanation:
mass of the man = 72 kg, mass of the equipment = 400 kg
total mass on top of the antenna = 72 + 400 = 472 kg
total weight of the man and equipment = total mass × acceleration due to gravity (g, 9.81 m/s²) = 472 × 9.81 = 4630.32 N
to calculate the ΔL ( how much the antenna was compressed), we use the formula below
E = stress/ strain = (F/A) / (ΔL/L) = FL / ΔLA where F is the force in Newton, A is the surface area of the circular face of the antenna in m²
E = 2.1 × 10^11 N/m² which is the young modulus of steel
A = πr² = 3.142 × (0.150²) = 0.071m²
make ΔL the subject of the formula
ΔL = FL / AE
substitute the values into the equation
ΔL = (4630.32 × 610) / ( 0.071 × 2.1 × 10^11) = 1.894 × 10^-4 m
Which of the following are criticism of kohlberg's theory of moral developmen?
Answer:
Model reasoning vs Moral behavior,cultural differences, bias
Explanation:
The answer for this case: Model reasoning vs Moral behavior,cultural differences, bias
Model reasoning
In particular usually we have a positive correlation between higher stages of reasoning and higher levels of moral behavior. But in some cases, some found we have situational factors are better predictors of moral behavior.
One example is when people prefer to help people with some disease because an institute with most prestigy says this condition is given by a problem of the society when the reality is not true.
Cultural differences
This item can create some problem since we can create problems when we need to establish if one statement is true or not.
Bias
This is one of the most common problems since we don't have a criteria to decide if we have bias or no.
Viscosity: Blood is about 4 to 5 times ____________ viscous than water. Viscosity of blood depends upon the amount of dissolved substances in the blood relative to the amount of fluid. Viscosity is ____________ if the amount of substances increases, the amount of fluid decreases, or both.
Answer:more, increased
Explanation:
Blood is four to five times more viscous than water.
Viscosity depends upon dissolved substances in blood and it increases as the amount is increases or if the fluid in the blood decreases.
Viscosity is the resistance offered by liquid to the flow and it also depends upon temperature and it decreases with increase in temperature.
What are the 3 energy-matter components and their percentage of the energy pie, and how do we come to know about each?
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.
The purpose of antilock brake systems is to prevent the brakes from becoming locked.Choose one True False
Answer:True
Explanation:
The above statement is true as an anti-lock braking system (ABS) is a vehicle protection system that helps the wheels on a vehicle to keep up tractive force with the road. ABS does not allow the brake to lock during braking and thus avoiding skidding which is necessary to avoid in wet roads otherwise it could be fatal for drivers.
It is widely used in vehicles to improve the safety measures and reduce road accidents.
active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zone?
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
After sliding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of +3.0 m/s. Miranda runs after her at a velocity of +4.2 m/s and hops on the inner tube.
How fast do the two of them slide across the snow together on the inner tube? Ashley's mass is 69 kg and Miranda's is 59 kg. Ignore the mass of the inner tube and any friction between the inner tube and the snow.
Answer:
Their combined velocity on the inner tube is +3.55 m/s
Explanation:
This question deals with the conservation of linear momentum. The total linear momentum in a closed system is conserved.
Therefore,
P_i = P_f
m₁ v₁ + m₂ v₂ = (m₁ + m₂) v₁₂
where
m₁ is the mass of Ashleym₂ is the mass of Mirandav₁ is Ashley's velocityv₂ is Miranda's velocityv₁₂ is their combined velocityTherefore,
v₁₂ = (m₁ v₁ + m₂ v₂) / (m₁ + m₂)
v₁₂ = ( (69 kg)(3 m/s) + (59 kg)(4.2 m/s) ) / (69 kg + 59 kg)
v₁₂ = +3.55 m/s
Therefore, Ashley and Miranda's combined velocity on the inner tube is +3.55 m/s.
The positive sign shows that the velocity is in the positive direction.
Two physics students, Will N. Andable and Ben Pumpiniron, are in the weightlifting room. Will lifts the 100-pound barbell over his head 10 times in one minute; Ben lifts the 100-pound barbell over his head 10 times in 10 seconds. Which student does the most work? ______________ Which student delivers the most power? ______________ Explain your answers.
