If 4.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.73 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer 1

Explanation:

Below is an attachment containing the solution.

If 4.00 103 Kg Of Gold Is Deposited On The Negative Electrode Of An Electrolytic Cell In A Period Of
Answer 2
Final answer:

To calculate the current, we first convert the mass of gold deposited to moles and then use the equivalence of one mole of gold requiring one faraday of charge. We calculate the total charge transferred and then find the average current by dividing this charge by the total time in seconds.

Explanation:

To find the current in the cell during the period when 4.00 × 10⁻³ kg of gold was deposited on the negative electrode, we need to calculate the total charge that resulted in the deposition of gold. First, we will convert the mass of gold to moles using the molar mass of gold, which is approximately 197 g/mol. Since gold ions carry one unit of positive charge, one mole of gold ions will require one faraday (96,485 C) to be reduced and deposited as gold metal.

After calculating the number of moles of gold, we can then determine the number of moles of electrons that moved through the cell using the equivalence of moles of electrons to moles of gold deposited. Multiplying this number by the charge per mole of electrons (1 faraday), we get the total charge moved. To find the current, we divide the total charge by the total time in seconds, and this gives us the average current over the 2.73 hours.


Related Questions

Three identical 4.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 4.0 m/s2 . What is the magnitude of the force exerted on the right cube by the middle cube in this case

Answers

Final answer:

The magnitude of the force exerted on the right cube by the middle cube, when three identical cubes accelerate together on a frictionless surface, is calculated using Newton's second law (F = ma) and is found to be 16.0 Newtons.

Explanation:

The student's question pertains to Newton's second law of motion and the concept of force and acceleration. The problem involves three identical cubes on a frictionless surface accelerating due to a force applied to the first cube. To find the magnitude of the force exerted on the right cube by the middle cube, we use the formula F = ma, where F is the force, m is the mass, and a is the acceleration.

Since the cubes are identical and accelerate together, each 4.0-kg cube has an acceleration of 4.0 m/s². Therefore, the force exerted by one cube on another can be found by multiplying one cube's mass by the given acceleration:

F = m × aF = 4.0 kg × 4.0 m/s²F = 16.0 N

The force exerted on the right cube by the middle cube is 16.0 Newtons.

Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 2 meters per second and the second car's velocity is 9 meters per second. At a certain instant, the first car is 8 meters from the intersection and the second car is 6 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer:
a -10
b -7
© -8.4

Answers

Answer:

Explanation:.

Given that

First car velocity is 2m/s

Second car velocity is 9m/s.

At a certain time the first car is 8m from intersection

And at the same time second car is 6m from intersection.

The rate of change of distance, i.e dx/dt, which is the speed of the car.

Using Pythagoras theorem, their distance apart is given as

Z²=X²+Y²

Z²=6²+8²

Z²=36+64

Z²=100

Z=√100

Z=10m

Let assume the direction,

Let assume the first car is moving in positive x direction, then

dx/dt=2m/s

And also second car will be moving in negative y direction

dy/dt=-9m/s

Now, to know dz/dt, let use the Pythagoras formulae above

x²+y²=z²

differentiate with respect to t

2dx/dt+2dy/dt=2dz/dt

Divide through by 2

dz/dt=dx/dt+dy/dt

dz/dt=2-9

dz/dt=-7m/s

The rate of change of distance between the two body is -7m/s

Option B is correct

Answer:

Rate of change of the distance between the cars = -7 m/s

Explanation:

Let the distance between the first car and the intersection be p = 8 meters

Let the distance between the second car and the intersection be q = 6 meters

velocity of the first car, dp/dt = -2 m/s

velocity of the second car, dq/dt = -9 m/s

We can get the distance between the two cars using pythagora's theorem

s² = p² + q²...................................(1)

s² = 8² + 6²

s = √(8² + 6²)

s = 10 m

Differentiating equation (1) through with respect to t

2s ds/dt = 2p dp/dt + 2q dq/dt

(2*10*ds/dt) = (2*8*(-2)) + (2*6*(-9))

20 ds/dt = -32 - 108

20 ds/dt = -140

ds/dt = -140/20

ds/dt = -7 m/s

Two bodies are falling with negligible air resistance, side byside, above a horizontal plane near Earth's surface. If one of thebodies is given an additional horizontal acceleration during itsdescent, it ........................
(a) strikes the plane at the same time as the other body
(b) has the vertical component of its velocity altered
(c) has the vertical component of its accelerationaltered
(d) follows a hyperbolic path
(e) follows a straight line path along the resultantacceleration vector

Answers

Answer:

(a) strikes the plane at the same time as the other body

Explanation:

Two bodies are falling with negligible air resistance, side by side, above a horizontal plane near Earth's surface. If one of the bodies is given an additional horizontal acceleration during its descent, it strikes the plane at the same time as the other body

Suppose an Atwood machine has a mass of m1 = 6.0 kg and another mass of m2 = 2.0 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.

