The Reynolds number for flow in the hose is 10.75 and the Reynolds number for flow in the nozzle is 32.88.
Given data are:
Radius of nozzle, r₁ = 0.290 cm,
Radius of garden hose, r₂ = 0.810 cm,
Flow rate through hose, Q = 0.420 L/s = 0.420 x 10⁻³ m³/s,
Viscosity of water, η = 1.005 x 10³ N/m²s
(a) Calculate the Reynolds number for flow in the hose.
The Reynolds number is given by the relation:
Re = ρvD/η
where,ρ = Density of fluid, v = Velocity of fluid, D = Diameter of the pipe,
where,D = 2r₂ = 2 x 0.810 cm = 1.620 cm = 0.01620 m
Density of water at 20°C, ρ = 998 kg/m³
Flow rate, Q = πr₂²v = π(0.810 cm)²v = π(0.00810 m)²v0.420 x 10⁻³ m³/s = π(0.00810 m)²v
∴ v = Q/πr₂² = 0.420 x 10⁻³ m³/s / π(0.00810 m)² = 0.670 m/s
Now,Re = ρvD/η= 998 kg/m³ x 0.670 m/s x 0.01620 m / (1.005 x 10³ N/m²s)= 10.75
(b) Calculate the Reynolds number for flow in the nozzle.
The Reynolds number is given by the relation:
Re = ρvD/η
where,D = 2r₁ = 2 x 0.290 cm = 0.580 cm = 0.00580 m, Density of water at 20°C, ρ = 998 kg/m³, Velocity of fluid (water) through the nozzle, v = ?
Let's assume the velocity of water through the nozzle is equal to the velocity of water through the garden hose, i.e.
v = 0.670 m/s
Then,Re = ρvD/η= 998 kg/m³ x 0.670 m/s x 0.00580 m / (1.005 x 10³ N/m²s)= 32.88
Therefore, the Reynolds number for flow in the nozzle is 32.88.
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A 13225 N car traveling at 42.0 km/h rounds a curve of radius 1.34×10 2 m. The acceleration of gravity is 9.81 m/s 2 . a) Find the centripetal acceleration of the car. Answer in units of m/s 2 . b) Find the force that maintains circular motion. Answer in units of N. c) Find the minimum coefficient of static friction between the tires and the road that will allow the car to round the curve safely.
a) Centripetal acceleration = 0.918 m/s²
b) Centripetal force = 1237.43 N
c) Minimum coefficient of static friction = 0.0935
a) To find the centripetal acceleration of the car, we use the formula for centripetal acceleration, a = v²/r, where v is the velocity and r is the radius of the curve. First, we need to convert the car's speed from km/h to m/s: 42.0 km/h = (42.0 × 1000 m) / (3600 s) = 11.7 m/s. Plugging the values into the formula,
we have a = (11.7 m/s)² / (1.34 × 10² m) ≈ 0.918 m/s².
b) The force that maintains circular motion is the centripetal force, which is given by F = ma, where m is the mass of the car. To find the mass, we divide the weight of the car by the acceleration due to gravity: m = 13225 N / 9.81 m/s² ≈ 1349.03 kg. Plugging in the values,
we have F = (1349.03 kg) × (0.918 m/s²) ≈ 1237.43 N.
c) The minimum coefficient of static friction, μs, can be determined by comparing the maximum static friction force, μsN, to the centripetal force. Since the car is in circular motion, the normal force N is equal to the weight of the car, 13225 N. Setting μsN = F,
we have μs(13225 N) = 1237.43 N. Solving for μs,
we find μs = 1237.43 N / 13225 N ≈ 0.0935.
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An important news announcement is transmitted by radio waves to people who are 300 km away and sitting next to their radios, and also by sound waves to people sitting 4.00 m from the newscaster in a newsroom. Who receives the news first? people in the newsroom both at the same time At = people next to their radios What is the difference in time At between each group of people receiving the news?
Who receives the news first and calculate the time difference between the two groups of people, we need to compare the speed of radio waves and sound waves.people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
The speed of light, which includes radio waves, is approximately 3.00 x 10^8 meters per second (m/s) in a vacuum. However, when radio waves travel through the Earth's atmosphere, they slow down slightly but the difference is negligible for this calculation.