Answer:
"Same"
"Ben Pumpiniron"
Explanation:
Given that
Will lifts 100 pound ,10 times in 1 minute
Ben lifts 100 pound ,10 times in 10 sec
As we know that work done given as
W= m g h
The height and mass are same for both therefore there will do the same work.
But we know that rate of work is known as power.Therefore Ben is more power .
Power
[tex]P=\dfrac{W}{t}[/tex]
t is less for Ben that is why power is more in Ben.
The answer will be "Same" and "Ben Pumpiniron".
Explanation:
Both Ben and Will are doing same amount of work. that is lifting 100 pounds of barbell 10 times.
They apply the same force to lift the same barbell up to same distance hence, they do equal work.
Yet Ben is more powerful than Will because rate of work done (Power) by Ben is higher than the that of Will.
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36.0 m. During the collision at the bottom of the elevator shaft, a 85.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
Answer:
a) I = -2257.6 Kg*m/s
b) F = -451,520N
Explanation:
part a.
we know that:
I = [tex]P_f-P_i[/tex]
where I is the impulse, [tex]P_f[/tex] the final momentum and [tex]P_i[/tex] the initial momentum.
so:
I = [tex]MV_f-MV_i[/tex]
where M is the mass, [tex]V_f[/tex] the final velocity and [tex]V_i[/tex] the initial velocity.
Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.
now, we will use the law of the conservation of energy:
[tex]E_i=E_f[/tex]
so:
mgh = [tex]\frac{1}{2}MV_i^2[/tex]
where g is the gravity and h the altitude. So, replacing values, we get:
(85kg)(9.8m/s^2)(36m)= [tex]\frac{1}{2}(85kg)V_i^2[/tex]
solving for [tex]V_i[/tex]:
[tex]V_i = 26.56m/s[/tex]
Then, replacing in the initial equation:
I = [tex]MV_f-MV_i[/tex]
I = [tex](85kg)(0m/s)-(85kg)(26.56m/s)[/tex]
I = -2257.6 Kg*m/s
Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.
part b.
we know that:
Ft = I
where F is the average force, t is the time and I is the impulse. So, replacing values, we get:
F(0,005s) = -2257.6 Kg*m/s
solving for F:
F = -451520N
Finally, the force is -451,520N, it is negative because it is upwards.
The position of a particle moving along a straight line at any time t is given by s(t)=t^3+9t^2-27.
Find the velocity of the particle at the time when the acceleration is zero.
Answer:
v= - 27 m/s
Explanation:
Given that
s= t³- 9 t²-27 ( Correct from sources)
As we know that velocity given as
[tex]v=\dfrac{ds}{dt}[/tex]
v=3 t ² - 18 t ------------1
As we know that acceleration given as
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]v=\dfrac{d^2s}{dt^2}[/tex]
v=3 t ² - 18 t
a=6 t -18
Given that acceleration is zero (a= 0 )
0 = 6 t - 18
t= 3 sec
Now by putting the values in the equation 1
v=3 t ² - 18 t m/s
v=3 x 3 ² - 18 x 3 m/s
v= 27 - 54 m/s
v= - 27 m/s
Given:
[tex]s(t)=t^3+9t^2-27[/tex]As we know,
The velocity:
→ [tex]v = \frac{ds}{dt}[/tex]
[tex]v = 3t^2 -18 t[/tex] ...(eqn 1)
and,
The acceleration:
→ [tex]a = \frac{dv}{dt}[/tex]
[tex]a = 6t -18[/tex]
When, a = 0
→ [tex]0 = 6t -18[/tex]
[tex]6t = 18[/tex]
[tex]t = \frac{18}{6}[/tex]
[tex]= 3 \ sec[/tex]
By putting the values in "eqn 1", we get
→ [tex]v = 3t^2-18t[/tex]
[tex]= 3\times 3^2-18\times 3[/tex]
[tex]= 27-54[/tex]
[tex]= -27 \ m/s[/tex]
Thus the above approach is appropriate.
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Calculate the work WAB done by the electrostatic force on a particle of charge q as it moves from A to B.
Express your answer in terms of some or all the variables E, q, L, and α.
