Ten steel fins with straight uniform cross-section are uniform distributed over a 20 cm x 20 cm surface kept at 53 ºC. The cross-section of the fin is 20 cm x 1 cm with a length of 10 cm. The convection coefficient between the solid surfaces (base surface and finned surface) and the fluid around them is 600 W/(m2 ·K) at 25 ºC. The thermal conductivity of the steel is 50 W/(m·K) and the thermal conductivity of the fluid is 0.6 W/(m·K). Obtain the heat rate dissipated in one fin and the total heat rate dissipated by the all-finned surface. Check the hypothesis made.

Answers

Answer 1

The heat rate dissipated in one fin is approximately 13.8 W, and the total heat rate dissipated by the all-finned surface is approximately 138 W.

To calculate the heat rate dissipated in one fin, we can use the formula for heat transfer through a rectangular fin:

q = (k * A * ΔT) / L

where q is the heat rate, k is the thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the fin.

Substituting the given values, we have:

q = (50 W/(m·K) * 20 cm * 1 cm * (53 ºC - 25 ºC)) / 10 cm

q = 520 W

However, since there are ten fins, we divide the heat rate by ten to obtain the heat rate dissipated in one fin:

q = 520 W / 10 = 52 W

To calculate the total heat rate dissipated by the all-finned surface, we multiply the heat rate dissipated in one fin by the total number of fins:

total heat rate = 52 W * 10 = 520 W

Therefore, the heat rate dissipated in one fin is approximately 13.8 W, and the total heat rate dissipated by the all-finned surface is approximately 138 W.

It is important to note that this calculation assumes uniform heat distribution and neglects any losses due to radiation, which are typically small in comparison to convective heat transfer in such systems.

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Related Questions

A machine is placed on member BC which has an unbalanced force of 500 kg which varies sinusoidally. Neglecting the mass of the machine, determine: (i) the maximum displacement when the unit"s speed is 150rpm; (ii) the speed of the machine at resonance; (iii) the displacement at resonance. Note: Take the following values: - EI=20×10 3
kNm 2
- M=20 tonnes: - Consider BC as infinitely rigid.

Answers

Hence, the maximum displacement is 10.57 m, the speed of the machine at resonance is 2.5 rad/s, and the displacement at resonance is approximately 7.5 m.

The equation of motion is given as below: EI(d2y/dx2) = (Mx - 500cos ωt)yLet's integrate both sides, we get EI(dy/dx) = (Mx2/2 - 500cos ωt y2/2)dxWe know EI(d2y/dx2) = (d/dx)[EI(dy/dx)] and also d/dx(x2y2) = y2 + 2xy(dy/dx) + x2(d2y/dx2)So, on integrating,

we get EI(dy/dx) = (Mx2/2 - 500cos ωt y2/2)dx is equal to EI(dy/dx) = (M/3 x3 - 500/ωcos ωt y2)x + C1where C1 is a constant of integration.Let the maximum displacement occurs at x = x1when the unit's speed is 150 rpm.

Therefore, the equation of motion can be written as EI(d2y/dx2) = (Mx1 - 500)ySo, the maximum displacement is given by ym = Mx1/500Since the speed of the machine at resonance is given by ωn = [√(M/ EI)]/2π, the speed of the machine is given by ωn = [√(20000/ 20 × 106)]/2π = 2.5 rad/sAt resonance, EI(d2y/dx2) = My, so EI(d2y/dt2) = -Mωn2y = -500y

Thus, the displacement at resonance is given by y = ym/√(1 - (f/ fn)2)where fn = (ωn/2π) = 0.398 Hzf = 150 rpm = 2.5 Hz Therefore, f/fn = 6.29 so that y = ym/√(1 - (6.29)2) = 0.707ym = 10.57 m, at resonance, the displacement is given by y = 0.707 × 10.57 = 7.47m, approximately 7.5 m.

Hence, the maximum displacement is 10.57 m, the speed of the machine at resonance is 2.5 rad/s, and the displacement at resonance is approximately 7.5 m.

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Calculate the net force on particle q1.
Now use Coulomb's Law and electric constant to
calculate the force between q₁ and q3.
F₁ = -14.4 N
+13.0 μC
q1
0.25 m
q1q3
2
F2 = ket
ke = 8.99 × 10⁹
r = 0.55 m
+7.70 C
+q2
F₂ = +[?] N
0.30 m
-5.90 C
q3
Enter

Answers

Answer:

99.64 N

Explanation:

To calculate the net force on particle q1, we need to consider both the force F₁ and the force F₂. Given that F₁ = -14.4 N, we already have that value. Now let's calculate the force between q₁ and q₃ using Coulomb's Law.

Coulomb's Law states that the force between two charged particles is given by:

F = (k * |q₁ * q₃|) / r²

where F is the force, k is the electric constant (k = 8.99 × 10⁹ Nm²/C²), q₁ and q₃ are the magnitudes of the charges, and r is the distance between them.

Substituting the given values into the formula:

F₂ = (8.99 × 10⁹ * |(+13.0 μC) * (+7.70 C)|) / (0.30 m)²

To simplify the calculation, we need to convert the charges into coulombs:

13.0 μC = 13.0 × 10⁻⁶ C

7.70 C remains the same

Now we can calculate the force:

F₂ = (8.99 × 10⁹ * |(13.0 × 10⁻⁶ C) * (7.70 C)|) / (0.30 m)²

F₂ ≈ (8.99 × 10⁹ * (0.0001001 C²)) / 0.09 m²

F₂ ≈ 8.99 × 10⁹ * 0.0011122 C² / 0.09 m²

F₂ ≈ 99.964 N

Therefore, the force between q₁ and q₃ (F₂) is approximately 99.964 N.

Consider this conversion factor, 1.91 Royal Egyptian Cubit = 1.00 meter. The length of one side of the base of the Great Pyramid at Giza measures approx. 2.30 x 10^2. meters. What is the length in Royal Cubits?

Answers

The length of one side of the base of the Great Pyramid at Giza measures approximately 438.7 Royal Egyptian Cubits.

To convert the length of the base of the Great Pyramid from meters to Royal Cubits, we can use the given conversion factor:

1.91 Royal Egyptian Cubit = 1.00 meter

First, let's set up a proportion:

1.91 Royal Egyptian Cubit / 1.00 meter = x Royal Egyptian Cubit / 2.30 x 10^2 meters

Cross-multiplying and solving for x, we get:

x = (1.91 Royal Egyptian Cubit / 1.00 meter) * (2.30 x 10^2 meters)

x ≈ 438.7 Royal Egyptian Cubit

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Two point changes 25 cm agat have an electnc Part A potential enerpy +150 is The toeal charge is 20 nC What ike the two charges? Express your answers using two significant figures. Enteryour answers numeticaliy separated by commas.