Answers

Answer:

Acceleration=[tex]4.9 /s^2[/tex]

Tension=29.4 N

Explanation:

We are given that

[tex]m_1=6 kg[/tex]

[tex]m_2=2 kg[/tex]

We have to find the magnitude of the acceleration of the two objects and the tension in the cord.

Tension, [tex]T=m_1(a+g)[/tex]

[tex]m_2g-T=m_2a[/tex]

Substitute the values

[tex]m_2g-m_1(a+g)=m_2a[/tex]

[tex]m_2g-m_1a-m_1g=m_2a[/tex]

[tex]g(m_2-m_1)=m_2a+m_1a=a(m_1+m_2)[/tex]

[tex]a=\frac{(m_2-m_1)g}{m_1+m_2}[/tex]

Substitute the values

[tex]a=\frac{(2-6)\times 9.8}{2+6}=-4.9m/s^2[/tex]

Where [tex]g=9.8m/s^2[/tex]

Hence, the magnitude of the acceleration of the two objects =[tex]4.9 m/s^2[/tex]

Substitute the values of a

[tex]T=m_1(a+g)=6(-4.9+9.8)=29.4 N[/tex]

A spring with a force constant of 5400 N/m and a rest length of 3.5 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m)? (Assume the rock is launched from ground height.) m

Answers

Answer:

5.51 m

Explanation:

From the question,

The energy used to stretch the spring = the potential energy of the rock.

(1/2)ke²  = mgh ................. Equation 1

Where k = spring constant, e = extension/compression, m = mass of the rock, g = acceleration due to gravity, h = height of the rock above the ground

make h the subject of the equation.

h = ke²/2mg ....................equation 2

Given: k = 5400 N/m, e = 1 m, m = 48 kg.

Constant: g = 9.8 m/s²

Substitute into equation 2

h = 5400(1²)/(2×48×9.8)

h = 5400/940.8

h = 5.51 m.

Hence the height of the rock = 5.51 m

Before entering the cyclotron, the particles are accelerated by a potential difference V. Find the speed v with which the particles enter the cyclotron.

Answers

Work = F.d= qE.d=q.V
Kenetic energy= 1/2. M.v^2
qV=(1/2)mv^2
v= sqrt(2qV/m)

If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 56.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answers

Answer:

Change in momentum =15.01kgm/s

Impulse applied by bat = 15.01Ns

Explanation:

Please see attachment below.

a proton is fired with a speed of 200,000m/s from the midpoint of the capacitor toward the positive plate. This speed is insufficient to reach the positive plate. What is the proton's speed as it collides with the negative plate

Answers

Answer: The proton speed = 3 × 10^5m/s

Explanation: The electric P.E change the proton if It can reach the positive plate.

The workdone

I have attached an image of the diagram showing the nature of this motion

Answer:

Protons speed = 2.96 x 10^(5) m/s

Explanation:

A) At closest point of approach to the positive plate, the proton came to rest momentarily.

Thus;

Loss in Kinetic Energy = Gain in Electric potential energy

Hence;

(1/2)(mv^(2)) = eΔV

So, ΔV = (mv^(2))/(2e)

Mass of proton = 1.673 × 10-27 kilograms

Proton elementary charge(e) = 1.6 x 10^(-19) coulumbs

And from the question v = 200,000 m/s

So, ΔV = [1.673 × 10^(-27) x 200000^(2)] / (2 x 1.6 x 10^(-19)) = 209 V

This is less than 250V which is half of the charge at the positive plate shown in the diagram.

Therefore, the speed is insufficient to reach the positive plate from P to Q.

B) Gain in KE = qΔV

Thus; 1/2mvf^(2) - 1/2mvi^(2) = eΔV

Where, vf is final velocity and vi is initial velocity.

So simplifying, we get;

vf^(2) - vi^(2) = (2eΔV)/m

So, vf = √[(2eΔV)/m) + (vi^(2))

= √[(2 x 1.6 x 10^(-19) x 250)/(1.673 × 10^(-27)) + (200,000^(2))

= 2.96 x 10^(5) m/s

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons transported from the positive electrode to the negative electrode from the time the battery is first used until it is essentially dead?

a. 3.7 x 10^18
b. 5.3 x 10^23
c. 4.4 x 10^22
d. 1.6 x 10^19
e. 1.2 x 10^23

Answers

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

[tex]I(t)=0.88e^{-t/6\times3600s}[/tex]

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

[tex]q=\int\limits {I} \, dt[/tex]

Here the limits of integration is from 0 to infinite. So,

[tex]q=\int\limits {0.88e^{-t/6\times3600s}}\, dt[/tex]

[tex]q=0.88\times(-6\times3600)(0-1)[/tex]

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

[tex]N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}[/tex]

N = 1.2 x 10²³

The dean of a university located near the ocean (who was not a science major in college) proposes building an infrared telescope right on campus and operating it in a nice heated dome so that astronomers will be comfortable on cold winter nights. Criticize this proposal, giving your reasoning.

