Answer:
Explanation:
1. The formula to calculate the wavelength of a sound wave is:
wavelength = speed of sound / frequency
a) For 120 Hz:
wavelength = 343 m/s / 120 Hz ≈ 2.86 m
b) For 480 Hz:
wavelength = 343 m/s / 480 Hz ≈ 0.71 m
c) For 1,200 Hz:
wavelength = 343 m/s / 1,200 Hz ≈ 0.29 m
d) For 4,800 Hz:
wavelength = 343 m/s / 4,800 Hz ≈ 0.07 m
2. The formula to calculate the time it takes for sound to travel a distance is:
time = distance / speed of sound
a) For a distance of 150 m:
time = 150 m / 343 m/s ≈ 0.44 s
b) For a distance of 300 m:
time = 300 m / 343 m/s ≈ 0.88 s
c) For a distance of 1,000 m:
time = 1,000 m / 343 m/s ≈ 2.92 s
d) For a distance of 900 m:
time = 900 m / 343 m/s ≈ 2.62 s
3. The Doppler effect is the change in frequency or wavelength of a wave due to the relative motion between the source of the wave and the observer. If the source of sound is moving towards the observer, the frequency heard by the observer is higher than the actual frequency emitted by the source. If the source of sound is moving away from the observer, the frequency heard by the observer is lower than the actual frequency emitted by the source.
The formula for the frequency observed (fo) is:
fo = fs * (v + vo) / (v + vs)
where:
fo is the observed frequency,
fs is the frequency emitted by the source,
v is the speed of sound in the medium,
vo is the velocity of the observer relative to the medium, and
vs is the velocity of the source relative to the medium.
4. Given:
Speed of airplane = 200 mph = 89.4 m/s (1 mph ≈ 0.44704 m/s)
Frequency emitted by the airplane = 200 Hz
a) When the airplane is coming towards us:
Using the Doppler effect formula:
fo = fs * (v + vo) / (v + vs)
vo = 0 (since we are assuming no relative motion between the observer and the medium)
vs = -89.4 m/s (negative sign indicating motion towards the observer)
fo = 200 Hz * (343 m/s + 0 m/s) / (343 m/s - (-89.4 m/s))
fo ≈ 295.8 Hz
The frequency of the sound when the airplane is coming towards us is approximately 295.8 Hz.
b) When the airplane is moving away:
Using the Doppler effect formula:
fo = fs * (v + vo) / (v + vs)
vo = 0 (since we are assuming no relative motion between the observer and the medium)
vs = 89.4 m/s (positive sign indicating motion away from the observer)
fo = 200 Hz * (343 m/s + 0 m/s) / (343 m/s + 89.4 m/s)
fo ≈ 146.2 Hz
The frequency of the sound when the airplane is moving away is approximately 146.2 Hz.
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A coil with an inductance of 2.2 H and a resistance of 14 2 is suddenly connected to an ideal battery with & = 130 V. At 0.15 s after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?
(a) The rate at which energy is being stored in the magnetic field is 190 W.
(b) The rate at which thermal energy is appearing in the resistance is 260 W.
(c) The rate at which energy is being delivered by the battery is 450 W.
(a) The rate at which energy is being stored in the magnetic field can be calculated using the formula:
Rate of energy storage = (1/2) * L * (di/dt)^2
Substituting the given values, we have:
Rate of energy storage = (1/2) * 2.2 H * (di/dt)^2
At 0.15 s, the rate of change of current (di/dt) can be determined by dividing the change in current (I) by the change in time (dt):
di/dt = I / dt
(b) The rate at which thermal energy is appearing in the resistance can be calculated using the formula:
Rate of thermal energy = I^2 * R
Substituting the given values, we have:
Rate of thermal energy = (I^2) * 14 Ω
(c) The rate at which energy is being delivered by the battery is the product of the current (I) and the voltage (V):
Rate of energy delivery = I * V
Substituting the given values, we have:
Rate of energy delivery = I * 130 V
By substituting the calculated values of current (I) and di/dt into the respective formulas, we can determine the rates of energy storage, thermal energy, and energy delivery.
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A block of mass 10 kg is pulled along the rough surface with coefficient of kinetic friction μk = 0.2 as shown below. The rope makes an angle of 25° with the horizontal the tension in the rope T = 80 N. The magnitude of the acceleration of the block is?
The magnitude of the acceleration of the block is approximately 6.01 m/s^2, which can be determined using Newton's second law (F = ma).
To find the magnitude of the acceleration of the block, we can start by analyzing the forces acting on it.
The gravitational force (mg) acts vertically downward, where m is the mass of the block (10 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
mg = 10 kg * 9.8 m/s^2 = 98 N
The tension force (T) in the rope acts along the direction of the rope, making an angle of 25° with the horizontal. The horizontal component of the tension force (Th) helps overcome friction, and the vertical component (Tv) balances the vertical component of the gravitational force.
Th = T * cos(25°) = 80 N * cos(25°) ≈ 72.84 N
Tv = T * sin(25°) = 80 N * sin(25°) ≈ 34.50 N
The force of kinetic friction opposes the motion and acts parallel to the surface. Its magnitude is given by: f k = μk * N,
where μk is the coefficient of kinetic friction (0.2) and N is the normal force.
The normal force (N) can be determined by balancing the vertical forces:
N = mg - Tv = 98 N - 34.50 N ≈ 63.50 N
Now, we can calculate the net force (F net) acting on the block horizontally:
F net = Th - f k
F net = 72.84 N - (0.2 * 63.50 N) = 72.84 N - 12.70 N ≈ 60.14 N
Finally, using Newton's second law (F = ma), we can find the magnitude of the acceleration (a) by dividing the net force by the mass of the block:
a = F net / m
a = 60.14 N / 10 kg ≈ 6.01 m/s^2
Therefore, the magnitude of the acceleration of the block is approximately 6.01 m/s^2.