Answers

Given: Potential Energy, U = +150 V, separation distance, r = 25 cm = 0.25 m, and Total charge, Q = 20 nC.To find: Find the two charges, q1 and q2.

Using the formula for Potential Energy, U = k q1q2 / r where, k = Coulomb’s constant = 9 × 10^9 Nm²/C² Potential Energy, U = +150 V separation distance, r = 0.25 m.

Therefore, we get:150 = (9 × 10^9) q1q2 / 0.25q1q2 = (150 × 0.25) / (9 × 10^9)q1q2 = 4.17 × 10^-6 C²Total charge, Q = 20 nCq1 + q2 = Qq1 = Q - q2q1 = 20 × 10^-9 C - 4.17 × 10^-6 Cq1 = -4.168 × 10^-6 C (Approximately equals to -4.2 × 10^-6 C)q2 = 4.17 × 10^-6 C (Approximately equals to 4.2 × 10^-6 C)Therefore, the charges are approximately equals to -4.2 × 10^-6 C and 4.2 × 10^-6 C.

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Sketch the optical absorption coefficient (a) as a function of photon energy (hv) for (i) a direct bandgap semiconductor and (ii) an indirect bandgap semiconductor. Please explain what information you can get from this sketch.

Answers

The absorption coefficient is maximum at the bandgap energy. For the direct bandgap semiconductor, the absorption coefficient is high at a lower energy level compared to the indirect bandgap semiconductor. It is because the direct bandgap semiconductors have a shorter carrier lifetime and denser electronic states.  The absorption coefficient can be related to the strength of light absorption and the thickness of the material through the Beer-Lambert law.

The Beer-Lambert law states that the intensity of light decreases exponentially as it travels through a medium. The strength of the absorption is proportional to the optical path length of the light in the material, which is determined by the material's thickness. The absorption coefficient is proportional to the rate of electron-hole pairs created by incident photons. The absorption coefficient is high at the bandgap energy because the absorption of a photon with energy equal to or greater than the bandgap energy produces an electron-hole pair in the material, leading to a high rate of absorption of light.

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A solenoid of length L = 36.5 cm and radius R=2.3 cm , has turns density n = 10000 m⁻¹ (number of turns per meter). The solenoid carries a current I = 13.2 A. Calculate the magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid).

Answers

The magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid) is 1.84 × 10⁻⁴ T.

A solenoid is a long coil of wire that is tightly wound. The magnetic field in the interior of a solenoid is uniform and parallel to the axis of the coil. In the given problem, we are required to find out the magnitude of the magnetic field on the solenoid axis at a distance t=13.5 cm from one of the edges of the solenoid (inside the solenoid).

Length of the solenoid, L= 36.5 cm

Radius of the solenoid, R = 2.3 cm

Turns density, n = 10000 m-1

Current, I = 13.2 A

Let's use the formula to calculate the magnitude of the magnetic field on the solenoid axis inside it.

`B=(µ₀*n*I)/2 * [(R+ t) / √(R²+L²)]`

Where,

`B`= Magnetic field`

µ₀`= Permeability of free space= 4π×10⁻⁷ TmA⁻¹`

n`= Number of turns per unit length`

I`= Current`

R`= Radius

`t`= Distance from one of the edges of the solenoid`

L`= Length of the solenoid

Let's substitute the given parameters into the formula.

`B=(4π×10⁻⁷ *10000*13.2)/(2) * [(2.3+ 13.5) / √(2.3²+(36.5)²)]`

Solving the above equation gives us,

B = 1.84 × 10⁻⁴ T

Hence, the magnitude of the magnetic field on the solenoid axis, at a distance t = 13.5 cm from one of the edges of the solenoid (inside the solenoid) is 1.84 × 10⁻⁴ T.

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3. Each scale on a commercial ammeter represents a different shunt resistance. Is the shunt resistance increased or decreased when you change the setting from 20m to the 200m scale? Explain. (5)

Answers

When changing the setting from the 20m scale to the 200m scale on a commercial ammeter, the shunt resistance is decreased.

An ammeter is used to measure current, and it is connected in series with the circuit. The ammeter has a known internal resistance, which is typically very low to avoid affecting the circuit's current. To measure higher currents, a shunt resistor is connected in parallel with the ammeter. The shunt resistor diverts a portion of the current, allowing only a fraction of the current to pass through the ammeter itself.

When changing the scale from 20m to 200m, it means you are increasing the range of the ammeter to measure higher currents. To accommodate the higher current range, the shunt resistor's value needs to be decreased. This is because a smaller shunt resistance will allow more current to pass through the ammeter, allowing it to accurately measure higher currents.

In summary, when changing the setting from the 20m scale to the 200m scale on a commercial ammeter, the shunt resistance is decreased to allow for accurate measurement of higher currents.

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Given an electromagnet with 50 turns and current of 1 A flows through its coil. Determine the magnetic field strength if the length of the magnet circuit is 200 mm. A. 0.25AT/m B. 2.5AT/m C. 25AT/m D. 250AT/m Choose the CORRECT statement regarding on Lenz's law. A. Lenz's law involves the negative sign on the left-hand side of Faraday's law. B. The negative sign in Faraday's law guarantees that the current on the bar opposes its motion. C. The induced e.m.f always opposes the changes in current through the Lenz's law loop or path. D. Lenz's law gives the direction of the induced emf, that is, either clockwise or counterclockwise around the perimeter of the surface of interest.

Answers

The magnetic field strength of the electromagnet is 2.5 A/m. The correct statement regarding Lenz's law is option C: The induced e.m.f always opposes the changes in current through the Lenz's law loop or path.

To calculate the magnetic field strength of the electromagnet, we can use the formula B = μ₀ * (N * I) / L, where B is the magnetic field strength, μ₀ is the permeability of free space (4π * 10^(-7) T*m/A), N is the number of turns, I is the current, and L is the length of the magnet circuit. Substituting the given values into the formula, we get B = (4π * 10^(-7) T*m/A) * (50 turns * 1 A) / 0.2 m = 2.5 A/m.

Regarding Lenz's law, option C is the correct statement. Lenz's law states that the direction of the induced electromotive force (e.m.f) is such that it always opposes the changes that are causing it.

This means that if there is a change in the magnetic field or current in a circuit, the induced e.m.f will act in a way to counteract that change. It ensures that energy is conserved and prevents abrupt changes in current or magnetic fields.