Answers

Answer:

The proposal is bad and has many mistakes in itself.

Explanation:

-Telescopes are preferably to be placed far from cities, ideally a remote place on a mountain or raised land. It should not be erected at sea level.

-It should not be erected at sea level. High humidity present at sea level clouds observations made as infrared observations are not possible at sea level.

-Dome should not be heated at all. Strong air currents are generated during heating which inturn ruins observations. Also, a heated dome will emit infrared radiation which ultimately swamps astronomical signals.

Final answer:

Coastal location and heated dome are unsuitable for infrared telescope due to atmospheric interference, light pollution, and dome-related issues. Consider remote location, dry climate, and remote operation for optimal observations. Consult experts and consider environmental impact.

Explanation:

Criticisms of Building an Infrared Telescope Near the Ocean in a Heated Dome:

While the intention might seem well-meaning, building an infrared telescope near the ocean in a heated dome has several drawbacks:

Unsuitable Location:

Atmospheric interference: Coastal areas have higher humidity and turbulent air, which negatively affects infrared observations due to water vapor absorption and image distortion.

Light pollution: City lights and nearby human activity create significant light pollution, impacting observations of faint infrared sources.

Dome Issues:

Cost: Building and maintaining a heated dome adds significant expense compared to an open-air observatory.

Heat distortion: Heating the dome creates air currents that can distort telescope observations.

Ventilation challenges: Maintaining controlled airflow and humidity within the dome can be complex and costly.

Alternatives:

Remote location: Building the telescope at a high-altitude, dry site with minimal light pollution would be more suitable for infrared observations.

Remote access and automation: Modern telescopes can be operated remotely, eliminating the need for astronomers to be physically present during observations.

Additional considerations:

Expertise: Consult with professional astronomers for advice on telescope placement and operation.

Environmental impact: Consider the potential ecological impact of the telescope and dome on the coastal environment.

Overall, while the concern for astronomer comfort is understandable, a coastal location with a heated dome is not optimal for an infrared telescope. Exploring alternative locations and remote operation technologies would be more effective and cost-effective for achieving high-quality scientific observations.

Three charged particles form a triangle: particle 1 with charge Q1 = 80.0 nC is at xy coordinates (0, 3.00 mm), particle 2 with charge Q2 is at (0, −3.00 mm), and particle 3 with charge q = 18.0 nC is at (4.00 mm, 0). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) −80.0 nC?

Answers

Final answer:

The net electrostatic force on particle 3 due to the other two particles depends on the charges of these particles. When Q2 = 80.0 nC, the forces from the other two particles cancel out and when Q2 = -80.0 nC, the forces add up.

Explanation:

The electrostatic force on a charge due to other charges can be determined using Coulomb’s law. For the setup in the question, the deletion force on particle 3 because of particles 1 and 2 can be obtained by vectorially adding the forces it experiences due to each of these particles separately. When Q2 = -80.0 nC, the forces that particle 3 experiences due to particle 1 and particle 2 are in the same direction this time, therefore they add up to give the net force on particle 3. We can determine the exact value by substituting the given values in the equation for Coulomb's Law.

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A 5 meter ladder leans against a wall. The bottom of the ladder is 1 meter from the wall at time t = 0 and slides away from the wall at 1 meters per second. Find the velocity of the top of the ladder at time t = 2.

Answers

Answer:

V = –0.89m/s

Explanation:

Please see attachments below.

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Final answer:

When the beads reach the ends of the rod, the angular speed of the system is approximately 5.76 rad/s.

Explanation:

To find the angular speed of the system at the instant the beads reach the ends of the rod, we can use the principle of conservation of angular momentum. Initially, the angular momentum of the system is given by Li = Irodωi + 2(Ibead)ωi, where Irod is the moment of inertia of the rod about its center, Ibead is the moment of inertia of each bead about the center, and ωi is the initial angular speed of the system. When the beads reach the ends of the rod, the moment of inertia of the system changes, but the angular momentum remains constant. So, we have Li = Irodωf + 2(Ibead)ωf, where ωf is the final angular speed of the system. We can rearrange this equation to solve for ωf. Given the values of Irod, Ibead, and ωi, we can substitute them into the equation to find ωf.