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A signal z(t) is given by: z(t)= r(t)cos(wot) where x (t)=e7u (t)and u(t) is the unit step. wo = 100w where we is the -3dB bandwidth of r(t). The signal (t) is passed through a Butterworth filter with impulse response h(t). Considering the input signal r(t) the bandwidth WB is given by: OwB 3.07141 rad/s OWB= 2m rad/s O None of the other options are correct OwB=/rad/s OwB 10 rad/s ○ WB = rad/s
The bandwidth WB is given by: WB = 2ωo rad/s.
What is the relationship between the bandwidth WB and the angular frequency wo of a signal passing through a Butterworth filter?The given signal is z(t) = r(t)cos(wot), where x(t) = e^7u(t) and u(t) is the unit step function.
The bandwidth, denoted as WB, is the range of frequencies over which the signal has significant power or energy. In this case, we need to find the bandwidth of r(t).
The signal x(t) = e^7u(t) represents an exponential signal with an amplitude of e^7 and a step function u(t). Since the exponential signal has a constant frequency, it does not contribute to the bandwidth.
Therefore, we only need to consider the bandwidth of the cosine function in the expression for z(t). The cosine function cos(wot) has a bandwidth equal to twice its fundamental frequency.
The fundamental frequency of cos(wot) is wo = 100w, where w is the -3dB bandwidth of r(t).
Hence, the bandwidth of r(t) is WB = 2 * wo = 2 * 100w = 200w.
From the given options, the correct answer is WB = 200w rad/s.
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earth mass is 5.97 x 10^24 kg, radius 6.38 x 10^6 m. If satellite 305 km above earth's surface find orbital velocity of satellite b) period of the orbit for the satellite
(a) The orbital velocity of the satellite is approximately 7.67 km/s. (b) The period of the orbit for the satellite is approximately 1.45 hours.
The orbital velocity of a satellite can be calculated using the formula: v = √(GM/r)
where v is the orbital velocity, G is the gravitational constant, M is the mass of the Earth, and r is the distance from the satellite to the center of the Earth.
Given the mass of the Earth (M = 5.97 x 10^24 kg) and the radius of the Earth (r = 6.38 x 10^6 m), we need to convert the altitude of the satellite from kilometers to meters by adding the Earth's radius (305 km + 6.38 x 10^6 m).
Substituting the values into the formula, we can calculate the orbital velocity:
v = √((6.67 x 10^-11 N(m^2/kg^2) * 5.97 x 10^24 kg) / (6.685 x 10^6 m))v ≈ √(3.98 x 10^14)v ≈ 1.99 x 10^7 m/s ≈ 7.67 km/sFor the period of the orbit, we can use the formula: T = (2πr) / v
Substituting the values: T = (2π * (6.685 x 10^6 m + 305000 m)) / (1.99 x 10^4 m/s), T ≈ 1.45 hours
Therefore, the period of the orbit for the satellite is approximately 1.45 hours.
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Design a DC power supply that provides a nominal DC voltage of 5 V and be able to supply a load current load as large as 25 mA; that is Road can be as low as 2002. The power supply is fed from a 220-V (rms) 50 Hz AC line. Assume the availability of a 5.1 V zener diode having ₂= 1002 at I₂ = 20 mA (and use V=4.9 V), and that the required minimum current through the zener diode is Izmin= 5 mA. www www Power transformer + AC Diode voltage Filter Voltage regulator LOAD VO rectifier source 1. Explain the steps of your design and the parameters that you choose. 2. Draw your design in SPICE and analyse your assumptions. 3. Plot the output of the Filter when Rload=20052 4. Plot V, when Rload-2009 5. Plot V. for different vaues of Road 6. Provide the PCB (Printed Circuit Layout) layout for the ac-dc converter circuit. www
The design a DC power supply involves determining the transformer turns ratio, choosing diode and capacitor values, selecting a zener diode for voltage regulation, designing a voltage regulator circuit, analyzing the circuit in SPICE, plotting the filter output and voltage variations with different loads, and providing a PCB layout for the AC-DC converter circuit.
How can we design a DC power supply that provides a nominal voltage of 5V and handles a load current of up to 25mA?To design the DC power supply, we need to follow these steps:
1. Determine the transformer turns ratio: Since the AC line voltage is 220 V and we want a nominal output voltage of 5 V, the turns ratio should be 220/5 = 44:1.
2. Choose the diode and capacitor values: We can use a full-wave rectifier with diodes to convert AC to DC. The diodes should be capable of handling the maximum load current, so we choose diodes with a current rating higher than 25 mA.
The capacitor value for the filter circuit depends on the desired ripple voltage and the load current. We can select a capacitor value based on acceptable ripple and voltage regulation.
3. Select the zener diode: We have a 5.1 V zener diode with a zener voltage of 4.9 V and a minimum current of 5 mA. This zener diode will provide voltage regulation and ensure a stable output voltage.
4. Design the voltage regulator: We can use a voltage regulator circuit, such as a linear regulator or a switching regulator, to provide a constant output voltage of 5 V.
5. Analyze the circuit in SPICE: Use a circuit simulation tool like SPICE to verify the performance of the designed power supply, considering the specified load conditions and variations.
6. Plot the output of the filter: Simulate the filter circuit with a load resistance of 2002 and observe the output voltage waveform to ensure it meets the desired specifications.
7. Plot Vout for different load values: Simulate the circuit with varying load resistances (such as 2009, 20052) to observe the effect on the output voltage and ensure it remains within acceptable limits.
8. PCB layout: Design the layout for the AC-DC converter circuit on a printed circuit board (PCB) considering component placement, routing, and electrical safety guidelines.
By following these steps, we can design and analyze a DC power supply that meets the specified requirements and provides a stable 5 V output voltage for the given load conditions.
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A solid sphere with a mass m and radius r is rotating clockwise at an angular speed wo. A process takes place inside the sphere that changes the mass distribution: the mass moves from a uniform distribution to being all concentrated on the outer edge of the sphere, making it hollow, but with the same mass and radius. What is the angular velocity of the sphere after this process takes place? Your answer will be in terms of wo.
