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

Answer 1

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.


Related Questions

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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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.

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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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.

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 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 circular coil that has N = 230 turns and a radius of r = 10.0 cm lies in a magnetic field that has a magnitude of B 0 = 0.0650 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B through the coil?

Answers

Answer:

Φ = 0.469 Wb

Explanation:

Given,

N = 230 turns

Radius, r = 10 cm

Magnetic field, B = 0.0650 T

Magnetic flux  = ?

now,

Φ = NBA

Φ = 230 x 0.0650 x π x r²

Φ = 230 x 0.0650 x π x 0.1²

Φ = 0.469 Wb

Hence, the magnetic flux is equal to Φ = 0.469 Wb

The magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m². when A circular coil that has N = 230 turns and a radius of r = 10.0 cm.

Given:

N = 230 turns

r = 10.0 cm = 0.1 m

B₀ = 0.0650 T

The magnetic flux through a coil can be calculated using the formula:

ΦB = B₀ × A × N,

The area of the coil, we can use the formula for the area of a circle:

A = π × r²,

Let's calculate the magnetic flux:

A = π × r² = 3.14159 × (0.1)² = 0.0314159 m²

ΦB = B₀ × A × N = 0.0650 × 0.0314159 × 230 = 0.04958735 T·m²

Therefore, the magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m².

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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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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.

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

A magnetic field has a magnitude of 1.2 \times 10^{-3} T, and an electric field has a magnitude of 4.6 \times 10^{3}N/C. Both fields point in the same direction. A positive 1.8 \mu C charge moves at a speed of 3.1 \times 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Answers

Answer: F = 113.4.[tex]10^{-3}[/tex]N

Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is

F = Fe + Fm

Force due to electric field

To determine this force:

Fe = q.E, where q is the charge and E is electric field.

Calculating:

Fe = q.E

Fe = 1.8.[tex]10^{-6}[/tex].4.6.[tex]10^{3}[/tex]

Fe = 8.28.[tex]10^{-3}[/tex]N

Force due to magnetic field: It can only happens when the charge is in movement, so

Fm = q.(v×B), where v represents velocity and B is magnetic field

The cross product indicates that force is perpendicular to the velocity and the field.

Calculating:

Fm = q.v.B.senθ

As θ=90°,

Fm = q.v.B

Fm =  1.8.[tex]10^{-6}[/tex].3.1.[tex]10^{6}[/tex].1.2.[tex]10^{-3}[/tex]

Fm = 6.696.[tex]10^{-3}[/tex]N

F, Fm and Fe make a triangle. So, using Pythagorean theorem:

F = [tex]\sqrt{Fe^{2} + Fm^{2} }[/tex]

F = [tex]\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }[/tex]

F = 113.4.[tex]10^{-3}[/tex]N

The net force acting on the charge is F = 113.4.[tex]10^{-3}[/tex]N

A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 487.0 mV 487.0 mV . If the number of turns in the primary coil is 2464 2464 , calculate the number of turns on the secondary coil of the adapter transformer.

Answers

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

    [tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]

where [tex]V_{s}[/tex] is the voltage induced in the secondary coil

           [tex]V_{p}[/tex] is the voltage in the primary coil

          [tex]N_{s}[/tex] is the number of turns of secondary coil

         [tex]N_{p}[/tex] is the number of turns of primary coil

From the given question,

    [tex]\frac{487*10^{-3} }{120}[/tex] = [tex]\frac{N_{s} }{2464}[/tex]

⇒    [tex]N_{s}[/tex] = [tex]\frac{2462*487*10^{-3} }{120}[/tex]

            = 9.999733

  ∴   [tex]N_{s}[/tex] = 10 turns

Answer:

607,145 turns

Explanation:

Output voltage, that is secondary voltage,Es = 120 volts

Input voltage, that is primary voltage, Ep = 487/1000 = 0.487 volts

Number of turns in secondary = Ns

Number of turns in primary, Np = 2464

∴ Es/Ep = Ns/Np

Ns = Es * Np/Ep = 1`20 X 2464/0.487 =  607,145 turns ( Step up transformer)

A blood-flow meter emits a 1.1-MHz ultrasound pulse to measure the speed of blood moving directly away from the meter. The meter’s sensor detects the pulse reflected back from the blood at a frequency 21 Hz lower than the emitted frequency. Take the speed of sound in the tissues to be 1475 m/s.