Using the given values: mass of the rod = 190 g = 0.19 kg, length of the rod = 43 cm = 0.43 m, mass of each bead = 38 g = 0.038 kg, distance of the beads from the center = 10 cm = 0.1 m, and initial angular speed = 12 rad/s, we can calculate the moments of inertia as follows:

Irod = (1/12)mrodLrod2 = (1/12)(0.19)(0.43)2 = 0.002196 kg.m2

Ibead = mbeadR2 = (0.038)(0.1)2 = 0.000038 kg.m2

Now, substituting the values into the equation Li = Irodωi + 2(Ibead)ωi, we have (0.002196)(12) + 2(0.000038)(12) = (0.002196 + 2(0.000038))ωf. Solving for ωf, we get ωf ≈ 5.76 rad/s.

A 4.9 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.6 − x^2) N, , where x is in meters and the initial position of the block is x=0(a) What is the kinetic energy of the block as it passes through x = 2.1 m?

(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.1 m?

Answers

Answer with Explanation:

We are given that

Mass of block=m=4.9 kg

Initial velocity, u=0

[tex]F=(2.6-x^2) i N[/tex]

Initial position, x=0

a.We have to find the kinetic energy of the blocks as it passes through x=2.1 m

Work done=Kinetic energy=[tex]\int_{0}^{2.1}(2.6-x^2) dx[/tex]

Kinetic energy of the block=[tex][2.6 x-\frac{x^3}{3}]^{2.1}_{0}[/tex]

Kinetic energy of the block=[tex]2.6\times 2.1-\frac{(2.1)^3}{3}-0=2.373 J[/tex]

Kinetic energy of the block=2.373 J

b.Initial kinetic energy of block=[tex]K_i=\frac{1}{2}(4.9)(0)=0[/tex]

According to work energy theorem

[tex]W=K_f-K_i[/tex]

[tex]2.373 =k_f-0[/tex]

[tex]k_f=2.373 J[/tex]

Hence, the maximum kinetic energy of the block =2.373 J

To practice Problem-Solving Strategy 25.1 Resistor Circuits. Find the currents through and the potential difference across each resistor in the circuit shown on the diagram (Figure 1) . Use the following values: E = 12.0V , R1 = 15.0Ω , R2 = 45.0Ω , R3 = 20.0Ω , and R4 = 25.0Ω .

Part A

Step by step, reduce the circuit to the smallest possible number of equivalent resistors in order to find the equivalent resistance Req of the entire circuit.

Express you answer in ohms to three significant figures.

Part B

Find Ieq, the current through the equivalent resistor.

Express your answer in amperes to three significant figures.

Answers

Answer:

I₁ = 0.32 A

I₂ = 0.16 A

I₃ = 0.16 A

I₄ = 0.16 A

Explanation:

Part A

The equivalent resistance of the circuit is

Req = R₁ + (R₂||(R₃ + R₄))

Req = 15 + (45||(20 + 25))

Req = 15 + (45||45) = 15 + ((45×45)/(45+45)) = 15 + 22.5 = 37.5 Ω

Part B

From Ohm's law,

V = IR

Ieq = V/(Req) = 12/(37.5) = 0.32 A

Part C

Current through R₁ is the same as Ieq as R₁ is directly in series with the voltage source.

I₁ = 0.32 A

Then, this current flows through the (R₂||(R₃ + R₄)) loop too as the entire loop is in series with R₁

This current is them split into two branches of R₂ and (R₃ + R₄), since these two branches have equal resistances (45 Ω and 45 Ω), 0.32 A is split equally between the R₂ and (R₃ + R₄) branch.

Current through R₂ (using current divider)

I₂ = (45/90) × 0.32 = 0.16 A

Current through (R₃ + R₄) = 0.16 A too.

And because the two resistors are in series, the same current flows through them.

I₃ = I₄ = 0.16 A

Final answer:

First, we resolve the parallel circuits using the formula for resistors in parallel to get the equivalent resistance, Rp. We then substitute this into the series circuits to get the total equivalent resistance, Req. Using Ohm's Law, we can calculate the equivalent current, Ieq.

Explanation:

To solve this, we first need to determine the equivalent resistance of the whole circuit. As it is a combination of parallel and series circuits, we have to start by replacing the resistors in parallel. Using the formula for resistors in parallel, 1/R = 1/R2 + 1/R3, the parallel combination of R2 and R3 can be replaced by one resistor with resistance Rp = 1/ ((1/R2) + (1/R3)). This gives Rp = 13.333Ω. Now we have a simple series circuit with R1, Rp, and R4. The equivalent resistance of the whole circuit is then Req = R1 + Rp + R4 = 15.0Ω + 13.333Ω + 25.0Ω = 53.333Ω.