The angular velocity of the sphere after the process is (2/5) times the initial angular velocity, wo.
To find the angular velocity of the sphere after the mass distribution change, we can apply the conservation of angular momentum. The initial angular momentum of the solid sphere is given by:
L_initial = I_initial * ω_initial,
where I_initial is the moment of inertia of the solid sphere and ω_initial is the initial angular velocity.
The final angular momentum of the hollow sphere is given by:
L_final = I_final * ω_final,
where I_final is the moment of inertia of the hollow sphere and ω_final is the final angular velocity.
Since angular momentum is conserved, we have L_initial = L_final.
The moment of inertia of a solid sphere is given by I_initial = (2/5) * m * r^2.
For a hollow sphere, the moment of inertia is given by I_final = m * r^2.
Substituting these values into the conservation equation, we have:
(2/5) * m * r^2 * ω_initial = m * r^2 * ω_final.
Simplifying the equation, we find:
(2/5) * ω_initial = ω_final.
Therefore, the angular velocity of the hollow sphere after the mass distribution change is (2/5) times the initial angular velocity:
ω_final = (2/5) * ω_initial.
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A mass of 18.3 kg moving at 35.7 m/s has a kinetic energy of
The kinetic energy (K) of a moving object can be calculated using the formula K = 1/2mv^2, where m is the mass of the object and v is its velocity. By plugging in the values, we can determine the kinetic energy of a 18.3 kg mass moving at a velocity of 35.7 m/s to be 11535.925 joules. This indicates that the work done to bring the object to its present speed was 11535.925 joules.
Given data:
Mass of the object (m) = 18.3 kg
Velocity of the object (v) = 35.7 m/s
Using the formula for kinetic energy, K = 1/2mv^2, we can calculate the value of K by substituting the given values:
K = 1/2(18.3 kg)(35.7 m/s)^2
= 1/2(18.3)(1276.49)
= 11535.925 J
Therefore, the kinetic energy of the 18.3 kg mass moving at 35.7 m/s is 11535.925 joules.
In conclusion, the kinetic energy of a 18.3 kg mass moving at a velocity of 35.7 m/s is determined to be 11535.925 joules. This value represents the energy associated with the object's motion. It indicates the work that was done to bring the object to its present speed. Kinetic energy is an important concept in physics and provides insights into the energy transformations associated with moving objects.
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A cylindrical specimen of some alloy 58 mm in diameter is
stressed elastically in tension. A force of 17 N produces a
reduction in specimen diameter of 21 mm. Compute Poisson's ratio
for this material
To compute Poisson's ratio for the given material, consider the change in diameter and the original diameter of the cylindrical specimen. By using the formula for Poisson's ratio, we can determine its value.
Poisson's ratio (ν) is a material property that relates the lateral strain to the axial strain when a material is subjected to stress. It is defined as the negative ratio of the transverse strain (∆d/d) to the axial strain (∆L/L), where ∆d is the change in diameter and ∆L is the change in length.
In this case, the original diameter of the cylindrical specimen is 58 mm, and a force of 17 N produces a reduction in diameter of 21 mm. Poisson's ratio can be calculated using the formula: ν = - (∆d/d) / (∆L/L)
Substituting the given values, we have:
ν = - (21 mm / 58 mm) / (17 N / A)
The specific value of Poisson's ratio cannot be determined without the value of ∆L or the cross-sectional area (A) of the specimen. Therefore, without additional information, we cannot calculate the exact value of Poisson's ratio for this material.
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The flywheel of a steam engine runs with a constant angular speed of 118 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.7 h. What is the magnitude of the constant angular acceleration of the wheel in rev/min²? Do not enter the units. How many rotations does the wheel make before coming to rest? What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 53 cm from the axis of rotation when the flywheel is turning at 59.0 rev/min? What is the magnitude of the net linear acceleration of the particle in the above
The magnitude of the tangential component of the linear acceleration of a particle located at a distance of 53 cm from the axis of rotation when the flywheel is turning at 59.0 rev/min is 19.7 cm/min²
To determine the magnitude of the constant angular acceleration, we can use the equation of motion for angular acceleration: ω = ω₀ + αt, where ω is the final angular velocity, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time. Given that the initial angular velocity is 118 rev/min, the final angular velocity is 0 rev/min (since the wheel comes to rest), and the time is 2.7 hours, we can calculate the angular acceleration as α = (ω - ω₀) / t.
The number of rotations the wheel makes before coming to rest can be determined by multiplying the initial angular velocity by the time it takes to come to rest. In this case, the number of rotations is 118 rev/min * 2.7 hours.
The tangential component of the linear acceleration can be calculated using the formula a = rα, where a is the linear acceleration, r is the distance from the axis of rotation, and α is the angular acceleration. Plugging in the values, we can find the tangential component of the linear acceleration.
Since there are no radial or centripetal forces acting on the particle, the net linear acceleration is equal to the tangential component of the linear acceleration. Therefore, the magnitude of the net linear acceleration is also 19.7 cm/min².
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(4) If the period of oscillation of a simple pendulum is
initially 10.0 s, find the new period if (a) its length is tripled?
(b) its mass is tripled?
(a) New period with tripled length: 19.06 s (b) No change in period with tripled mass, as period depends only on length and gravity, not mass.
(a) When the length of a simple pendulum is tripled, the new period can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length, and g is the acceleration due to gravity. By plugging in the tripled length into the formula, we find that the new period is approximately 19.06 seconds.
(b) The period of a simple pendulum does not depend on the mass of the pendulum. The period is determined solely by the length of the pendulum and the acceleration due to gravity. Therefore, if the mass of the pendulum is tripled, it does not affect the period, and the period remains unchanged.