Answer:
[tex]W_{AB} = EqL[/tex]
Explanation:
The work done by the electrostatic force is
[tex]W_{AB} = \int\limits^A_B {\vec{F}(x)} \, d\vec{x}[/tex]
where F can be calculated by Coulomb's Law:
[tex]\vec{F} = \frac{1}{4\pi \epsilon_0}\frac{q_0q_1}{x^2}[/tex]
We can express this equation by the variables given in the question.
Electric field is denoted as E.
[tex]\vec{F} = \vec{E}q = Eq~(+\^i)[/tex]
The distance, x, is given as L. If B is greater than A, the work done is positive. Else, work is negative.
[tex]W_{AB} = \int\limits^A_B {\vec{E}q} \, d\vec{x} = Eq(B-A) = EqL[/tex]
The work done on a charge q moving from point A to B in an electrostatic field E can be calculated using the formula: WAB = - q * E * L * cos(α). The maximum work is done when the charge is moving parallel to the electric field.
Explanation:The work done on a charge, q, moving under an electrostatic field, E, from point A to B - referred to as WAB - can be calculated using the formula: WAB = - q * E * L * cos(α), where E is the electric field intensity, q is the charge, L is the displacement, and α is the angle between the electric field vector and the displacement.
This equation dictates that the work done is inversely proportional to the cosine of the angle between the electric field and the displacement direction - implying that the maximum work is done when the charge is moving parallel to the electric field direction(alpha=0).
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You attach a block to the bottom end of a spring hanging vertically. You slowly let the block move down and find that it hangs at rest with the spring stretched by 18.0 cm. Next, you lift the block back up and release it from rest with the spring unstretched. What maximum distance does it move down
Answer:36 cm
Explanation:
Given
At Equilibrium extension a is 18 cm
suppose m is the mass of block
thus
[tex]ka=mg[/tex]
[tex]k=\frac{mg}{a}[/tex]
[tex]k=\frac{mg}{0.18}[/tex]
Now if the block is released from un-stretched position
conserving Energy and supposing block moves x units down
[tex]\frac{1}{2}kx^2=mgx[/tex]
[tex]x=\frac{2mg}{k}[/tex]
[tex]x=2mg\times \frac{0.18}{mg}[/tex]
[tex]x=0.36 \approx 36 cm[/tex]
If an astronaut landed on a planet that had the same radius as Earth but 4 times its mass, then the astronaut's weight on the planet would be:
Answer:
4 times
Explanation:
[tex]M[/tex] = mass of the earth
[tex]R[/tex] = radius of the earth
[tex]g_{e}[/tex] = acceleration due to gravity on earth
acceleration due to gravity on the earth is given as
[tex]g_{e} =\frac{GM}{R^{2}}[/tex]
[tex]w_{e}[/tex] = weight of the astronaut on earth
weight of the astronaut on earth is given as
[tex]w_{e} = m g_{e} = \frac{GMm}{R^{2}}[/tex]
[tex]M_{p}[/tex] = mass of the planet = [tex]4 M[/tex]
[tex]R_{p}[/tex] = radius of the planet = R
[tex]g_{p}[/tex] = acceleration due to gravity on earth
acceleration due to gravity on the planet is given as
[tex]g_{p} =\frac{GM_{p}}{R_{p}^{2}}\\g_{p} = \frac{4GM}{R^{2}}\\g_{p} = 4 g_{e}[/tex]
[tex]w_{p}[/tex] = weight of the astronaut on planet
weight of the astronaut on planet is given as
[tex]w_{p} = m g_{p}\\w_{p} = m (4) g_{e}\\w_{p} = 4 w_{e}[/tex]
hence the weight of the astronaut on the planet is four times.
Final answer:
An astronaut would weigh 4 times more on a planet with the same radius as Earth but 4 times its mass, due to the direct relationship between mass and gravitational force.
Explanation:
To understand how an astronaut's weight would change on a planet with the same radius as Earth but 4 times its mass, we need to consider the universal law of gravitation. The formula to calculate gravitational force (which determines weight) is F = G (m1m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two masses (the radius of the planet, in this case).
Since the planet has 4 times the mass of Earth but the same radius, applying these values to the formula shows that the astronaut's weight would be 4 times greater on this new planet compared to Earth. The increase in mass directly increases the gravitational force, while the radius remains constant, leading to an increase in weight.
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle of elevation changing when the balloon is at an altitude of 2500 feet?