In part B, we apply Ohm's Law: I = E/R to get the equivalent current through the circuit. Using our values for E (12V) and Req (53.333Ω), we find Ieq = 0.225A.

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( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration would the force give to an object with a mass of (m2-m1)

Answers

Answer:

[tex]4.725 m/s^{2}[/tex]

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then [tex]a=\frac {F}{m}[/tex]  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula [tex]m=\frac {F}{a}[/tex]

For [tex]m1= \frac {F}{13.5}[/tex] and [tex]m2=\frac {F}{3.5}[/tex] hence [tex]F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}[/tex]

31. Three long, straight, parallel wires all lie in the yz plane and each carries a current of 20 A in the positive z direction. The two outer wires are each 4.0 cm from the center wire. What is the magnitude of the magnetic force on a 50-cm length of either of the outer wires

Answers

Final answer:

The magnitude of the magnetic force on a wire can be calculated using the formula F = B * I * L. In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law. Plugging in the values will give you the magnitude of the magnetic force on the wire.

Explanation:

The magnitude of the magnetic force on a wire can be calculated using the formula:

F = B * I * L

Where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the wire.

In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law:

B = (μ0 * I) / (2π * r)

Where μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

Plugging in the values, the magnetic force per meter on either of the outer wires is:

F = (2 * 10-7 Tm/A * 20 A * 0.5 m) / (2π * 0.04 m)

Calculating this will give you the magnitude of the magnetic force on the wire.

car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant

Answers

Answer:

0.308 m/s2 at an angle of 13.5° below the horizontal

Explanation:

The parallel acceleration to the roadway is the tangential acceleration on the rise.

The normal acceleration is the centripetal acceleration due to the arc. This is given by

[tex]a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072[/tex]

The tangential acceleration, from the question, is

[tex]a_T = 0.300[/tex]

The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by

[tex]a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2[/tex]

[tex]a = 0.308[/tex]

The angle the resultant makes with the horizontal is given by

[tex]\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400[/tex]

[tex]\theta=13.5[/tex]

Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc

Current passes through a solution of sodium chloride. In 1.00 s, 2.68×1016Na+ ions arrive at the negative electrode and 3.92×1016Cl− ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

Answers

Explanation:

Given that,

Number of sodium ions at the negative electrode, [tex]Na^+=2.68\times 10^{16}[/tex]

Number of chloride ions at the positive electrode, [tex]Cl^-=3.92\times 10^{16}[/tex]

(a) The current flowing in the circuit is due to the positive as well as negative charges such that total charge becomes:

[tex]Q=(Na^++Cl^-)e[/tex]

[tex]Q=(2.68\times 10^{16}+3.92\times 10^{16})(1.6\times 10^{-19})[/tex]

Q = 0.01056 C

The current is given by :

[tex]I=\dfrac{Q}{t}[/tex]

[tex]I=\dfrac{0.01056}{1}=10.56\ mA[/tex]

So, the current passing between the electrodes is 10.56 mA.

(b) The direction of electric current is towards negative electrodes.

Explanation:

(a)   First, we will calculate the charge of sodium ions as follows.

              q = ne

                  = [tex]2.68 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]

                  = [tex]4.288 \times 10^{-3} C[/tex]

Now, charge of chlorine ions is calculated as follows.

            q' = ne

                = [tex]3.92 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]

                = [tex]6.272 \times 10^{-3} C[/tex]

Hence, the current will be calculated as follows.

             i = [tex]\frac{q}{t} + \frac{q'}{t}[/tex]

               = [tex]\frac{4.288 \times 10^{-3} C}{1.00} + \frac{6.272 \times 10^{-3} C}{1.00}[/tex]

               = [tex]10.56 \times 10^{-3} A[/tex]

               = 10.56 mA

Therefore, current passing between the electrodes is 10.56 mA.

(b)   Since, positive ions are moving towards the negative electrode. And, current is the flow of ions or electrons therefore, the direction of current is towards the negative electrode.

A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 107 Ohms · m.


(a) If the potential difference between the inner and outer surfaces of a cell membrane is 80 mV, how much current flows through a square area of membrane 1.2 µm on a side?

_____________A


(b) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease or remain the same?

By what factor?

Answers

Explanation:

Below is an attachment containing the solution.

When a car drives through the Earth's magnetic field, an emf is induced in its vertical 60-cm-long radio antenna. Part A If the Earth's field (5.0×10−5 T) points north with a dip angle of 38∘, which direction(s) will the car be moving to produce the maximum emf induced in the antenna?