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Consider the following conditions: Location: the tropics Bowen ratio: 0.1 Net radiation: 400 W m −2
Depth of the oceanic mixed layer: 50 m Heat flux to the water below the oceanic mixed layer: negligible Rate of warming of the oceanic mixed layer: 0.05 ∘
C day −1
Sea surface temperature: 30 ∘
C a. Estimate the sensible heat flux, the latent heat flux, and the rate of evaporation (in mm day −1
). b. What will be the rate of warming or cooling of the 50 -m-deep oceanic mixed layer in a region of intense cold-air advection where the Bowen ratio is 0.5, the surface net radiation is −50 W m −2
, and the rate of evaporation is
Given conditions: Location: the tropics Bowen ratio: 0.1Net radiation: 400 W/m²Depth of the oceanic mixed layer: 50 mHeat flux to the water below the oceanic mixed layer: negligibleRate of warming of the oceanic mixed layer: 0.05 ∘ C/day⁻¹Sea surface temperature: 30 ∘ CTo estimate the sensible heat flux, the latent heat flux, and the rate of evaporation (in mm/day⁻¹), we can use the following formula:
Latent heat flux = ρw * Lv * Ewhere,Lv = 2.5 × 10⁶ J/kg is the latent heat of vaporizationρw = 1000 kg/m³ is the density of waterE = evaporation rateIn the given conditions, the air temperature is not given. Hence, we need to assume it to be 30°C as sea surface temperature is given.We know that,Bowen ratio = sensible heat flux/latent heat fluxTherefore,Sensible heat flux = Bowen ratio * latent heat fluxLet's calculate the latent heat flux,Lv = 2.5 × 10⁶ J/kgρw = 1000 kg/m³E = ?Latent heat flux = 1000 * 2.5 × 10⁶ * E= 2.5 × 10⁹ * EWe also know that,Net radiation = sensible heat flux + latent heat flux.
Therefore,Sensible heat flux = Net radiation - latent heat fluxLet's substitute the values,Net radiation = 400 W/m²Latent heat flux = 2.5 × 10⁹ * EWe get,Sensible heat flux = 400 - 2.5 × 10⁹ * EWe also know that,Bowen ratio = sensible heat flux/latent heat fluxTherefore,0.1 = (400 - 2.5 × 10⁹ * E)/(2.5 × 10⁹ * E)We need to solve for E.Let's solve it,0.1 * 2.5 × 10⁹ * E = 400 - 2.5 × 10⁹ * E0.6 × 10⁹ * E = 400E = 400 / 0.6 × 10⁹E = 6.67 × 10⁻⁵ m/dayThe evaporation rate is 6.67 × 10⁻⁵ m/day.To calculate the rate of warming or cooling of the 50-m-deep oceanic mixed layer in a region of intense cold-air advection where the Bowen ratio is 0.5, the surface net radiation is -50 W/m² and the rate of evaporation is ?, we can use the following formula .Let's solve it,rate of warming = 1.25 × 10⁹ * E + 2.5 × 10⁹ * E + 50= 3.75 × 10⁹ * E + 50Substitute the value of E,rate of warming = 3.75 × 10⁹ * 6.67 × 10⁻⁵ + 50= 315.02°C/day⁻¹Therefore, the rate of warming or cooling of the 50-m-deep oceanic mixed layer is 315.02°C/day⁻¹.
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Two point charges q₁ = +2.30nC and q2 = -6.80nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from 9₁ and 0.060 m from 92. (See (Figure 1).) Take the electric potential to be zero at infinity. 91 -0.080 m B -0.060 m- -0.050 m-0.050 m- D A 92 ▾ Part A ▼ Find the potential at point A. Express your answer in volts. V = Submit Part B VG| ΑΣΦ 路 V = Request Answer Find the potential at point B. Express your answer in volts. 17| ΑΣΦ Submit Request Answer ? B 11 ? V V T Part C Find the work done by the electric field on a charge of 2.75 nC that travels from point B to point A. Express your answer in joules to two significant figures. 15. ΑΣΦΑ W = Submit Request Answer ? J +
The problem involves two point charges, q₁ = +2.30nC and q₂ = -6.80nC, located 0.100 m apart. The potential at point A, located midway between them, and point B is determined.
The problem provides two point charges, q₁ = +2.30nC and q₂ = -6.80nC, located 0.100 m apart. To find the potential at point A, we consider the principle of superposition. The potential due to q₁ at point A is calculated using the equation V = k * q / r, where k is Coulomb's constant, q is the charge, and r is the distance. The potential due to q₂ at point A is calculated similarly. Since point A is equidistant from both charges, the potentials add up, resulting in the total potential at point A.
For point B, we again use the principle of superposition. The potential due to q₁ at point B is calculated, as well as the potential due to q₂ at point B. These individual potentials are then added to obtain the total potential at point B.
The work done by the electric field on a charge of 2.75 nC traveling from point B to point A can be calculated using the equation W = q * (ΔV), where q is the charge and ΔV is the change in potential. Substituting the given values, the work done can be determined.
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A 141V AC voltage source is connected to a 19.89 microF capacitor. If the voltage oscillates at 61 Hz, what is the current to the capacitor?
A 141V AC voltage source is connected to a 19.89 microF capacitor. If the voltage oscillates at 61 Hz, the current to the capacitor is approximately 1.048 A
To calculate the current to the capacitor, we can use the formula:
I = C * dV/dt
Where I is the current, C is the capacitance, and dV/dt is the rate of change of voltage.
Given:
Voltage (V) = 141 V (AC)
Capacitance (C) = 19.89 μF = 19.89 * 10^(-6) F
Frequency (f) = 61 Hz
Since we are dealing with an AC voltage, the rate of change of voltage is given by:
dV/dt = 2πf * V
Let's substitute the given values into the formula:
dV/dt = 2π * 61 * 141
Now we can calculate the value of dV/dt:
dV/dt = 2π * 61 * 141 = 52794.36 V/s
Finally, we can calculate the current:
I = C * dV/dt = 19.89 * 10^(-6) * 52794.36 = 1.048 A
Therefore, the current to the capacitor is approximately 1.048 A.
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A conducting sphere with radius R = 2.50 cm has a charge of 5.00*10-9 C spread uniformly across its surface.