On the other hand, the speed of sound depends on the medium through which it travels. In dry air at room temperature, the speed of sound is approximately 343 meters per second (m/s).
First, let's calculate the time it takes for the radio waves to travel a distance of 300 km:
Time taken by radio waves = Distance / Speed
= 300,000 m / (3.00 x 10^8 m/s)
≈ 1.00 x 10^(-3) seconds (or 1 millisecond)
Next, let's calculate the time it takes for sound waves to travel a distance of 4.00 meters:
Time taken by sound waves = Distance / Speed
= 4.00 m / 343 m/s
≈ 0.0117 seconds (or 11.7 milliseconds)
Therefore, people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
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A friend in another city tells you that she has two organ pipes of different lengths, one open at both ends, the other open at one end only. In addition she has determined that the beat frequency caused by the second lowest frequency of each pipe is equal to the beat frequency caused by the third lowest frequency of each pipe. Her challenge to you is to calculate the length of the organ pipe that is open at both ends, given that the length of the other pipe is 140 m
The length of the organ pipe that is open at both ends is also 140 m.
To solve this problem, let's denote the length of the pipe that is open at both ends as L1 and the length of the pipe that is open at one end as L2. We are given that L2 is 140 m.
The beat frequency is caused by the interference between two sound waves with slightly different frequencies. In this case, we are comparing the second lowest frequency of each pipe.
The fundamental frequency (first harmonic) of a pipe open at both ends is given by:
f1 = v / (2L1)
where v is the speed of sound.
The second lowest frequency (second harmonic) of a pipe open at both ends is given by:
f2 = 2f1 = 2v / (2L1) = v / L1
The fundamental frequency (first harmonic) of a pipe open at one end is given by:
f3 = v / (4L2)
The second lowest frequency (second harmonic) of a pipe open at one end is given by:
f4 = 3f3 = 3v / (4L2)
Given that the beat frequency caused by f2 and f3 is equal to the beat frequency caused by f4, we can set up the following equation:
|f2 - f3| = |f4|
Substituting the expressions for f2, f3, and f4, we have:
|v / L1 - v / (4L2)| = 3v / (4L2)
Simplifying:
|4L2 - L1| = 3L1
Now we can substitute L2 = 140 m:
|4(140) - L1| = 3L1
Simplifying further
560 - L1 = 3L1
4L1 = 560
L1 = 140 m
Therefore, the length of the organ pipe that is open at both ends is also 140 m.
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Problem 2.0 (25 Points) Five years ago, when the relevant cost index was 135, a nuclear centrifuge cost $32,000. The centrifuge had a capacity of separating 1250 gallons of ionized solution per hour. Today, it is desired to build a centrifuge with capacity of 3500 gallons per hour, but the cost index now is 270. Assuming a power-sizing exponent to reflect economies of scale, x, of 0.72, use the power-sizing model to determine the cost (expressed in today's dollars) of the new reactor.
The cost (expressed in today's dollars) of the new reactor would be $85,237.74 given that the cost of a nuclear centrifuge five years ago is $32,000.
The relevant cost index was 135. The capacity of separating ionized solution per hour = 1250 gallons Power-sizing exponent to reflect economies of scale, x, of 0.72
Desired to build a centrifuge with a capacity of 3500 gallons per hour
The cost index now is 270.The power sizing model is given as,C₁/C₂ = (Q₁/Q₂) ^ x Where,C₁ = Cost of the first centrifuge C₂ = Cost of the second centrifuge Q₁ = Capacity of the first centrifuge Q₂ = Capacity of the second centrifuge X = power-sizing exponent
Substitute the given values, For the first centrifuge,C₁ = $32,000Q₁ = 1250 gallons C₂ = ?Q₂ = 3500 gallons x = 0.72
Now, substitute the given values in the power-sizing model,C₁/C₂ = (Q₁/Q₂) ^ x32000/C₂ = (1250/3500) ^ 0.72C₂ = $32000/(0.357)^0.72C₂ = $85,237.74
Thus, the cost (expressed in today's dollars) of the new reactor would be $85,237.74.