Answer:
0.0031792338 rad/s
Explanation:
[tex]\theta[/tex] = Angle of elevation
y = Height of balloon
Using trigonometry
[tex]tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta[/tex]
Differentiating with respect to t we get
[tex]\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta[/tex]
Now, with the base at 200 ft and height at 2500 ft
The hypotenuse is
[tex]h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft[/tex]
Now y = 2500 ft
[tex]cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974[/tex]
[tex]\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s[/tex]
The angle is changing at 0.0031792338 rad/s
The rate of change of the angle of elevation in this problem would be calculated using the principles of related rates problems in calculus. This involves setting up an equation relating the quantities of interest (here, the altitude of the balloon and the angle of elevation seen by the observer), and differentiating this equation with respect to time. The precise calculation could not be completed as the question lacked certain specific data.
Explanation:Rate of change of the angle of elevation is being calculated in this question. This question falls into the category of related rates problems in calculus, where the rate of change of one quantity (here, the altitude or the height to which the balloon has ascended) affects the rate of change of another quantity (the angle of elevation seen by the observer on the ground). However, the provided reference information does not appear to give specific details needed to answer this question.
Generally, you would use the properties of a right angled triangle and the concept of derivatives to set up an equation relating the quantities and find the value. For example, if we consider a right triangle where the observer and the balloon at a particular moment form the ends of a line representing the hypotenuse, and the angle of elevation to the balloon as θ, then tan(θ) = (altitude of the balloon) / (distance of observer from the launch point). And you'd differentiate this equation with respect to time to find the rate of change of the angle θ. However, again, the specific calculation may require more detailed information not provided here.
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A 4.00-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70.0 kg construction worker stands at the far end of the beam. What is the magnitude of the gravitational torque about the point where the beam is bolted into place?
Answer:
12544 N.m
Explanation:
given,
length of beam = 4 m
mass of steel = 500 Kg
mass of worker = 70 Kg
torque at hinge = ?
weight of the beam will at the center of mass which is at 4/2 = 2 m distance form the hinge.
distance of worker = 4 m
torque acting at hinge=
= Mg x 2 + mg x 4
= 500 x 9.8 x 2 + 70 x 9.8 x 4
= 12544 N.m
hence, torque experience at the hinge is equal to 12544 N.m
Compare the Lagrangian method with the Newton-Euler method in detail.
The comparison of the Lagrangian and Newton methods are explained. Explanation:
The Newton-Euler Method is derived by Newton's Second Law of Motion, that describes the dynamic systems in terms of force and momentum.
It deals with concentration of particles to calculate the overall diffusion and convection of a number of particles.
The Lagrangian Method the dynamic behavior is described in terms of work and energy.
It deals with individual particles to calculate the trajectory of each particle separately.
The Lagrangian method and the Newton-Euler method are two approaches used to solve problems in classical mechanics involving Newton's laws of motion. The Newton-Euler method relies directly on Newton's second law and free-body diagrams, while the Lagrangian method is based on the principle of least action and the Euler-Lagrange equations. The choice between the two methods often depends on the complexity and nature of the physical system in question.
Explanation:When comparing the Lagrangian method with the Newton-Euler method, it is important to understand that both approaches are used to solve complex problems in classical mechanics, which often involve the application of Newton's laws of motion. The Newton-Euler method is more traditional and is founded upon direct application of Newton's second law, F=ma (force equals mass times acceleration), and the related concepts for rotary motion. This method involves creating a free-body diagram, identifying all the forces acting on the object, and applying Newton's second law to find the accelerations and subsequently the positions and velocities of the object in question.
Contrastingly, the Lagrangian method is a more modern approach, which is grounded in the principle of least action. Instead of focusing on forces, it involves the calculation of the Lagrangian, which is the difference between an object's kinetic and potential energies, and applying the Euler-Lagrange equations to find the equations of motion. This method is particularly powerful in systems where the forces are conservative and can be derived from a potential energy; the Lagrangian method is also more convenient when dealing with complex constraints and coordinate systems that are not Cartesian.
While the Newton-Euler method is practical and straightforward, especially in simple scenarios with few forces or when numerical solutions are required, the Lagrangian approach offers more flexibility and can simplify calculations in systems with symmetries or non-standard geometries. Understanding the application of these methods is an integral part of a problem-solving procedure when using Newton's laws of motion, and both methods reinforce concepts useful across various areas of physics.