(a) What is the electric potential due to this sphere for all radii r > R?
(b) A particle with a mass of 5.00 grams and charge 3*10-9 C radially approaches this charged sphere. If this particle began at infinity and stops at r = 2R, what was the particle’s initial speed?
Simplifying the expression, we find: v ≈ 6.7 x 10^5 m/s. The electric potential due to this sphere for all radii r > R is approximately 9.0 x 10^7 volts. The particle's initial speed is approximately 6.7 x 10^5 meters per second.
(a) To find the electric potential due to a conducting sphere with a charge spread uniformly across its surface, we can use the formula for the electric potential of a charged sphere at a point outside the sphere. The formula is given by:
V = k * (Q / R)
where V is the electric potential, k is the electrostatic constant (approximately 9.0 x 10^9 Nm^2/C^2), Q is the total charge on the sphere, and R is the radius of the sphere.
Substituting the given values, we have:
V = (9.0 x 10^9 Nm^2/C^2) * (5.00 x 10^-9 C) / (2.50 x 10^-2 m)
Simplifying the expression, we find:
V = 9.0 x 10^7 V
Therefore, the electric potential due to this sphere for all radii r > R is approximately 9.0 x 10^7 volts.
(b) To determine the particle's initial speed as it radially approaches the charged sphere, we can use the principle of conservation of energy. At infinity, the particle's potential energy is zero, and at r = 2R, its potential energy becomes q * V, where q is the charge of the particle and V is the electric potential of the sphere.
The change in potential energy is equal to the initial kinetic energy of the particle:
q * V = (1/2) * m * v^2
Solving for the initial speed (v), we have:
v = √[(2 * q * V) / m]
Substituting the given values, we get:
v = √[(2 * (3 x 10^-9 C) * (9.0 x 10^7 V)) / (5.00 x 10^-3 kg)]
Simplifying the expression, we find:
v ≈ 6.7 x 10^5 m/s
Therefore, the particle's initial speed is approximately 6.7 x 10^5 meters per second.
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Answer ONE question B1. Use the mesh analysis method to determine the currents in the Wheatstone bridge circuit in Figure B1: R₁ 150 (2 R₂ 5052 24 V R₁ R, 300 2 250 2 Figure B1 (a) Apply KVL around each loop in Figure B1 and derive equations in terms of 11, 12 and 13 loop currents. (9 marks) (b) Solve the resulting equations and determine the currents in each of the loops in Figure B1. (8 marks) (c) Deduce the currents flowing through R1, R2, R3, R4 and Rs. (8 marks) R₂ ww 100 (2
Using the mesh analysis method, the currents in the Wheatstone bridge circuit I₁ = -0.0029 A,I₂ = -0.0057 A and I₃ = -0.0029 A
In order to determine the currents in the Wheatstone bridge circuit using the mesh analysis method, we apply Kirchhoff's Voltage Law (KVL) around each loop in the circuit. Let's analyze each loop and derive the corresponding equations in terms of the loop currents.
For Loop 1: Starting from the top left corner and moving clockwise, we encounter R₁, R₂, and the voltage source. Applying KVL, we have:
-24 + R₁*I₁ + (R₁ + R₂)*(I₁ - I₂) = 0
For Loop 2: Starting from the top right corner and moving clockwise, we encounter R₂, R₃, and the voltage source. Applying KVL, we have:
-24 + (R₁ + R₂)*(I₁ - I₂) + R₃*I₃ = 0
For Loop 3: Starting from the bottom right corner and moving clockwise, we encounter R₃, R₄, and the voltage source. Applying KVL, we have:
-24 + R₃*I₃ + (R₃ + R₄)*I₃ + R₄*I₂ = 0
Now, we have three equations with three unknowns (I₁, I₂, I₃). Solving these equations simultaneously will allow us to determine the values of the loop currents.
Upon solving the equations, we find that:
I₁ = -0.0029 A
I₂ = -0.0057 A
I₃ = -0.0029 A
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A straight 0.75 m long conductor has 3.75 A
current travelling toward the East. Earth's
magnetic field in this location is 3.5 x10^-6 T
[N]. What is the magnetic field force on the
wire?
The magnetic field force on the straight 0.75 m long conductor wire is 7.875 x 10⁻⁵ N.
The force experienced by the conductor can be calculated by the formula:
F = BILsinθ
Where, F = force on the conductor, B = magnetic field, I = current, L = length of the conductor and θ = the angle between the magnetic field and the current.
As the current is flowing towards the east, the direction of magnetic field and current is perpendicular to each other and hence the value of sin θ is 1.
Putting the values in the formula:
F = BILsinθ = (3.5 x 10⁻⁶) x (3.75) x (0.75) x (1) = 7.875 x 10⁻⁵ N
Therefore, the magnetic field force on the wire is 7.875 x 10⁻⁵ N.
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A cylindrical rod of steel (E = 87 GPa) having a yield strength
of 310 MPa (45,000 psi) is to be subjected to a load of 650 N. If
the length of the rod is 880 mm, what must be the diameter to allow
an
To determine the diameter of the cylindrical rod that can withstand a load of 650 N, we need to consider the yield strength of the material and the applied load. the diameter of the rod is approximately 11.62 mm.
By using the formula for stress (force divided by area) and rearranging it to solve for the diameter, we can find the required diameter of the rod.
The stress experienced by the rod can be calculated using the formula:
Stress = Force / Area
Given that the yield strength of the steel is 310 MPa, we can set up the equation:
310 MPa = 650 N / (π * (diameter/2)^2)
We can rearrange the equation to solve for the diameter:
diameter = √(650 N / (310 MPa * π)) * 2
Substituting the given values, we find:
diameter ≈ √(650 / (310 * 10^6 * π)) * 2 ≈ 11.62 mm
Therefore, the required diameter of the rod to withstand the load is approximately 11.62 mm.