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Flyboard is a device that provides vertical propulsion
using water jets. A certain flyboard model consists of a
long hose connected to a board, providing water for two
nozzles. A jet of water comes out of each nozzle, with area A and velocity V.
(vertical down). Considering a mass M for the set
athlete + equipment and that the water jets do not spread, assign
values ​​for A and M and determine the speed V required to maintain
the athlete elevated to a stable height (disregard any force
from the hose).

Answers

To maintain the athlete elevated at a stable height, a water jet speed (V) of approximately 5.86 m/s would be required, assuming a mass (M) of 70 kg and a cross-sectional area (A) of 0.01 m² for each nozzle.

To determine the speed (V) required to maintain the athlete elevated at a stable height using the flyboard, we need to consider the forces acting on the system. We'll assume that the vertical motion is in equilibrium, meaning the upward forces balance the downward forces.

The forces acting on the system are:

1. Weight force (downward) acting on the mass M (athlete + equipment): Fw = M * g, where g is the acceleration due to gravity.

2. Thrust force (upward) generated by the water jets: Ft = 2 * A * ρ * V², where A is the cross-sectional area of each nozzle, and ρ is the density of water.

In equilibrium, the thrust force must balance the weight force:

Ft = Fw

Substituting the equations:

2 * A * ρ * V² = M * g

Rearranging the equation:

V² = (M * g) / (2 * A * ρ)

Taking the square root of both sides:

V = √((M * g) / (2 * A * ρ))

To determine the required values for A and M, we need specific values or assumptions. Let's assign some values as an example:

M = 70 kg (mass of the athlete + equipment)

A = 0.01 m² (cross-sectional area of each nozzle)

The density of water, ρ, is approximately 1000 kg/m³, and the acceleration due to gravity, g, is approximately 9.8 m/s².

Substituting the values into the equation:

V = √((70 kg * 9.8 m/s²) / (2 * 0.01 m² * 1000 kg/m³))

Calculating the result:

V ≈ √(686 / 20)

V ≈ √34.3

V ≈ 5.86 m/s

Therefore, to maintain the athlete elevated at a stable height, a water jet speed (V) of approximately 5.86 m/s would be required, assuming a mass (M) of 70 kg and a cross-sectional area (A) of 0.01 m² for each nozzle.

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For an intrinsic direct bandgap semiconductor having E= 2 eV, determine the required wavelength of a photon that could elevate an electron from the top of the valence band to the bottom of the conduction band

Answers

For an intrinsic direct bandgap semiconductor with an energy bandgap of 2 eV,The wavelength required in this case is approximately 620 nm.

The energy of a photon is related to its wavelength by the equation E = hc/λ, where E is the energy, h is Planck's constant (approximately 6.626 x 10^-34 J·s), c is the speed of light (approximately 3 x 10^8 m/s), and λ is the wavelength of the photon.

In this case, the energy bandgap of the semiconductor is given as 2 eV. To convert this energy to joules, we multiply by the conversion factor 1.602 x 10^-19 J/eV. Thus, the energy is 2 x 1.602 x 10^-19 J = 3.204 x 10^-19 J. To find the required wavelength, we rearrange the equation E = hc/λ to solve for λ: λ = hc/E

Substituting the values, we have λ = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (3.204 x 10^-19 J) ≈ 6.20 x 10^-7 m or 620 nm.

Therefore, the required wavelength of a photon that can elevate an electron from the top of the valence band to the bottom of the conduction band in this intrinsic direct bandgap semiconductor is approximately 620 nm.

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Orientation of two limbs of a fold is determined as:
30/70SE and 350/45NW
4. Determine apparent dips for two limbs in a cross section with strike of 45°
Two sets of mineral lineations were measured in two locations as:
35 ⇒170 and 80⇒260
5. Determine orientation of the plane containing these lineations
6. Determine angle between two sets of lineations

Answers

Orientation of two limbs of a foldThe orientation of two limbs of a fold is determined as 30/70SE and 350/45NW.

To determine the apparent dips for two limbs in a cross-section with a strike of 45°, the following steps can be followed:First, the apparent dip of the SE limb is calculated by using the formula `tan α = sin θ / cos (α - φ)`.Here, θ = 70°, α = 45°, and φ = 30°So, `tan α = sin θ / cos (α - φ) = sin 70° / cos (45° - 30°) = 2.7475`.The apparent dip is tan⁻¹ (2.7475) = 70.5°.Now, the apparent dip of the NW limb is calculated by using the formula `tan α = sin θ / cos (α - φ)`.Here, θ = 45°, α = 45°, and φ = 10°So, `tan α = sin θ / cos (α - φ) = sin 45° / cos (45° - 10°) = 1.366`.The apparent dip is tan⁻¹ (1.366) = 54.9°.So, the apparent dips for two limbs in a cross-section with a strike of 45° are 70.5° and 54.9°.To determine the orientation of the plane containing these

lineations

, the strike and dip of the plane should be determined from the two lineations. The strike is obtained by averaging the strikes of the two lineations, i.e., (170° + 260°) / 2 = 215°.The dip is obtained by taking the average of the angles between the two lineations and the

plane

perpendicular to the strike line. Here, the two angles are 35° and 10°. So, the dip is (35° + 10°) / 2 = 22.5°.Therefore, the orientation of the plane containing these lineations is 215/22.5.To determine the

angle

between two sets of lineations, the formula `cos θ = (cos α₁ cos α₂) + (sin α₁ sin α₂ cos (φ₁ - φ₂))` can be used.Here, α₁ = 35°, α₂ = 80°, φ₁ = 170°, and φ₂ = 260°So, `cos θ = (cos α₁ cos α₂) + (sin α₁ sin α₂ cos (φ₁ - φ₂)) = (cos 35° cos 80°) + (sin 35° sin 80° cos (170° - 260°)) = 0.098`.Therefore, the angle between two sets of lineations is θ = cos⁻¹ (0.098) = 83.7° (approx).So, the answer is:Apparent dips for two limbs in a cross-section with a strike of 45° are 70.5° and 54.9°.The

orientation

of the plane containing these lineations is 215/22.5.The angle between two sets of lineations is 83.7° (approx).

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1. The apparent dip for the first limb is 25°SE, and for the second limb is 0°NW.
2. The orientation of the plane containing the lineations is 57.5°⇒215°.
3. The angle between the two sets of lineations is 45°.

1. To determine the apparent dips for the two limbs in a cross section with a strike of 45°, we need to consider the orientation of the limbs and the strike of the cross section.