At what speed, in centimeters per second, is the blood moving?

Answers

Answer:

V = 2.8cm/s

Explanation:

Please see attachment below.

This problem involves the concept of doppler effect.

Answer:

The speed at which the blood is flowing in cm/s = 1.4 cm/s

Explanation:

emitted frequency ( f ) = 1.1 * 10^6 Hz

detected frequency ( F ) = 21 Hz

speed of sound in tissues ( c ) = 1475 m/s

speed ( V ) = ?

To calculate for speed of blood flowing we apply the detected frequency formula :

F = [tex]\frac{2fV}{c}[/tex]

21 = ( 2* 1100000* V ) / 1475

therefore V = (21 * 1475) / (2 * 1100000)

                V = 30975 / 2200000

                 V = 0.0140 m/s = 1.4 cm/sec

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

Final answer:

The acceleration at the end of a test tube 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm is calculated to be 15051.2 m/s², demonstrating the significant centrifugal forces generated by such devices.

Explanation:

To calculate the acceleration at the end of a test tube that is 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm, we first need to convert the rotational speed to radians per second. The formula to convert revolutions per minute (rpm) to radians per second (ω in rad/s) is ω = (2π×rpm)/60. Thus, ω = (2π× 3700)/60 = 387.98 rad/s. Next, we use the formula for centripetal acceleration, a = ω²×r, where r is the radius of the circle (distance from the center of rotation to the point of interest) in meters. Given that r = 10 cm = 0.1 m, the acceleration a = (387.98)^2× 0.1 = 15051.2 m/s².

This centripetal acceleration is much larger than Earth's gravitational acceleration, indicating the extreme forces at play in a centrifuge's operation, which is crucial for its role in laboratory settings for sedimentation of materials.

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

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:

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

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.

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]

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.

Use the exact values you enter to make later calculations.You measure the potential difference across a capacitor at different times while it's charging and record the following results.Voltage (V) Time (s)1.4790 0.015843.0000 0.036894.5210 0.066516.0210 0.11700

The final voltage the capacitor reaches after you go get some coffee is 6.500 V.
(a) Determine the time constant from the slope.

Answers

Answer:

0.0800 is time constant for slope.

Explanation:

See attached pictures for explanation.

Final answer:

To find the time constant from the slope, use the formula RC = -1/slope, where R is the resistance and C is the capacitance. Calculate the slope by taking the ratio of voltage change to time change between two data points.

Explanation:

To determine the time constant from the slope, we can use the formula:

RC = -1/slope

where R is the resistance in the circuit and C is the capacitance of the capacitor.

In this case, since we are only given the voltage and time data, we need to find the slope by taking the ratio of the change in voltage to the change in time between any two points.

Let's take the first and second data points:

Slope = (V2 - V1) / (t2 - t1)

          = (3.0000 V - 1.4790 V) / (0.036894 s - 0.015843 s)

Now, calculate the slope:

Slope = (1.5210 V) / (0.021051 s)

Slope ≈ 72.27 V/s

Once we have the slope, we can plug it into the formula RC = -1/slope to find the time constant:

RC = -1 / (72.27 V/s)

Calculate RC:

RC ≈ -0.0138 s/V

So, the time constant (τ) is approximately 0.0138 seconds per volt (s/V). This value represents the product of resistance and capacitance in the circuit.

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The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)

(a) The pool is completely filled with water.
______m below ground level

(b) The pool is filled halfway with water.
______m below ground level

Answers

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

Final answer:

The apparent depth of a swimming pool is measured by considering the water's refractive index. With the pool completely filled, the bottom appears to be 2.25m deep. When halfway filled, the pool appears to be 2.625m deep.

Explanation:

When light travels from a medium with a high refractive index to one with a lower refractive index, the light is refracted, or bent, making objects appear closer than they actually are. We can calculate this apparent depth by using the formula d' = d / n, where d' is the apparent depth, d is the actual depth, and n is the refractive index.