When rock retains its new shape (without fracturing) after stress has been removed, it has undergone ____________________ ____________________.
Answer:
Elastic deformation
Explanation:
When rock retains its new shape (without fracturing) after stress has been removed, it has undergone elastic deformation.
Elastic deformation is the process in which the object under temporary stress regains its original shape after the stress is removed. The stress must be low and not strong enough that it can create fracture and permanent damage to the object. for example, we can stretch a rubber band by applying some force and when we remove the force, the rubber band comes back to its original length and shape.
If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the cell potential at 25.0°C. The standard reduction potentials are: $$ Eo = 0.80 V $$ Eo = 0.00 V
Answer: The cell potential is 0.712 V
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, silver will undergo reduction reaction will get reduced. Hydrogen will undergo oxidation reaction and will get oxidized
Half reactions for the given cell follows:
Oxidation half reaction: [tex]H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V[/tex]
Reduction half reaction: [tex]Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V[/tex] ( × 2 )
Net reaction: [tex]H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.80-(0.00)=0.80V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ag^{+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.80 V
n = number of electrons exchanged = 2
[tex][Ag^{+}]=0.0115M[/tex]
[tex][H^{+}]=0.355M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V[/tex]
Hence, the cell potential is 0.712 V
The cell potential of the cell is 0.71 V.
The overall equation of the redox reaction is;
2Ag^+(aq) + H2(g) ----> 2H^+(aq) + Ag(s)
The E°cell = E°cathode - E°anode
E°cell = 0.80 V - 0.00 V = 0.80 V
Using the Nernst equation;
Ecell = E°cell - 0.0592/n logQ
Where n = 2
Q = [H^+]^2/[Ag^+]^2 = (0.355 M)^2/(0.0115 M)^2
Q = 952.7
Substituting values;
Ecell = 0.80 V - 0.0592/2 log ( 952.7)
Ecell = 0.71 V
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An 880 kg cannon at rest fires a 12.4 kg cannonball forward at 540 m/s. What is the recoil velocity of the cannon? (Unit m/s)
Answer:
7.61 m/s backwards
Explanation:
Initial momentum = final momentum
0 = (880 kg) v + (12.4 kg) (540 m/s)
v = -7.61 m/s
The cannon's recoil is 7.61 m/s backwards.
Answer:
The recoil velocity vector of the cannon is [tex](7.609,0,0)\frac{m}{s}[/tex]
Explanation:
We can solve this problem by applying the Momentum Conservation Principle.
The principle of conservation of momentum states that when you have an isolated system with no external forces, we can use the following equation to calculate the final velocity of one object.
[tex]m1.v1=m2.v2[/tex] (I)
Where ''[tex]m1[/tex]'' and ''[tex]v1[/tex]'' are the mass and velocity of the first object.
And where ''[tex]m2[/tex]'' and ''[tex]v2[/tex]'' are the mass and velocity of the second object.
The momentum is a vectorial magnitude.
If we use the equation (I) with the data given :
[tex](880kg).v1=(12.4kg).(540\frac{m}{s})[/tex]
[tex]v1=7.609\frac{m}{s}[/tex]
If we considered as negative the sense of the velocity vector from the cannonball, the cannon's velocity vector will have the same direction but opposite sense that the cannonball's velocity vector (It will be positive).
We can give it a vectorial character like this :
[tex]v1=(7.609,0,0)\frac{m}{s}[/tex]
The velocity vector will be entirely in the x-axis.
What is the relationship between height and gravitational potential energy of water behind a dam?
Water is contained in a tidal basin behind a dam. The water has a depth h at high tide and zero at low tide. The gravitational potential energy of the water stored in the basin between a high tide and a low tide is proportional to
1. h^(1/2)
2. h
3. h^2
4. h^3
Answer:
3. [tex]h^2[/tex]
Explanation:
Energy of volume contained in a dam is given by
[tex]E=\dfrac{1}{2}A\rho gh^2[/tex]
where,
A = Horizontal surface area of the barrage
[tex]\rho[/tex] = Density of water
g = Acceleration due to gravity = 9.81 m/s²
h = Vertical tidal range,
It can be seen that
[tex]E\propto h^2[/tex]
Hence, the energy is proportional to [tex]h^2[/tex]
Final answer:
The gravitational potential energy of water behind a dam is directly proportional to the height of the water column, using the formula PE = mgh. For a tidal basin, the GPE is proportional to the average height (h/2), therefore, the correct relationship is proportional to the height h. So the correct option is 2.