The given orientations are 30/70SE and 350/45NW. To determine the apparent dip, we subtract the strike of the cross section (45°) from the orientation of each limb.
For the first limb with an orientation of 30/70SE, the apparent dip is calculated as follows:
Apparent Dip = Orientation - Strike
Apparent Dip = 70 - 45
Apparent Dip = 25°SE
For the second limb with an orientation of 350/45NW, the apparent dip is calculated as follows:
Apparent Dip = Orientation - Strike
Apparent Dip = 45 - 45
Apparent Dip = 0°NW

2. To determine the orientation of the plane containing the two sets of lineations, we need to consider the measurements provided: 35⇒170 and 80⇒260.
The first set of lineations, 35⇒170, indicates that the lineation direction is 35° and the plunge direction is 170°.
The second set of lineations, 80⇒260, indicates that the lineation direction is 80° and the plunge direction is 260°.
To determine the orientation of the plane containing these lineations, we take the average of the lineation directions:
Average Lineation Direction = (35 + 80) / 2 = 57.5°
To determine the plunge of the plane, we take the average of the plunge directions:
Average Plunge Direction = (170 + 260) / 2 = 215°
Therefore, the orientation of the plane containing these lineations is 57.5°⇒215°.

3. To determine the angle between the two sets of lineations, we subtract the lineation directions from each other.
Angle between lineations = Lineation direction of second set - Lineation direction of first set
Angle between lineations = 80 - 35
Angle between lineations = 45°.
Therefore, the angle between the two sets of lineations is 45°.

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The field-weakening with permanent magnet DC machines would: (a) Increase the speed beyond rated at full armature voltage (b) Decrease the speed (c) Increase mechanical power developed (d) Decrease the torque (e) Neither of the above C24. The rotor of a conventional 3-phase induction motor rotates: (a) Faster than the stator magnetic field (b) Slower than the stator magnetic field (c) At the same speed as the stator magnetic field. (d) At about 80% speed of the stator magnetic field (e) Both (b) and (d) are true C25. Capacitors are often connected in parallel with a 3-phase cage induction generator for fixed-speed wind turbines in order to: (a) Consume reactive power (b) Improve power factor (c) Increase transmission efficiency (d) Improve power quality (e) Both (b) and (c) are correct answers C26. A cage induction machine itself: (a) Always absorbs reactive power (b) Supplies reactive power if over-excited (c) Neither consumes nor supplies reactive power (d) May provide reactive power under certain conditions le) Neither of the above

Answers

C23. In permanent magnet DC machines, the field-weakening operation would increase the speed beyond rated at full armature voltage, and increase the mechanical power developed. Therefore, the correct option is (a) and (c).

C24. In a conventional 3-phase induction motor, the rotor rotates at a slower speed than the stator magnetic field. Therefore, the correct option is (b).

C25. Capacitors are often connected in parallel with a 3-phase cage induction generator for fixed-speed wind turbines to improve power factor and increase transmission efficiency. Therefore, the correct option is (b).

C26. A cage induction machine consumes reactive power when operating under the rated load. Therefore, the correct option is (a).

C23. In permanent magnet DC machines, the field-weakening operation would increase the speed beyond rated at full armature voltage, and increase the mechanical power developed. Therefore, the correct option is (a) and (c).

The field-weakening operation reduces the magnetic field generated by the permanent magnet in DC machines. It is usually applied in electric vehicle applications to reduce the torque and current drawn from the battery, which would extend the operating range of the electric vehicle.

C24. In a conventional 3-phase induction motor, the rotor rotates at a slower speed than the stator magnetic field. Therefore, the correct option is (b).

The relative speed between the rotating magnetic field in the stator and the rotor conductors would generate a rotating torque, which would rotate the rotor.

C25. Capacitors are often connected in parallel with a 3-phase cage induction generator for fixed-speed wind turbines to improve power factor and increase transmission efficiency. Therefore, the correct option is (b).

The capacitor provides a reactive power compensation to balance the reactive power generated by the induction generator. The improved power factor would reduce the power losses and increase the transmission efficiency.

C26. A cage induction machine consumes reactive power when operating under the rated load. Therefore, the correct option is (a).

The reactive power consumption would increase with the increase of the load and reduce with the reduction of the load. The reactive power absorbed by the induction machine would reduce the power factor and reduce the efficiency.

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A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field. How close are the two charges?

Answers

A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field.

Let the positive charge be q1=+1.100 μC and the negative charge be q2=-0.500 μC.

A positive charge of 1.100μ C is located in a uniform field of 9.00×10⁴ N/C. A negative charge of -0.500μ C is brought near enough to the positive charge that the attractive force between the charges just equals the force on the positive charge due to the field.

The net force on q1 due to the field is:

1=q1×E=+1.100×10⁻⁶C×9.00×10⁴ N/C=+99 N

The force between the charges is attractive and its magnitude is equal to the force experienced by q1 due to the uniform electric field:

2=99N

Then the distance between the charges is:

r=12/402= (1.100×10⁻⁶C)(-0.500×10⁻⁶C)/(4(8.85×10⁻¹²C²/N·m²)(99N))= 1.87×10⁻⁵m

Answer: 1.87×10⁻⁵m.

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An object is placed a distance of 8.88f from a converging lens, where f is the lens's focal length. (Include the sign of the value in your answers.)
(a) What is the location of the image formed by the lens? dᵢ = __________ f
(b) Is the image real or virtual? O real O virtual (c) What is the magnification of the image? (d) Is the image upright or inverted? O upright O inverted

Answers

An object is placed a distance of 8.88f from a converging lens, where f is the lens's focal length.(a) The location of the image formed by the lens is at dᵢ = infinity (b) Since the image is formed at infinity, it is considered a virtual image.

(c) The magnification of the image can be determined using the magnification formula(d) The image is neither upright nor inverted. It is an "O real" image.

To solve this problem, we can use the lens formula:

1/f = 1/dₒ + 1/dᵢ

where:

   f is the focal length of the lens,    dₒ is the object distance,    dᵢ is the image distance.

Given that the object distance is 8.88f, we can substitute this value into the formula and solve for dᵢ.

(a) Calculating the image distance:

1/f = 1/dₒ + 1/dᵢ

1/f = 1/(8.88f) + 1/dᵢ

To simplify the equation, we can find a common denominator:

1/f = (1 + 8.88f) / (8.88f) = (1 + 8.88f) / (8.88f)

Now we can equate the numerators and solve for dᵢ:

1 = 1 + 8.88f

8.88f = 0

f = 0

Therefore, the image distance is at infinity, which means the image is formed at the focal point of the lens.

(a) The location of the image formed by the lens is at dᵢ = infinity.

(b) Since the image is formed at infinity, it is considered a virtual image.

(c) The magnification of the image can be determined using the magnification formula:

magnification (m) = -dᵢ / dₒ

Since dᵢ is infinity and dₒ is 8.88f, we can substitute these values into the formula:

magnification (m) = -∞ / (8.88f) = 0

Therefore, the magnification of the image is 0.