Answers

Answer:

Explanation:

The magnetic field is in north south direction . In order to cut lines of forces to the maximum extent  , car has to move in east- west direction ie towards east or towards west to produce maximum emf.

emf produced = B L V

where B is horizontal component of earth magnetic field

L is length of rod

v is velocity of car carrying antenna rod .

Final answer:

The car should be moving east or west to produce the maximum emf induced in the antenna.

Explanation:

When a car drives through the Earth's magnetic field, the maximum emf induced in its vertical radio antenna occurs when the car is moving in a direction perpendicular to the magnetic field. In this case, the Earth's magnetic field points north with a dip angle of 38°. To produce the maximum emf, the car should be moving east or west.

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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.

What is the electric potential at the center of the sphere?

Express your answer with the appropriate units.

Answers

Answer:

  V_inside = 36 V

Explanation:

Given  

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

Required

We are asked to calculate the potential at the centre of the sphere  

Solution

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r)                                          (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside                               (2)

Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

V_inside = (1.2 m )/(0.600)*18

               = 36 V

  V_inside = 36 V

Answer:

36 V

Explanation:

The solid conducting sphere is a positive charge

and has radius R₁ = 0.6m

at a point R₂ = 1.20 m, the electric potential V = 18.0 V

V, electric potential = K q/R where k  = 1/4 πε₀

V is inversely proportional to R

V₁ = electric potential at the center

V₂ = electric potential at 1.2 m

then

V₁ /V₂ = R₂ / R₁

V₁ = V₂ ( R₂ / R₁) = 18.0 V ( 1.2 / 0.6 ) = 36 V

A string is stretched to a length of 212 cm and both ends are fixed. If the density of the string is 0.02 g/cm, and its tension is 357 N, what is the fundamental frequency? Course hero

Answers

The fundamental frequency of a string fixed at both ends is f₁ = c/2L, where 'c' is calculated from the tension and linear mass density of the string. Using the given tension and density, one can find the wave speed 'c' first, then substitute into the formula to get the fundamental frequency.

To calculate the fundamental frequency of a string fixed at both ends, you can use the formula for the fundamental frequency of a string, which is given by f1 = c/2L, where f1 is the fundamental frequency, c is the speed of the wave on the string, and L is the length of the string.

To find the speed of the wave on the string, we use the formula c = \/(T/μ), where c is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string. Substituting the given values, c = \/(357 N / (0.02 g/cm * 100 cm/m)) = \/(357 / 0.0002 kg/m), we can calculate c and then use the result to find f1.

In a dentist's office, an X-ray of a tooth is taken using X-rays that have a frequency of 9.81 1018 Hz. What is the wavelength in vacuum of these X-rays?

Answers

Answer:

[tex]\lambda=3.056\times10^{-11}m[/tex]

Explanation:

The equation that involves frequency (f) and wavelength ([tex]\lambda[/tex]) of a wave is [tex]v=\lambda f[/tex], where v is the speed of the wave. X-rays are a type of electromagnetic wave, so in vacuum it moves at the speed of light c, which means that our wavelength will be:

[tex]\lambda=\frac{c}{f}=\frac{299792458m/s}{9.81\times10^{18}Hz}=3.056\times10^{-11}m[/tex]

Final answer:

To calculate the wavelength of an X-ray with a frequency of 9.81 x 1018 Hz, we substitute the given frequency into the formula λ = c/f. The wavelength of these X-rays is approximately 3.06 x 10-11 meters.

Explanation:

In Physics, we can find the wavelength of an X-ray given its frequency using the formula: λ=c/f, where 'λ' is the wavelength, 'c' is the speed of light (approximately 3.0 x 108 m/s in a vacuum), and 'f' is the frequency. Here, the frequency of the X-rays is given as 9.81 x 1018 Hz.

Substituting the values into the formula, we have: λ = (3.0 x 108) / (9.81 x 1018) = ~3.06 x 10-11 meters. The result indicates that the wavelength of these X-rays is incredibly small, which is consistent with our understanding of X-rays as high energy radiation with very short wavelengths and high frequencies.

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Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 4J.
a)This describes an elastic collision (and it could NOT be inelastic).
b)This describes an inelastic collision (and it could NOT be elastic).
c)This is NEITHER an elastic collision nor an inelastic collision
d)This describes a collision that is EITHER elastic or inelastic, but more information is required to determine which.