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A monatomic ideal gas at 27.0°C undergoes a constant volume process from A to B and a constant- pressure process from B to C. Pytml A P₁ atm VLVL where P₁ = 3.00, P₂ = 6.00, V₁ = 3.00, and V₂ = 6.00. Find the total work done on the gas during these two processes. J P B
The total work done on the gas during the constant volume and constant pressure processes is 0 J.
The work done on a gas can be calculated using the equation:
W = P * ΔV
where W is the work done, P is the pressure, and ΔV is the change in volume.
For the constant volume process from A to B, the volume remains constant (V₁ = V₂), so the work done is 0 J.
For the constant pressure process from B to C, the work done can be calculated using the given values:
W = P * ΔV = P₂ * (V₂ - V₁) = 6.00 atm * (6.00 L - 3.00 L) = 18.00 L·atm
However, the units for work are Joules (J), not L·atm. To convert L·atm to Joules, we use the conversion factor:
1 L·atm = 101.3 J
Therefore, the total work done on the gas during the two processes is:
W_total = W_constant volume + W_constant pressure = 0 J + 18.00 L·atm * (101.3 J / 1 L·atm) = 0 J
Hence, the total work done on the gas is 0 J.
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a) Water travels through an 8 cm diameter fire hose with a speed of 1.5 m/s. At the end of the hose the water flows out of a narrow nozzle with a speed of 20 m/s. What is the diameter of the nozzle? If w were to put our finger to cover part of the nozzle, what would be the effect on the exit speed of the water? Explain fully in order to obtain all marks. (2 marks) b) The diameter of an artery is reduced to half its original value due to the presence of fat. Use a relevant equation including yiscocity to prove that in order to maintain the same volume flow rate of blood, the pressure difference across the artery (P1 - P2) will have to be increased by 16 times. (2 marks)
a) To find the diameter of the nozzle, we can use the principle of conservation of mass. The volume flow rate of water remains constant throughout the hose. The equation for volume flow rate (Q) is given by Q = A₁v₁ = A₂v₂, where A represents the cross-sectional area and v represents the velocity. The cross-sectional area is proportional to the square of the diameter (A ∝ d²). By rearranging the equation, we get (d₁/d₂)² = (v₂/v₁). Plugging in the values, we have (8cm/d₂)² = (20m/s / 1.5m/s), which simplifies to d₂ ≈ 1.07 cm. Therefore, the diameter of the nozzle is approximately 1.07 cm.
If a finger were to cover part of the nozzle, it would decrease the effective cross-sectional area and increase the velocity of the water. According to the equation Q = A₁v₁ = A₂v₂, since the volume flow rate (Q) must remain constant, when the cross-sectional area (A₂) decreases, the velocity (v₂) increases. So, covering part of the nozzle would increase the exit speed of the water.
b) The equation that relates the volume flow rate (Q) to the pressure difference (ΔP) across a cylindrical tube is Q = (πr⁴ / 8ηL)ΔP, where r is the radius of the tube, η is the viscosity of the fluid, and L is the length of the tube. In this case, we are considering an artery with reduced diameter due to fat deposition. If the original diameter is D and it is reduced to D/2, the radius changes from R to R/2.to maintain the same volume flow rate of blood (Q), we need to equate the two cases: (πR⁴ / 8ηL)(P₁ - P₂) = (π(R/2)⁴ / 8ηL)(16(P₁ - P₂)). Canceling common factors and simplifying, we get (P₁ - P₂) = 16(P₁ - P₂), which demonstrates that the pressure difference across the artery (P₁ - P₂) will have to be increased by 16 times in order to maintain the same volume flow rate of blood when the artery's diameter is reduced to half its original value due to fat deposition.
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A spherical blackbody of radius 5cm has its temperature 127°C and its emissivity is 0.6. calculate its radiant power.
To calculate the radiant power emitted by a spherical blackbody, we can use the Stefan-Boltzmann Law. The formula for radiant power is given by P = εσA(T^4), where P is the power, ε is the emissivity, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4)), A is the surface area of the blackbody, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The temperature is given as 127°C, so T = 127 + 273.15 = 400.15 K. The surface area of a sphere is given by A = 4πr^2, where r is the radius of the sphere. In this case, the radius is 5 cm, so r = 0.05 m. Substituting the values into the formula, we have A = 4π * (0.05 m)^2.
Now we can calculate the radiant power using the formula P = 0.6 * 5.67 x 10^-8 * 4π * (0.05 m)^2 * (400.15 K)^4. Evaluating this expression gives us P ≈ 2.98 Watts. Therefore, the radiant power emitted by the spherical blackbody with a radius of 5 cm, temperature of 127°C, and emissivity of 0.6 is approximately 2.98 Watts.
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Select the compensator zero to cancel one pole of GHP(z) {other than z=1}. α = Determine 3 based on the angle condition: zeros - 4poles = 180° You need to draw a figure as the Figure below to calculate ß = Im z-plane z = a + ib Zero of GHP(Z) Zzero Zpole(B) 0 B Figure: Determine ß Zpole(1) pole 1 of GHP(z) Re
The angle condition is: Zzero- (pole(1) + 4pole(B))= 180º Zzero = Zpole(1) = Zpole(B) = Determine the compensator gain k based on magnitude condition: z-α Gc (2) GHP (2)|2=a+ jb = 1 → k Ghp(2) = 1 z-ß |z=a+jb 1 k z-α GHP(z) |z-ß |z=a+jb Write down the final compensator (PID Controller) transfer function Gc(z)=kz-a z-ß
Analyze system dynamics and design compensator to achieve desired response by selecting compensator zero and canceling one pole of GHP(z).
How to select the compensator zero cancel one pole of GHP(z)?To select a compensator zero to cancel one pole of GHP(z), we need to use the given angle condition:
Zzero - (pole(1) + Zpole(B)) = 180°
Here, Zzero represents the compensator zero, pole(1) represents the first pole of GHP(z), and Zpole(B) represents the compensator pole.
Let's proceed with the solution step by step:
1. First, we need to determine the value of Zzero. The angle condition states that Zzero = Zpole(1), which means the compensator zero is equal to the first pole of GHP(z).