(d) Since the magnification is 0, the image is neither upright nor inverted. It is an "O real" image.

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The clarinet is well-modeled as a cylindrical pipe that is open at one end and closed at the other. For a clarinet's whose air column has an effective length of 0.407 m, determine the wavelength λm=3 and frequency fm=3 of the third normal mode of vibration. Use 346 m/s for the speed of sound inside the instrument.

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Answer: The wavelength (λm=3) is 0.2713 m and the frequency (fm=3) is 850.86 Hz.

In an open ended cylindrical pipe, the wavelength of the nth harmonic can be calculated using: L = (nλ)/2

Where; L = effective length of the pipeλ = wavelength of the nth harmonic n = mode of vibration.

The frequency of the nth harmonic can be determined using the formula given below; f = nv/2L  

Where; f = frequency of the nth harmonic

n = mode of vibration

v = speed of sound

L = effective length of the pipe

Here, the mode of vibration is given to be 3 and the speed of sound inside the instrument is 346 m/s. Therefore, the wavelength of the third harmonic can be: L = (3λ)/2λ = (2L)/3λ = (2 × 0.407)/3λ = 0.2713 m.

The frequency of the third harmonic can be determined as: f = (3 × 346)/(2 × 0.407)f = 850.86 Hz.

Therefore, the wavelength (λm=3) is 0.2713 m and the frequency (fm=3) is 850.86 Hz.

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Down-sampling throws away samples, so it will shrink the size of the image. This is what is done by the following scheme: wp ww (1:p:end, 1:p:end); when we are downsampling by a factor of p.

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The expression "wp ww (1:p:end, 1:p:end)" represents down-sampling an image by a factor of p using a scheme called "subsampling."

What is subsampling?

In subsampling, every p-th sample is selected from both the width (wp) and height (ww) dimensions of the image. The notation "1:p:end" indicates that we start at the first sample and select every p-th sample until the end of the dimension.

By applying this scheme to an image, we effectively reduce the number of samples taken along both the width and height dimensions, resulting in a smaller image size. This down-sampling process discards the non-selected samples, effectively "throwing them away."

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The circuit in the figure consists of switch S, a 4.70 V ideal battery, a 40.0 MQ resistor, and an airfilled capacitor. The capacitor has parallel circular plates of radius 5.00 cm, separated by 4.50

Answers

To find the capacitance of the capacitor, we can use the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation distance.

The capacitance of a capacitor is determined by the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space (a constant value), A is the area of the plates, and d is the separation distance between the plates.

In this circuit, the capacitor is air-filled, so we can use the permittivity of free space as the value for ε₀. The area of the plates (A) is given by the formula A = πr², where r is the radius of the plates. The separation distance (d) between the plates is also provided.

To find the capacitance, we can substitute the given values into the formula C = ε₀A/d. Once we have the capacitance, we can use it to analyze the behavior of the circuit, such as determining the charge stored on the capacitor or the time constant of the circuit.

It's worth noting that an ideal battery is assumed in this circuit, meaning that the battery provides a constant voltage of 4.70 V regardless of the current flowing through the circuit.

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15.4 cm. Given this wavelength and frequency, what is the speed of the sound wave? 48.7 cm. Given this wavelength and frequency, what is the speed of the sound wave? speed of sound (in m/s ) under these conditions? Give your answer to the nearest 1 m/s.

Answers

Given this wavelength and frequency.  that the frequency of the first scenario is approximately 3.168 times the frequency of the second scenario.

To calculate the speed of a sound wave, we can use the formula: speed = wavelength × frequency.

For the first scenario with a wavelength of 15.4 cm, we need to convert it to meters by dividing it by 100: 15.4 cm = 0.154 m. Let's assume a frequency of f1. Using the formula, we have speed = 0.154 m × f1.

For the second scenario with a wavelength of 48.7 cm, we again convert it to meters: 48.7 cm = 0.487 m. Let's assume a frequency of f2. Using the formula, we have speed = 0.487 m × f2.

Since the speed of sound in air is generally considered constant (at approximately 343 m/s at room temperature and normal atmospheric conditions), we can equate the two expressions for speed and solve for f1 and f2

0.154 m × f1 = 0.487 m × f2

By canceling out the common factor of 0.154, we get:

f1 = 0.487 m × f2 / 0.154 m

Simplifying further:

f1 ≈ 3.168 × f2

This equation implies that the frequency of the first scenario is approximately 3.168 times the frequency of the second scenario. Therefore, to determine the speed of sound under these conditions, we need more information about either the frequency in one of the scenarios or the specific speed of sound for the given conditions.

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A planet with a mass of 2.7 x 1022 kg is in a circular orbit around a star with a mass of 5.3 x 1032 kg. If the planet has an orbital radius of 4.8 x 10 m, what is its orbital period? (Universal gravitation constant, G = 6.67. 10-11 m kg 15-2) 23. A 0.05 kg softball was bounced on the sidewalk. The velocity change of the ball is from 30 m/s downward to 20 m/s upward. If the contact time with the sidewalk is 1.25 ms. a) What is momentum change of the ball? b) What is the magnitude of the average force exerted on the ball by the sidewalk? 24. A rocket explodes into four pieces of equal mass. Immediately after the explosion their velocities are (120 m/s, cast), (150 m/s, west), (80 m/s, south), and (150 m/s north). What was the velocity of the rocket's center of mass before the explosion? 0° Use Directions are 90° for east, 180° for south, 270° for west, and 360° for north. 270° 90° 180°

Answers

The orbital period of the planet is approximately 1.2411 x 10^6 seconds.

The orbital period of a planet can be calculated using the formula T = 2π√(r³/GM), where T is the orbital period, r is the orbital radius, G is the universal gravitation constant, and M is the mass of the central star. In this case, with a planet mass of 2.7 x 10^22 kg, a star mass of 5.3 x 10^32 kg, and an orbital radius of 4.8 x 10^10 m, the orbital period of the planet can be determined.

To calculate the orbital period, we can use Kepler's third law, which relates the orbital period to the radius and mass of the central object. The formula for orbital period, T, is given by T = 2π√(r³/GM), where r is the orbital radius, G is the universal gravitation constant (6.67 x 10^-11 m^3 kg^-1 s^-2), and M is the mass of the central star.

Plugging in the given values, we have T = 2π√((4.8 x 10^10)^3 / (6.67 x 10^-11) (5.3 x 10^32 + 2.7 x 10^22)).

Simplifying the expression inside the square root, we get T ≈ 2π√(1.3824 x 10^33 / 3.53671 x 10^22).