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54. An extra-solar planet orbits the distant star Pegasi 51. The planet has an orbital velocity of 2.3 X 10 m/s and an orbital radius of 6.9 X 10° m from the centre of the star. Determine the mass of the star. (6.2)
The mass of the star Pegasi 51 is approximately 3.76 x [tex]10^30[/tex] kilograms.
To determine the mass of the star, we can make use of the orbital velocity and radius of the planet. According to Kepler's laws of planetary motion, the orbital velocity of a planet depends on the mass of the star it orbits and the distance between them. By using the formula for orbital velocity, V = sqrt(GM/r), where V is the velocity, G is the gravitational constant, M is the mass of the star, and r is the orbital radius, we can solve for the mass of the star.
Given that the orbital velocity (V) is 2.3 x [tex]10^4[/tex] m/s and the orbital radius (r) is 6.9 x 10^10 m, we can rearrange the formula to solve for M:
M = V² * r / G
Plugging in the given values and the gravitational constant (G ≈ 6.67430 x 10^-11 m^3/kg/s^2), we can calculate the mass of the star:
M = (2.3 x [tex]10^4[/tex]m/s)²* (6.9 x [tex]10^10[/tex] m) / (6.67430 x[tex]10^-^1^1[/tex] m[tex]^3[/tex]/kg/[tex]s^2[/tex])
Calculating the expression gives us a value of approximately 3.76 x 10^30 kilograms for the mass of the star Pegasi 51.
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If this wave is traveling along the x-axis from left to right
with a displacement amplitude of 0.1 m in the y direction, find the
wave equation for y as a function of x and time t.
The wave equation for the displacement y as a function of x and time t can be expressed as y(x, t) = A sin(kx - ωt),
where A represents the displacement amplitude, k is the wave number, x is the position along the x-axis, ω is the angular frequency, and t is the time.
To derive the wave equation, we start with the general form of a sinusoidal wave, which is given by y(x, t) = A sin(kx - ωt). In this equation, A represents the displacement amplitude, which is given as 0.1 m in the y direction.
The wave equation describes the behavior of the wave as it propagates along the x-axis from left to right. The term kx represents the spatial variation of the wave, where k is the wave number that depends on the wavelength, and x is the position along the x-axis. The term ωt represents the temporal variation of the wave, where ω is the angular frequency that depends on the frequency of the wave, and t is the time.
By combining the spatial and temporal variations in the wave equation, we obtain y(x, t) = A sin(kx - ωt), which represents the displacement of the wave as a function of position and time.
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"The charges and coordinates of two charged particles held fixed
in an xy plane are q1 = 2.22 μC,
x1 = 4.01 cm, y1 = 0.369 cm
and q2 = -4.12 μC, x2 =
-2.11 cm, y2 = 1.39 cm. Find the
(a) magnitude
The magnitude of the force between the two charged particles is approximately [tex]1.03 \times 10^{-3} N[/tex].
To find the magnitude of the force between two charged particles, we can use Coulomb's law, which states that the force between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for the magnitude of the force is given by:
F = k |q₁ × q₂| / r²
where:
F is the magnitude of the force,
k is the electrostatic constant (k = 8.99 × 10⁹ N m²/C²),
|q₁ × q₂| is the absolute value of the product of the charges, and
r² is the square of the distance between the charges.
q₁ = 2.22 μC = 2.22 × 10⁻⁶ C
q₂ = -4.12 μC = -4.12 × 10^-6 C
x₁ = 4.01 cm = 4.01 × 10⁻² m
y₁ = 0.369 cm = 0.369 × 10⁻² m
x₂ = -2.11 cm = -2.11 × 10⁻² m
y₂ = 1.39 cm = 1.39 × 10⁻² m
Calculating the distance between the charges using the Pythagorean theorem:
r [tex]= \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}[/tex]
= [tex]\sqrt{((-2.11 \times 10^{-2} m - 4.01 \times 10^{-2} m)^2 + (1.39 \times 10^{-2} m - 0.369 \times 10^{-2} m)^2)}[/tex]
≈ 0.0634 m
F = F = k |q₁ × q₂| / r²
[tex]= (8.99 \times 10^9 Nm^2/C^2) \times |2.22 \times 10^{-6} C \times -4.12 * 10^{-6} C| / (0.0634 m)^2[/tex]
[tex]\approx 1.03 \times 10^{-3} N[/tex].