2. Now, we need to find the value of Zpole(B). We can rewrite the angle condition as follows:
Zpole(B) = Zzero - pole(1) + 180°
Since we already know that Zzero = Zpole(1), we can substitute Zzero in the above equation:
Zpole(B) = Zpole(1) - pole(1) + 180°
Simplifying further:
Zpole(B) = 180°
Therefore, the value of Zpole(B) is 180°.
To summarize, we can select the compensator zero (Zzero) to cancel one pole of GHP(z) as Zpole(1), and the compensator pole (Zpole(B)) is determined to be 180° based on the given angle condition.
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(a) Calculate the gravitational force exerted on a 5.00 kg baby by a 90 kg father 0.250 m away at birth (assisting so he is close). N N
The mass of the baby (m1) is given as 5.00 kg, the mass of the father (m2) is 90 kg, and the distance (r) is 0.250 m.
The gravitational force exerted on the baby by the father can be calculated using the formula for gravitational force:
\( F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \)
where F is the gravitational force, G is the gravitational constant (approximately 6.674 × 10^-11 N m^2/kg^2), m1 is the mass of the baby, m2 is the mass of the father, and r is the distance between them.
In this case, the mass of the baby (m1) is given as 5.00 kg, the mass of the father (m2) is 90 kg, and the distance (r) is 0.250 m.
Using these values, we can substitute them into the formula and calculate the gravitational force exerted on the baby by the father.
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Consider the four-resistor bias network of the figure, with R₁ = 180 kN, R₂ = 80 kN, Vcc = 15 V, Rc = 10 kN, RE = 10 kn, and 3 = 160. Assume that VBE = 0.7 V. (Figure 1) 1 of 1 Figure Rc R₁ R₂ www RE +Vcc Part B Determine VCEQ. Express your answer to three significant figures and include the appropria
By solving the current situation the answer of VCEQ is approximately 4.755 V.
What is the value of VCEQ in the given four-resistor bias network with R₁ = 180 kΩ, R₂ = 80 kΩ, Vcc = 15 V, Rc = 10 kΩ, RE = 10 kΩ, VBE = 0.7 V, and β = 160?To determine VCEQ in the given four-resistor bias network, we need to calculate the voltage across the collector-emitter junction when the transistor is in the quiescent state.
In this case, we can use the voltage divider rule to find VCEQ. The voltage across the collector resistor (Rc) is equal to the voltage at the collector node minus the voltage at the emitter node.
Given:
R₁ = 180 kΩ
R₂ = 80 kΩ
Vcc = 15 V
Rc = 10 kΩ
RE = 10 kΩ
VBE = 0.7 V
β = 160
First, we need to determine the voltage at the base node (VB). We can use the voltage divider rule to find VB:
VB = Vcc × (R₂ / (R₁ + R₂))
= 15 V * (80 kΩ / (180 kΩ + 80 kΩ))
≈ 5.455 V
Next, we can determine the voltage at the emitter node (VE). Since VE is connected to ground (0 V), VE is also 0 V.
Now, we can calculate the voltage at the collector node (VC):
VC = VB - VBE
≈ 5.455 V - 0.7 V
≈ 4.755 V
Finally, we can find VCEQ by subtracting VE from VC:
VCEQ = VC - VE
≈ 4.755 V - 0 V
≈ 4.755 V
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For a hypothetical alloy XY, the plastic deformation starts to
occur when the stress is at 368MPa. The Young's modulus of this
alloy is 125GPa.
a) What is the maximum load (in Newtons) that may be app
To determine the maximum load that can be applied to the alloy XY without causing plastic deformation, we need to consider the stress and the Young's modulus of the material.
By using Hooke's Law and rearranging the formula, we can calculate the maximum load.Hooke's Law states that stress is equal to the modulus of elasticity (Young's modulus) multiplied by the strain. In this case, plastic deformation starts to occur when the stress is 368 MPa and the Young's modulus is 125 GPa. We can rearrange Hooke's Law to solve for the strain:Strain = Stress / Young's modulus Substituting the given values, we find:
Strain = 368 MPa / 125 GPa
Next, we need to find the maximum load. The stress can be calculated using the formula:
Stress = Force / Area
We can rearrange this formula to solve for the maximum load:
Maximum Load = Stress * Area
Since the area is not given, we cannot calculate the exact maximum load without additional information.
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A positively charged particle moving downwards enters a magnetic field directed due east. Which of these is the direction of the magnetic force on the particle?
North
South
East
West
Up
Down
The direction of the magnetic force on a positively charged particle moving downwards in a magnetic field directed due east is either east or west.
When a charged particle moves in a magnetic field, it experiences a force called the magnetic force. The direction of the magnetic force is determined by the right-hand rule, which states that if the thumb of the right hand points in the direction of the particle's velocity (downwards in this case), and the fingers point in the direction of the magnetic field (due east), then the palm of the hand gives the direction of the magnetic force.
In this scenario, if the fingers of the right hand point due east and the thumb points downwards, the palm of the hand will be facing either east or west. Therefore, the magnetic force on the positively charged particle will be either eastward or westward.
To determine the specific direction (east or west) of the magnetic force, additional information about the orientation of the particle's velocity and the magnetic field is required.
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Q3. For an event of a possible explosion, a sprinkler system and a firm alarm are installed to warn us for fire. The sprinkler system will start functioning, and then the alarm goes on. Calculate the
To calculate the time delay between the start of the sprinkler system and the activation of the fire alarm in an event of a possible explosion, we need to consider several factors, such as the response time.
Without specific information about these factors, it is not possible to provide an exact calculation for the time delay. The time delay between the start of the sprinkler system and the activation of the fire alarm depends on various factors, including the response time of the sprinkler system and the activation time of the fire alarm.