Explanation:
The relationship between the height and gravitational potential energy (GPE) of water behind a dam is that GPE is directly proportional to the height of the water column. The formula to calculate the GPE is PE = mgh, where PE is the potential energy, m is the mass of the water, g is the acceleration due to gravity, and h is the height above the reference point.
When we consider a tidal basin with water depth h at high tide transitioning to zero at low tide, the average height of water behind the wall is h/2. Since mass (m) is the product of the water's density (assumed to be 1,000 kg/m³ for water) and volume—which, in turn, is the area of the water surface (A) multiplied by its height (h)—the GPE of water stored between a high tide and low tide is proportional to the average height, thus proportional to h. Therefore, the correct answer is option 2, which states that the gravitational potential energy of the water stored in the basin between high tide and low tide is proportional to h.
When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?
Answer:
Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.
Explanation:
We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.
According to the Ohm's law, we find its resistance:
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{120}{20}[/tex]
[tex]R=6\ \Omega[/tex]
So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.
Here the value of frequency is the number of times the current changes its direction or the polarity in one second.
The control of the AC power frequency is managed by power plant generators, which use turbine controls to ensure electricity alternates at a specific frequency, such as the standard 60 Hz in North America. Voltage and current in AC power cross zero 120 times a second due to this oscillation.
Alternating Current (AC) electric power is characterized by a voltage that alternates in polarity from positive to negative, resulting in a sinusoidal waveform. The frequency of this oscillation in North America is 60 times per second (60 Hz). This oscillation means that the voltage swings from a peak value to the opposite peak value, crossing zero in the process, twice every cycle. Therefore, for a 60 Hz AC supply, voltage and current cross zero 120 times a second.
The control of the frequency at which this oscillation occurs is managed by the generator at the power plant. In the case of a hydro-electric dam, turbine controls adjust the rotational frequency to produce the desired AC frequency. For example, in North America, these turbines are regulated to rotate in such a way that the electricity they generate alternates at 60 Hz. This regulated rotation is what ensures the consistency of the 120 volts at 60 cycles per second.
The power consumption of an electrical device such as a light bulb in an AC circuit also swings. Peak power is calculated by multiplying peak current by peak voltage. Given that the root mean square (RMS) voltage is 120 volts for a standard AC outlet, the peak voltage is actually higher, around 170 volts. Thus, for a 60-watt light bulb, the power consumption will pulse from zero up to its peak value 120 times per second, averaging out to the rated 60 watts.
An object falls freely from rest on a planet where the acceleration due to gravity is it is on Earth. In the first 5 seconds it falls a distance of twice as much as
A) 250 m.
B) 500 m.
C) 150 m.
D) 100 m. E) none of these
Answer:
A. The object falls a distance of 250 m
Explanation:
Hi there!
In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.
The equation of height of a falling object is the following:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the object after a time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).
Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.
Then, the height of the object after 5 s will be :
y = 1/2 · 2 · g · t² (notice that the acceleration due to gravity is 2 · g)
y = g · t²
y = -10 m/s² · (5 s)²
y = -250 m
The object falls a distance of 250 m.
The object falling on this other planet with the same gravity as Earth falls a distance closest to 250m in 5 seconds, according to the standard formula and doubling the Earth result.
Explanation:Given the planet where the object is falling has the same acceleration due to gravity as Earth, we can use the formula for the distance fallen, d (in meters), over a certain time, t (in seconds), assuming the object falls from rest. This formula is d = 0.5 * g * t^2, where g is the acceleration due to gravity. For Earth, g is approximately 9.8 m/s^2.
When t=5 seconds on Earth, using the given formula, d = 0.5 * 9.8 * (5^2) = 122.5 meters. Since the object on the other planet falls twice the distance in the same time, the object falls 2 * 122.5 = 245 meters. Thus, Choice A) 250 m is the closest to the calculated value.