Answers

Answer:

B

Explanation:

Newton's third law of motion states that to every action there is equal an opposite reaction. Momentum is always conserved provided there is no net force on the body. Considering the experiment: the momentum is conserved as expected but the  kinetic energy is not conserved meaning that the collision is not elastic; some energy is converted into internal energy. When the collision is elastic the total kinetic energy before will equal total kinetic energy after and when the body stick together, they move with a common velocity and that described a perfectly inelastic collision.  5J ≠ 4J

Final answer:

The scenario describes an inelastic collision because while the momentum is conserved, the kinetic energy decreases from 5J to 4J, indicating that not all kinetic energy is conserved in the collision.

Explanation:

When examining collisions in physics, an elastic collision is one in which both momentum and kinetic energy are conserved, whereas an inelastic collision is one where momentum is conserved but kinetic energy is not. Given that in the scenario the total momentum of the system remains the same before and after the collision, that part of the conservation law is satisfied in both cases. However, since the kinetic energy of the system decreases from 5J to 4J, this means that some of the kinetic energy has been transformed into other forms of energy, like heat or sound, which typically happens during an inelastic collision.

Therefore, the correct answer is (b): This describes an inelastic collision (and it could NOT be elastic) since the total kinetic energy of the system is not the same before and after the collision. An elastic collision would have required that the kinetic energy also remain constant.

(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

Answers

Answer with Explanation:w

a.We are given that

Potential difference, V=48 V

[tex]R_1=24\Omega[/tex]

[tex]R_2=96\Omega[/tex]

Equivalent resistance when R1 and R2 are connected in series

[tex]R_{eq}=R_1+R_2[/tex]

Using the formula

[tex]R_{eq}=24+96=120\Omega[/tex]

We know that

[tex]I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A[/tex]

In series combination, current passing through each resistor is  same and potential difference across each resistor is different.

Power, P=[tex]I^2 R[/tex]

Using the formula

Power,[tex]P_1=I^2R_1=(0.4)^2\times 24=3.84 W[/tex]

Power, [tex]P_2=I^2 R_2=(0.4)^2(96)=15.36 W[/tex]

b.

In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..

Current,[tex]I=\frac{V}{R}[/tex]

Using the formula

[tex]I_1=\frac{V}{R_1}=\frac{48}{24}=2 A[/tex]

[tex]I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A[/tex]

[tex]P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W[/tex]

[tex]P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W[/tex]

Final answer:

When two resistors are in series, the current through each is the same and is calculated by dividing the total voltage by the total resistance. The power through each resistor is the current squared times the resistance of each resistor. When resistors are in parallel, the total resistance decreases and the power increases. The current remains the same.

Explanation:

When the 24-ohm and 96-ohm resistors are connected in series, the total resistance can be calculated using the formula R = R1 + R2. Here, R1 is the resistance of the first resistor (24 ohms) and R2 is the resistance of the second resistor (96 ohms). R equals 120 ohms. The current, I, is calculated using Ohm's Law, I = V / R. Therefore, the current flowing through the circuit is I = 48.0 V / 120 Ω = 0.4 A. The power, P, through each resistor is calculated using the formula P = I^2 * R. Therefore, for the 24-ohm resistor, P is (0.4 A)^2 * 24 Ω = 3.84 W, and for the 96-ohm resistor, P is (0.4 A)^2 * 96 Ω = 15.36 W. When the resistors are connected in parallel, the total resistance is calculated using the formula 1/R = 1/R1 + 1/R2. The current through each resistor remains the same (0.4 A) and the power for each resistor can be calculated in the same way as described above.

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Considering the form of the Gibb's Free Energy and the condition about it that some process be spontaneous, if some system has the following properties, what would be the result?

Answers

Answer:

A) Always Spontaneous.

B) Spontaneous at lower temperatures.

C) Spontaneous at higher temperatures.

D) Never Spontaneous.

Explanation:

The change in Gibb's free energy for a process, ΔG, is given by

ΔG = ΔH - TΔS

where ΔH is the change in enthalpy for the process

T = absolute temperature in Kelvin of the process

ΔS = change in enthalpy of the process.

And for a process to be spontaneous, its change in Gibb's free energy must be negative.

If it is positive, then the reaction isn't spontaneous.

So, we consider the conditions given one by one.

A) The enthalpy change is negative and the entropy change is positive.

ΔH = -ve, ΔS = +ve

ΔG = ΔH - TΔS = (negative number) - T(positive number) = (negative number - negative number) = negative number (since T isn't negative for these processes.)

Hence, the process with these conditions is always spontaneous.

B) The enthalpy change is negative and the entropy change is negative.

ΔH = negative, ΔS = negative

ΔG = ΔH - TΔS

For this relation, ΔG can only be positive if the numerical value of the ΔH (without the sign) is greater than TΔS.

This will happen mostly at low temperatures as low T, helps to reduce the numerical value of TΔS, thereby making ΔH (without the negative sign) the bigger number and subsequently makethe overall expression negative and the process, spontaneous.