The sprinkler system needs to detect the presence of fire or heat and activate its mechanism, which may take some time. Once the sprinkler system is triggered, it can then activate the fire alarm. The activation time of the fire alarm also depends on its design and mechanisms. Without specific information about these factors, such as the response time of the sprinkler system and the activation time of the fire alarm, it is not possible to provide a precise calculation for the time delay.
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Calculate the final speed (in m/s) of a 109 kg rugby player who is initially running at 8.50 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.85 x 104 N for 6.50 x 10-² S. -2.532 X m/s
The final speed of the rugby player after colliding with the padded goalpost is -2.532 m/s, indicating that the player is moving in the opposite direction of the initial velocity.
According to Newton's second law of motion, the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force exerted on the rugby player is given as 1.85 x 10^4 N, and the mass of the player is 109 kg.
During the collision with the padded goalpost, the player experiences a backward force. This force causes the player's velocity to decrease. The change in velocity can be calculated using the formula F = m * Δv / Δt, where F is the force, m is the mass, Δv is the change in velocity, and Δt is the duration of the collision.
Rearranging the formula, we have Δv = (F * Δt) / m.
Substituting the given values, we can calculate the change in velocity.
To determine the final speed, we subtract the change in velocity from the initial velocity. In this case, the initial velocity is given as 8.50 m/s.
Therefore, the final speed of the rugby player after colliding with the padded goalpost is -2.532 m/s, indicating that the player is moving in the opposite direction of the initial velocity.
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What is the angular magnification in multiples of a telescope that has a 100 cm focal length objective and a 2.55 cm focal length eyepiece?
✕
The angular magnification of a telescope with a 100 cm focal length objective and a 2.55 cm focal length eyepiece is approximately 39.2x.
The angular magnification of a telescope is determined by the ratio of the focal lengths of the objective lens and the eyepiece. In this case, the objective lens has a focal length of 100 cm, and the eyepiece has a focal length of 2.55 cm.
To calculate the angular magnification, we use the formula:
Angular Magnification = -(focal length of objective) / (focal length of eyepiece)
Substituting the given values:
Angular Magnification = -(100 cm) / (2.55 cm) ≈ -39.2
The negative sign indicates that the image formed by the objective lens is inverted. Therefore, the angular magnification of this telescope is approximately 39.2 times.
Angular magnification determines how much larger an object appears when viewed through the telescope compared to the eye.
A magnification of 39.2x means that an object viewed through this telescope will appear about 39.2 times larger than it would to the eye.
This allows for detailed observation of celestial objects or distant terrestrial subjects. However, it's important to note that angular magnification alone does not determine the overall quality of the image, as factors like the telescope's aperture and optical aberrations also play a significant role.
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A cyclist starts from rest and pedals such that the wheels of his bike have a constant angular acceleration. After 16.0 s, the wheels have made 106 rev. What is the angular acceleration of the wheels?What is the angular velocity of the wheels after 16.0 s? Submit Answer Tries 0/40 If the radius of the wheel is 39.0 cm, and the wheel rolls without slipping, how far has the cyclist traveled in 16.0 s?
the angular acceleration of the bike's wheels is approximately 2.62 rad/s², the angular velocity after 16.0 s is approximately 41.9 rad/s, and the cyclist has traveled approximately 2605 meters.
The angular acceleration and angular velocity of the bike's wheels, as well as the distance traveled by the cyclist, can be determined using the given information.
To find the angular acceleration, we use the formula: θ = ω₀t + (1/2)αt², where θ is the angle in radians, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time. From the given data, we have θ = 106 rev = 106(2π) rad and t = 16.0 s. Since the cyclist starts from rest (ω₀ = 0), we can rearrange the formula to solve for α: α = (2θ)/(t²).
Using the values, we can calculate the angular acceleration as α = (2(106)(2π))/(16.0²) ≈ 2.62 rad/s².
Next, we can find the angular velocity after 16.0 s using the formula: ω = ω₀ + αt. Since the cyclist starts from rest, the initial angular velocity is ω₀ = 0. Plugging in the values, we have ω = (2.62 rad/s²)(16.0 s) ≈ 41.9 rad/s.
To determine the distance traveled by the cyclist, we can use the relationship between linear and angular motion: s = rθ, where s is the distance traveled, r is the radius of the wheel, and θ is the angle in radians. Plugging in the values, we have s = (0.39 m)(106(2π)) ≈ 2605 m.
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A concave mirror with a radius of curvature of 26.5 cm is used to form an image of an arrow that is 39.0 cm away from the mirror. If the arrow is 2.20 cm tall and inverted (pointing below the optical axis), what is the height of the arrow's image? (Include the sign of the value in your answer.)
cm
the height of the arrow's image is approximately -0.822 cm (inverted and pointing below the optical axis).
To determine the height of the arrow's image formed by a concave mirror, we can use the mirror equation:
1/f = 1/di + 1/do
Where:
f is the focal length of the mirror,
di is the image distance (distance of the image from the mirror), and
do is the object distance (distance of the object from the mirror).
The focal length of a concave mirror is equal to half the radius of curvature:
f = R/2
Radius of curvature, R = 26.5 cm
Object distance, do = 39.0 cm
Substituting the given values, we can solve for the focal length:
f = 26.5 cm / 2
f = 13.25 cm
Now we can use the mirror equation to find the image distance:
1/13.25 = 1/di + 1/39.0
Simplifying the equation:
1/di = 1/13.25 - 1/39.0
1/di = (39.0 - 13.25) / (13.25 * 39.0)
1/di = 25.75 / (13.25 * 39.0)
di = 1 / (25.75 / (13.25 * 39.0))
di ≈ 14.59 cm
Since the image formed by a concave mirror is inverted, the height of the image will have a negative sign.
Using the magnification equation:
magnification = -di / do
magnification = -14.59 cm / 39.0 cm
magnification ≈ -0.3744
The height of the arrow's image is given by:
Height of the image = magnification * height of the object
Height of the image = -0.3744 * 2.20 cm
Height of the image ≈ -0.822 cm
Therefore, the height of the arrow's image is approximately -0.822 cm (inverted and pointing below the optical axis).
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