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A drone traveling horizontally at 110 m/s over flat ground at an elevation of 3000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by x = 110 t , y = − 4.9 t 2 + 3000 , t ≥ 0 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.
Answer:
The horizontal distance of the target should be 2721,4 meters.
Explanation:
First of all we need to find the time that the emergency package hits the ground after the moment of release:
y=0 (because when it hits the ground it is on the level of 0m);
[tex]0=-4,9*t^2+3000\\t=24,74[/tex]
The emergency package hits the ground after 24,74 seconds from release.
Lets assume that package preserves his 110 m/s horizontal speed during the free fall. The targets horizontal distance is:
[tex]110*24,74=2721,4[/tex]
2721,4 meters
Because of its center-surround organization, a neuron that has its entire receptive field exposed to bright light will:
a. fire slowly until the light turns off, then begin firing rapidly.
b. maintain the same rate of firing as if there was no light presented.
c. stop firing entirely.
d. fire rapidly.
Answer:
b. maintain the same rate of firing as if there was no light presented.
Explanation:
The neuron will maintain the same rate of firing as if there was no light presented.
Receptive filed of neuron is place on its sensory surface generally on the back of eye that an stimulus must reach to activate the neuron. The bright light cannot change the rate of firing of the neuron.
Neurons are messengers of information. A neuron that has its full receptive field exposed to strong light will maintain the same rate of firing as if there was no light present because of its center-surround arrangement.
What are neurons?Neurons are messengers of information. They transfer information between different parts of the brain and between the brain and the rest of the nervous system through electrical impulses and chemical signals.
A neuron that has its full receptive field exposed to strong light Hence it will maintain the same rate of firing as if there was no light present because of its center-surround arrangement.
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Galileo found that a ball rolling down one inclined plane would roll how far up another inclined plane?
A) the ball would not roll up the other plane at all
B) to nearly its original height
C) to about one quarter its original height
D) to nearly twice its original height
E) to nearly half its original height
Answer: B.) To nearly its original height
Explanation:
Because if you have a ball than the inclined plane is heading up you would get almost the same height as it started.
Galileo found that a ball rolling down one inclined plane then, the ball will reach about the original height. Hence, option B is correct.
What is momentum?The momentum is the result of a particle's velocity and its mass. Force and motion, meaning it has both magnitude and the direction.
According to Isaac Newton's second equation of motion, the force acting on a particle equals the time rate of increase of momentum.
The impulse, which is the product of the force and the intervals (the impulse), is equal to the difference in momentum, according to Newton's 2nd law, if a steady force operates on a particle for a specific amount of time.
On the other hand, a particle's momentum is the amount of time needed for a consistent action to fight it to rest.
Because if you were to throw a ball while the inclined plane was moving upward, you would nearly reach its starting height.
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Read the false statement. In an atom, protons and electrons are in the nucleus, which is surrounded by neutrons. Which option rewords the false statement so it is true?
1. In an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
2. In an atom, electrons and neutrons are in the nucleus, which is surrounded by protons.
3. In an atom, electrons and neutrons are in the nucleus, which is surrounded by photons.
4. In an atom, electrons and photons are in the nucleus, which is surrounded by neutrons.
Answer: A.
In an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
Explanation:
It must first be noted that an atom consists of protons, neutrons and electrons.
The proton and neutron is contained in the atom while the electron is found in the outer most shell of the atoms. It can be concluded then that in an atom, protons and neutrons are in the nucleus, which is surrounded by electrons.
Answer:
A
Explanation:
Two men decide to use their cars to pull a truck stuck in the mud. They attach ropes and one pulls with a force of 615 N at an angle of 31◦with respect to the direction in which the truck is headed, while the other car pulls with a force of 961 N at an angle of 25◦with respect to the same direction. What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N.
Answer:
1398.12 N
Explanation:
We define the x-axis in the direction parallel to the movement of the truck on and the y-axis in the direction perpendicular to it.
x-components of the ropes forces
T₁x = 615N*cos31°=527.1579 N :Tension in direction x of the rope of the car 1
T₂x= 961 N*cos25°=870.96 N :Tension in direction x of the rope of the car 2
Net forward force exerted on the truck in the direction it is headed (Fnx)
Fnx = T₁x + T₂x
Fnx = 527.1579 N + 870.96 N
Fnx = 1398.12 N