(a) If the pool is completely filled with water, for a person looking from above ground, the bottom of the pool appears to be closer than it actually is. Substituting the given values into the formula, we get the apparent depth: d' = 3.00m / 1.333 = 2.25 m below ground level.

(b) If the pool is halfway filled with water, the apparent depth of the water is calculated in the same way. However, the depth beneath the water is below the refractive index border and is not subject to refraction. Therefore, the apparent total depth of the partially filled pool is the sum of the actual depth of the air part (1.50m) and the apparent depth of the water part (1.50m / 1.333 = 1.125m). This gives us 2.625m.

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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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You are to connect resistors R1 and R2, with R1 > R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.

Answers

Answer:

Parallel, R2, R1, series

Explanation:

From Ohm's law, V=IR hence making current the subject of the formula then [tex]I=\frac {V}{R}[/tex]

Since R1>R2 eg 4>2 then current for individual connection of R2 will be greater than that for R1 for example, assume V is 8 then if R2 we will have 8/2=4 A but for R1 we shall have 8/4=2 A.

When in parallel, the equivalent resistance will be given by [tex]\frac {1}{R1}+\frac {1}{R2}[/tex] for example here it will be 1/4+1/2=3/4. Still taking V of 8 then I= 8/(3/4)=10.667 A

When in series connection, equivalent resistance is given by adding R1 and R2 hence using the same figures we shall have 4+2=6 hence I=8/6=1.33A

We can conclude that the arrangement of current from greatest will be parallel, R2, R1, series

According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".

Parallel and Series connection

By using Ohm's law

Voltage (V) = Current (I) × Resistance (R)

or,

→ I = [tex]\frac{V}{R}[/tex]

Since R1 > R2

Let, Current (V) = 8 then the resistance will be:

R2 = [tex]\frac{8}{2}[/tex] = 4 A

and,

R1 = [tex]\frac{8}{2}[/tex] = 4 A

When the resistance is in parallel connection,

= [tex]\frac{1}{R1} + \frac{1}{R2}[/tex]

= [tex]\frac{1}{4} +\frac{1}{2}[/tex]

By taking L.C.M, we get

= [tex]\frac{1+2}{4}[/tex] = [tex]\frac{3}{4}[/tex] and,

The current be:

→ I = [tex]\frac{8}{\frac{3}{4} }[/tex] = 10.667 A

When the resistance is in series connection,

= R1 + R2

= 4 + 2

= 6 and,

The current be:

→ I = [tex]\frac{8}{6}[/tex] = 1.33 A

Thus the response above is appropriate.

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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 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 current of 8 A exists in a copper (Cu) wire which has a diameter of 5 mm. What is the current density? Each atom of copper contributes one conduction electron, and the average thermal speed r k T m of an electron is 2.4 × 106 m/s . The mass

Answers

Explanation:

Below is an attachment containing the solution.

You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1- e^{ \frac{-\left(t\right)}{RC} } \right). You curve fit the data using the inverse exponent function Y=A \cdot \left(1- e^{-\left(Cx\right)} \right)+B and LoggerPro gives the following values for A, B, and C. A = 4.211 ± 0.4211 B = 0.1699 ± 0.007211 C = 1.901 ± 0.2051 What is the time constant for and its uncertainty?

Answers

Answer:

The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s

Explanation:

If we make a comparison we have to:

y = A*(1-e^-(C*x)) + B

If the time remains constant we have to:

t = R*C = 1/C

In this way we calculate the time constant and its uncertainty. this will be equal to:

t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s

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

a carbon steel ball with a 30mm diameter is pressed against a flat carbon steel plate with a force of 20n. calculate the diameter of the circular contact area and the maximum pressure that occurs at the center of the contact area

Answers

Final answer:

The diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.

Explanation:

To calculate the diameter of the circular contact area, we need to determine the radius of the contact area first. The radius can be found by dividing the diameter of the ball by 2. In this case, the radius is 15 mm.

The area of the circular contact area can be calculated using the formula A = πr^2, where A is the area and r is the radius. The maximum pressure at the center of the contact area can be found by dividing the force applied on the plate by the area of the contact area.

Using these formulas, the diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.

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