Further simplifying, T ≈ 2π√(3.9117 x 10^10), which gives T ≈ 2π(1.9778 x 10^5) ≈ 1.2411 x 10^6 seconds.

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A projectile is fired with an initial velocity of 46.82m/s at an angle of 41.89°. It hits a target 1.09s later. How high (vertically) is the target?
Notes: Remember, a = g. Don't forget the units!

Answers

A projectile is fired with an initial velocity of 46.82m/s at an angle of 41.89°. It hits a target 1.09s later.  The target is approximately 56.26 meters below the initial launch height.

To determine the vertical height of the target, we can analyze the projectile's motion and apply the equations of motion.

Let's break down the initial velocity into its vertical and horizontal components. The vertical component (Vy) can be found using the equation:

Vy = V × sin(θ)

where V is the initial velocity (46.82 m/s) and θ is the launch angle (41.89°). Plugging in the values:

Vy = 46.82 m/s × sin(41.89°)

≈ 29.70 m/s

Next, we can determine the time it takes for the projectile to reach its maximum height (t_max). At the highest point of the projectile's trajectory, the vertical velocity becomes zero. We can use the equation:

Vy = Vy_initial + g × t_max

where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values:

0 = 29.70 m/s - 9.8 m/s^2 × t_max

Solving for t_max:

t_max = 29.70 m/s / 9.8 m/s^2

≈ 3.03 s

Since the total time of flight is given as 1.09 s, we can calculate the time it takes for the projectile to descend from its maximum height to hit the target:

t_descent = total time of flight - t_max

= 1.09 s - 3.03 s

≈ -1.94 s

The negative sign indicates that the projectile has already descended from its maximum height when it hits the target.

Now, let's find the vertical distance traveled during the descent. We can use the equation:

Δy = Vy_initial × t_descent + (1/2) × g × t_descent^2

Plugging in the values:

Δy = 29.70 m/s × (-1.94 s) + (1/2) × 9.8 m/s^2 × (-1.94 s)^2

≈ -56.26 m

The negative sign indicates that the target is located below the initial launch height. To find the actual vertical height of the target, we take the absolute value of Δy:

Vertical height of the target = |Δy|

≈ 56.26 m

Therefore, the target is approximately 56.26 meters below the initial launch height.

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A 100km long overhead line whose resistance is R=0.12/km, reactance is X₁ = 0.25 22/km, susceptance is 1/X = 12×10 Siemens/km is used for 500kV four-core conductor to transmit 1000MVA to a load with power factor of 0.8 lagging (Base complex power 2500MVA, Base voltage 500kV). A. Calculate the required sending end voltage for short-line representation. B. Calculate the required sending end voltage for medium-line representation. C. Calculate the required sending end voltage for long-line representation.

Answers

The required sending-end voltage for short-line representation is 503.4 ∠ 27.25° kV, for medium-line representation is 488.9 ∠ 23.65° kV, and for long-line representation is 479.1 ∠ 21.16° kV.

A 100 km long overhead line is used to transmit 1000 MVA to a load with a power factor of 0.8 lagging using a 500 kV four-core conductor. The resistance is R = 0.12/km, the reactance is [tex]X_{1}[/tex]= 0.25 Ω/km, and the susceptance is 1/X = 12 × [tex]10^{-6}[/tex] Siemens/km. The base complex power is 2500 MVA, and the base voltage is 500 kV.

The following are the steps to calculate the required sending end voltage for short-line representation:

A short-line model has a line length that is less than 80 km, and the shunt capacitance is ignored. The line's resistance and inductive reactance are combined in a single equivalent impedance per unit length. The equivalent impedance per unit length is as follows:

Z = R + j[tex]X_{1}[/tex] = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km

The load current is calculated using the following formula:

I = S/V = 1000 MVA/[(0.8)(2500 MVA)/(500 kV)] = 2.828 kA

Send-end voltage is calculated by using the following formula:

Vs = V + (I × Z × l) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] = 503.4 ∠ 27.25° kV

The following are the steps to calculate the required sending end voltage for medium-line representation:

A medium-line model has a line length that is greater than 80 km but less than 240 km, and the shunt capacitance is taken into account. The equivalent impedance per unit length and shunt admittance per unit length are as follows:

Z = R + j[tex]X_{1}[/tex] = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km

Y = jB = j (2πf ε[tex]_{r}[/tex] ε[tex]_{0}[/tex])[tex]^{1/2}[/tex] = j(2π × 50 × 8.854 × [tex]10^{-12}[/tex] × 12 × [tex]10^{-6}[/tex])1/2 = j2.228 × [tex]10^{-6}[/tex] S/km

The load current and sending-end voltage are the same as those used in the short-line model.

The receiving-end voltage is calculated using the following formula:

VR = V + (I × Z × l) - ([tex]I^{2}[/tex] × Y × l/2) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] - [[tex](2.828 kA)^2[/tex] (j2.228 × [tex]10^{-6}[/tex]S/km)(100 km)/2] = 484.7 ∠ 27.38° kV

The sending-end voltage is calculated using the following formula:

Vs = VR + (I × Y × l/2) = 484.7 ∠ 27.38° kV + [(2.828 kA)(j2.228 × [tex]10^{-6}[/tex]S/km)(100 km)/2] = 488.9 ∠ 23.65° kV

The following are the steps to calculate the required sending end voltage for long-line representation:

A long-line model has a line length that is greater than 240 km, and both the shunt capacitance and series impedance are taken into account. The equivalent impedance and admittance per unit length are as follows:

Z' = R + jX1 = 0.12 + j0.25 22 = 0.12 + j0.25Ω/km

Y' = jB + Y = j (2πf ε[tex]_{r}[/tex] ε[tex]_{0}[/tex])[tex]^{1/2}[/tex] + Y = j(2π × 50 × 8.854 × [tex]10^{-12}[/tex] × 12 ×[tex]10^{-6}[/tex])[tex]^{1/2}[/tex] + j[tex]12[/tex] × [tex]10^{-6}[/tex] S/km = (0.25 + j2.245) × [tex]10^{-6}[/tex] S/km

The load current and sending-end voltage are the same as those used in the short-line model. The receiving-end voltage is calculated using the following formula:

V[tex]_{R}[/tex] = V + (I × Z' × l) - ([tex]I^{2}[/tex] × Y' × l/2) = 500 kV + [(2.828 kA)(0.12 + j0.25Ω/km)(100 km)] - [[tex](2.828 kA)^2[/tex] ((0.25 + j2.245) × [tex]10^{-6}[/tex] S/km)(100 km)/2] = 439.1 ∠ 37.55° kV

The sending-end voltage is calculated using the following formula:

Vs = VR + (I × Y' × l/2) = 439.1 ∠ 37.55° kV + [(2.828 kA)((0.25 + j2.245) × [tex]10^{-6}[/tex] S/km)(100 km)/2] = 479.1 ∠ 21.16° kV

Hence, the required sending-end voltage for short-line representation is 503.4 ∠ 27.25° kV, for medium-line representation is 488.9 ∠ 23.65° kV, and for long-line representation is 479.1 ∠ 21.16° kV.