C) The enthalpy change is positive and the entropy change is positive.

ΔH = positive, ΔS = positive.

ΔG = ΔH - TΔS

For this to be spontaneous,

TΔS > ΔH

And the one thing that favours this is high temperatures for the process.

At high temperatures, TΔS gives a much larger number which would drive the overall expression towards the negative sign, thereby making the process spontaneous.

D) The enthalpy change is positive and the entropy change is negative.

ΔH = positive, ΔS = negative.

ΔG = ΔH - TΔS

The signs on these state functions mean that the change in Gibb's free energy will always be positive for this set of conditions (since the temperature can't go as low as being negative).

A process with these conditions, is never spontaneous.

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown. g

Answers

Answer:

a) 138 ft

b) 4.62 s

c) 1.375 s

d) 168.25 ft

Explanation:

The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as

h = f(t) = -16t² + 44t + 138

with t in seconds, and h in feet

a) The bridge's height above the water.

At t=0 s, the rock is at the level of the Bridge's height.

At t = 0,

h = 0 + 0 + 138 = 138 ft

b) How many seconds after being thrown does the rock hit the water?

The rock hits the water surface when h = 0 ft. Solving,

h = f(t) = -16t² + 44t + 138 = 0

-16t² + 44t + 138 = 0

Solving this quadratic equation,

t = 4.62 s or t = -1.87 s

Since time cannot be negative,

t = 4.62 s

c) How many seconds after being thrown does the rock reach its maximum height above the water?

At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.

At maximum height,

(dh/dt) = f'(t) = (df/dt) = 0

h = f(t) = -16t² + 44t + 138

(dh/dt) = (df/dt) = -32t + 44 = 0

32t = 44

t = (44/32)

t = 1.375 s

d) What is the rock's maximum height above the water?

The maximum height occurs at t = 1.375 s,

Substituting this for t in the height equation,

h = f(t) = -16t² + 44t + 138

At t = 1.375 s, h = maximum height = H

H = f(1.375) = -16(1.375²) + 44(1.375) + 138

H = 168.25 ft

Hope this Helps!!!

. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.

Answers

Answer: 9.59° and 350.41°

Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as

F =qvB sin x

Where x is the angle between the strength of magnetic field and velocity of the charge.

q = 1.609×10^-19 C

v = 4×10³ m/s

B = 1.25 T

F = 1.40×10^-16 N

By substituting the parameters, we have that

1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx

sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25

sin x = 1.40×10^-16 /8.045*10^(-16)

sin x = 0.1666

x = 9.59°

The value of sin x is positive in first and fourth quadrant.

Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x

Hence = 360 - 9.59 = 350.41°

Final answer:

The angle between the velocity of the electron and the magnetic field is approximately 6.27° or 173.73°.

Explanation:

To find the angle between the velocity of the electron and the magnetic field, we can use the formula:

F = q(vsinθ)B

Where F is the magnetic force, q is the charge of the electron, v is the velocity of the electron, θ is the angle between the velocity and the magnetic field, and B is the magnetic field strength.

In this case, the magnetic force is given as 1.40 × 10^-16 N, the charge of an electron is 1.6 × 10^-19 C, the velocity of the electron is 4.00 × 10³ m/s, and the magnetic field strength is 1.25 T.

Plugging in these values, we can solve for θ:

1.40 × 10^-16 N = (1.6 × 10^-19 C)(4.00 × 10³ m/s)(sinθ)(1.25 T)

Solving for sinθ:

sinθ = (1.40 × 10^-16 N) / [(1.6 × 10^-19 C)(4.00 × 10³ m/s)(1.25 T)]

sinθ ≈ 0.109

Taking the inverse sine of 0.109, we find that θ ≈ 6.27° or θ ≈ 173.73°.

a bag of ice cubes absorbs 149,000 J of heat, which causes its temperature to increase by 5.23 degrees celsius. what is the mass of the ice in the bag?
Ice=c=2000J/(kg*c) (unit=kg)

Answers

Answer:

14.3kg

Explanation:

Given parameters:

Quantity of heat = 149000J

Change in temperature = 5.23°C

specific heat of the ice = 2000J/kg°C

Unknown:

Mass of the ice in the bag = ?

Solution:

The heat capacity of a substance is given as:

            H = m c Ф

H is the heat capacity

m is the mass

c is the specific heat

Ф is the temperature change;

  since m is the unknown, we make it the subject of the expression;

                    m = H/ mФ

                  m = [tex]\frac{149000}{2000 x 5.23}[/tex]  = 14.3kg

Answer: 14.2447

Explanation:

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