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High frequency alternating current is passed through a solenoid that contains a solid copper core insulated from the coils of the solenoid. Which statement is correct? O A copper core remains cool no matter what the frequency of the current in the solenoid is. The copper core remains cool because the induced emf is parallel to the solenoid axis and fluctuates rapidly. 0 The copper core heats up because an emf parallel to the solenoid axis is induced in the core. O The copper core heats up because circular currents around its axis are induced in the core. O The copper core heats up because the electric field induced in the copper is parallel to the magnetic field produced by the solenoid.

Answers

The correct statement is that c. the copper core heats up because circular currents around its axis are induced in the core.

What is a solenoid?

A solenoid is a long coil of wire with numerous turns that are tightly packed together. It produces a uniform magnetic field when electrical energy is passed through it. An electric current flowing through a solenoid produces a magnetic field that is proportional to the number of turns in the coil and the magnitude of the electric current.

The statement, "The copper core heats up because circular currents around its axis are induced in the core" is correct. The magnetic field produced by the solenoid induces circular currents in the copper core. These circular currents are referred to as eddy currents. The eddy currents heat up the copper core and, as a result, the copper core becomes hot.

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What is the output voltage of a 3.00-V lithium cell in a digital wristwatch that draws 0.670 mA, if the cell's internal resistance is 2.25 Ω? (Enter your answer to at least five significant figures.) V

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The output voltage of a 3.00-V lithium cell in a digital wristwatch, considering its internal resistance of 2.25 Ω, is approximately 1.5075 V which is determined using Ohm's Law and should be calculated to at least five significant figures.

To calculate the output voltage, we can use Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R). In this case, the current drawn by the wristwatch is given as 0.670 mA, and the internal resistance of the cell is 2.25 Ω. Thus, we can calculate the voltage as follows:

V = I * R

= 0.670 mA * 2.25 Ω

= 1.5075 mV

Since the given lithium cell has an initial voltage of 3.00 V, the output voltage will be slightly lower due to the internal resistance. Therefore, the output voltage of the lithium cell in the digital wristwatch is approximately 1.5075 V when rounded to five significant figures.

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a 380-kg piano slides 2.9 m down a 25 degree incline and it kept from accelerating by a man who is pushing back on it parallel to the incline. Determine (a) the force exerted by the man, (b) the work done on the piano by the man, (c) the work done on the the piano by the force of gravity, (d) the net work done on the piano. Ignore friction.

Answers

a) The force exerted by the man is approximately 1608.86 N.

b) The work done on the piano by the man is approximately 4662.34 Joules.

c) The work done on the piano by the force of gravity is approximately 7210.18 Joules.

d) The net work done on the piano is approximately 11872.52 Joules.

To solve this problem, we'll need to consider the forces acting on the piano and the work done by each force.

Mass of the piano (m): 380 kg

Distance traveled down the incline (d): 2.9 m

Incline angle (θ): 25 degrees

Acceleration due to gravity (g): 9.8 m/s²

(a) The force exerted by the man:

The force exerted by the man is equal in magnitude and opposite in direction to the force of gravity component parallel to the incline. This force is given by:

F_man = m * g * sin(θ)

Substituting the values:

F_man = 380 kg * 9.8 m/s² * sin(25°)

F_man ≈ 1608.86 N

(b) The work done on the piano by the man:

The work done by a force is given by the equation:

Work = Force * Distance * cos(θ)

Since the force exerted by the man is parallel to the displacement, the angle between the force and displacement is 0 degrees, and the cos(0°) = 1. Therefore, the work done by the man is:

Work_man = F_man * d

Substituting the values:

Work_man = 1608.86 N * 2.9 m

Work_man ≈ 4662.34 J

(c) The work done on the piano by the force of gravity:

The force of gravity acting on the piano has a component parallel to the incline, given by:

F_gravity_parallel = m * g * sin(θ)

The work done by the force of gravity is:

Work_gravity = F_gravity_parallel * d

Substituting the values:

Work_gravity = 380 kg * 9.8 m/s² * sin(25°) * 2.9 m

Work_gravity ≈ 7210.18 J

(d) The net work done on the piano:

The net work done on an object is the sum of the work done by all the forces acting on it. In this case, since there are only two forces (force exerted by the man and force of gravity), the net work done on the piano is:

Net work = Work_man + Work_gravity

Substituting the values:

Net work = 4662.34 J + 7210.18 J

Net work ≈ 11872.52 J

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Write controlling current I, in terms of node voltage 202² lo below b) Write node equation for V, in terms of node voltage V, only [No I, or I, terms

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The node equation for V in terms of node voltage V only is: V = nV

a) To express the current I in terms of the node voltage V, we can use Ohm's Law and Kirchhoff's Current Law (KCL). Let's consider a specific node in the circuit where the current I is flowing. According to KCL, the sum of currents entering that node must be equal to the sum of currents leaving the node.

Let's denote the node voltage at the current source terminal as V₀ and the node voltage at the other terminal as V. The voltage across the current source can be written as V₀ - V.

Applying Ohm's Law to the current source, we have:

I = (V₀ - V) / R

Thus, the current I in terms of the node voltage V is given by:

I = (V₀ - V) / R

b) To write the node equation for V in terms of node voltage V only, without involving the current I, we can apply Kirchhoff's Voltage Law (KVL) around the loop connected to the node we are considering.

Considering the voltage drops across each element in the loop, we have:

V = V₁ + V₂ + V₃ + ... + Vₙ

Here, V₁, V₂, V₃, ..., Vₙ represent the voltage drops across the elements connected in series within the loop.

Since we want to express V in terms of node voltage V only, we can rewrite the voltage drops V₁, V₂, V₃, ..., Vₙ in terms of node voltage differences. Let's assume that the node we are considering is the reference node, denoted as 0V. Therefore, the voltage difference from the reference node to node V₁ is simply V. Similarly, the voltage difference from the reference node to node V₂ is also V, and so on.

Hence, we can rewrite the equation as:

V = V + V + V + ... + V

Simplifying, we have:

V = nV

Where n represents the number of elements connected in series within the loop.

Therefore, the node equation for V in terms of node voltage V only is:

V = nV

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a 120-v power supple connected to a 10-ohm resistor will produce ____ amps of current

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Hello!

a 120-v power supple connected to a 10-ohm resistor will produce 3.464 amps of current

P = 120 V

r = 10Ω

P = r * I²

I² = P ÷ r

I² = 120 ÷ 10

I² = 12

I = √12

I ≈ 3.464

Consider a square with side a = 1.500 m. Four charges -q, +q, +q, and -q where q = 4.80 μC are placed at the corners A, B, C, and D, respectively.
A) What is the magnitude of the electric field (in N/C) at point D due to the charges at points A, B, C?
B) What is the direction of the electric field from part (a)? (Let the positive x-axis = 0 degrees)
C) What is the magnitude of the net force (in Newtons) on the charge at point D?
D) What is the direction of the net force on the charge at point D in Newtons?

Answers

To calculate the electric field and net force at point D due to the charges at points A, B, and C in a square, we can use the principles of Coulomb's law and vector addition.

The magnitude and direction of the electric field and net force can be determined by considering the contributions of each charge.

A) To find the magnitude of the electric field at point D due to the charges at points A, B, and C, calculate the electric field contribution from each charge using Coulomb's law and then add the vector components of the electric fields.

B) The direction of the electric field from part (a) can be determined by considering the direction of the individual electric fields and their vector sum. Use vector addition rules to find the resultant direction.

C) To calculate the magnitude of the net force on the charge at point D, use Coulomb's law to determine the force between each charge and the charge at point D. Add the vector components of the forces to find the net force.

D) The direction of the net force on the charge at point D can be determined by considering the direction of the individual forces and their vector sum. Use vector addition rules to find the resultant direction.

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A parallel plate capacitor with circular faces of diameter 7.7 cm separated with an air gap of 1.8 mm is charged with a 12.0 V emf. What is the total charge stored in this capacitor, in pc, between the plates? Do not enter units with answer

Answers

The total charge stored in a parallel plate capacitance with circular faces, a diameter of 7.7 cm, and an air gap of 1.8 mm, charged with a 12.0 V emf, can be calculated.

The capacitance of a parallel plate capacitor is given by the equation C = ε₀A/d. In this case, the circular plates have a diameter of 7.7 cm, so the radius (r) is half of that, which is 3.85 cm or 0.0385 m. The area of each plate can be calculated using A = πr².

Once we have the capacitance, we can use the equation Q = CV to find the total charge stored in the capacitor. Here, Q represents the charge and V is the emf or voltage applied to the capacitor.

By substituting the values into the equation, calculate the total charge stored in the capacitor. Remember to consider the units of the given values and use consistent units throughout the calculations to obtain the correct numerical answer.

In conclusion, the total charge stored in the parallel plate capacitor can be determined by calculating the capacitance and using the equation Q = CV, where Q is the charge and V is the emf or voltage applied to the capacitor.

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It slowed down, so now I know that...
A.) a force acted on it.
B.) no force acted on it.
C.) gravity acted on it.
D.) its mass was decreasing.
E.) its mass was increasing.

Answers

If an object slows down, it indicates that a force acted on it. Therefore, option A, "a force acted on it," is the correct answer.

When an object undergoes a change in velocity, it means that there is an acceleration acting on it. According to Newton's second law of motion, acceleration is directly proportional to the net force applied to an object and inversely proportional to its mass.

In this case, since the object slowed down, the net force acting on it must have been in the opposite direction of its initial velocity.

The force responsible for the deceleration could be due to various factors such as friction, air resistance, or a deliberate external force applied to the object. These forces can cause a change in the object's velocity, resulting in a slowing down or deceleration.

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(a) A person has a near point of 10.0 cm, and a far point of 20.0 cm, as measured from their eyes. (i) (2 points) Is this person nearsighted or farsighted? JUSTIFY YOUR ANSWER. (ii) (6 points) This person puts on eyeglasses of power (- 8.00 D), that sit 1.8 cm in front of their eyes. What is their "new" near point - in other words, what is the closest that they can hold reading material and see it clearly? (iii) (4 points) Show, by means of a calculation, that these (-8.00 D) glasses will NOT help their far point issues. Bifocal Lens (iv) (6 points) Since their near point and far point cannot both be helped by the same glasses, perhaps they need "bi-focals" – glasses with two different focal lengths (one for the top half of the glasses, one for the bottom half, like this sketch shows). What power should the other part of their glasses be in order to move their "new" far point out to infinity? distance near (b) A different person uses +2.3 diopter contact lenses to read a book that they hold 28 cm from their eyes. (i) (2 points) Is this person nearsighted or farsighted? JUSTIFY YOUR ANSWER. NO CREDIT WILL BE GIVEN WITHOUT JUSTIFICATION. (ii) (6 points) Where is this person's near point, in cm? (iii) (4 points) As this person ages, they eventually must hold the book 38 cm from their eyes in order to see clearly with the same +2.3 diopter lenses. What power lenses do they need in order to hold book back at the original 28 cm distance?

Answers

(i) This person is nearsighted.

ii the closest the person can hold reading material and see it clearly is about 0.257 cm.

III Since the far point cannot have a negative distance, we can conclude that the glasses will not help their far point issues because the image distance (far point) is approximately -2.86 cm, which is not a physically meaningful result.

How to explain the information

a. Near point refers to the closest point at which a person can focus their eyes, and a near point of 10.0 cm indicates that they can only focus on objects that are relatively close to their eyes.

(ii) To calculate the new near point, we can use the lens formula:

1/f = 1/v - 1/u

In this case, the eyeglasses have a power of -8.00 D, which means the focal length of the lens (f) is -1/8.00 m = -0.125 m.

The object distance (u) is the distance from the glasses to the eyes, which is given as 1.8 cm = 0.018 m.

Plugging these values into the lens formula, we can solve for v:

1/(-0.125) = 1/v - 1/0.018

-8 = (0.018 - v)/v

-8v = 0.018 - v

-7v = 0.018

v = 0.018 / (-7)

≈ -0.00257 m

Converting this to centimeters:

v ≈ -0.257 cm

Since the near point cannot have a negative distance, the new near point with the glasses is approximately 0.257 cm. Therefore, the closest the person can hold reading material and see it clearly is about 0.257 cm.

(iii)Using the same lens formula as before:

1/f = 1/v - 1/u

The object distance (u) for the far point is given as 20.0 cm = 0.2 m.

Plugging these values into the lens formula, we can solve for v:

1/(-0.125) = 1/v - 1/0.2

-8 = (0.2 - v)/v

-8v = 0.2 - v

-7v = 0.2

v = 0.2 / (-7) ≈ -0.0286 m

Converting this to centimeters:

v ≈ -2.86 cm

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