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Define pyroelectric coefficient along with its formula. Find the pyroelectric coefficient of a chip, if its area is 10 cm² and is heated from 10 °C to 15 °C in 5 minutes to obtain a current of 10pA?
The pyroelectric coefficient is a material property that quantifies the change in polarization per unit temperature change in a pyroelectric material.
It describes the sensitivity of a material to temperature variations and is typically denoted by the symbol "p" or "p_e". The pyroelectric coefficient is measured in units of C/m²·K.
The formula for the pyroelectric coefficient is given by:
p = ΔP / ΔT
where:
p is the pyroelectric coefficient,
ΔP is the change in electric polarization,
and ΔT is the change in temperature.
To find the pyroelectric coefficient of the chip in question, we need to know the change in electric polarization and the change in temperature. However, the given information only provides the area of the chip, the change in temperature (10°C to 15°C), and the resulting current (10pA). Without additional information about the material or its properties, it is not possible to calculate the pyroelectric coefficient in this case. The pyroelectric coefficient is specific to the material being used, and additional material-specific data is required to determine its value accurately.
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A 326-g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s . If the total energy of the system is 5.83 J , find (a) the maximum speed of the object.
To find the maximum speed of the object in simple harmonic motion, we can use the equation for total energy, which is given by the sum of the kinetic and potential energies.
Total energy (E) = Kinetic energy + Potential energy
The kinetic energy of an object executing simple harmonic motion can be expressed as (1/2)mv^2, where m is the mass of the object and v is its velocity. The potential energy of the system is given by (1/2)kA^2, where k is the spring constant and A is the amplitude of the motion. In this case, we are given the total energy E = 5.83 J and the mass m = 326 g = 0.326 kg.
Using the formula for period, T = 2π√(m/k), we can solve for k. Rearranging the equation, we get: k = (4π^2 * m) / T^2 Now that we have the value of k, we can find the amplitude A. Total energy (E) = Kinetic energy + Potential energy 5.83 J = (1/2)mv^2 + (1/2)kA^2 Since the object is at its maximum speed at the amplitude, we can assume the velocity at that point is v = vmax. Now we can substitute the value of k we found earlier into the equation: By substituting the given values of E, m, T, and solving for vmax, you can find the maximum speed of the object.
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Explain each of the following cases of magnification. magnification (M) M>1, M<1 and M=1 explain how you can find the image of a faraway object using a convex lens. Where will the image be formed?
What lens is used in a magnifying lens? Explain the working of a magnifying lens.
Magnification (M) refers to the degree of enlargement or reduction of an image compared to the original object. When M > 1, the image is magnified; when M < 1, the image is reduced; and when M = 1, the image has the same size as the object.
To find the image of a faraway object using a convex lens, a converging lens is typically used. The image will be formed on the opposite side of the lens from the object, and its location can be determined using the lens equation and the magnification formula.
A magnifying lens is a convex lens with a shorter focal length. It works by creating a virtual, magnified image of the object that appears larger when viewed through the lens.
1. M > 1 (Magnification): When the magnification (M) is greater than 1, the image is magnified. This means that the size of the image is larger than the size of the object. It is commonly observed in devices like magnifying glasses or telescopes, where objects appear bigger and closer.
2. M < 1 (Reduction): When the magnification (M) is less than 1, the image is reduced. In this case, the size of the image is smaller than the size of the object. This type of magnification is used in devices like microscopes, where small objects need to be viewed in detail.
3. M = 1 (Unity Magnification): When the magnification (M) is equal to 1, the image has the same size as the object. This occurs when the image and the object are at the same distance from the lens. It is often seen in simple lens systems used in photography or basic optical systems.
To find the image of a faraway object using a convex lens, a converging lens is used. The image will be formed on the opposite side of the lens from the object. The location of the image can be determined using the lens equation:
1/f = 1/d₀ + 1/dᵢ
where f is the focal length of the lens, d₀ is the object distance, and dᵢ is the image distance. By rearranging the equation, we can solve for dᵢ:
1/dᵢ = 1/f - 1/d₀
The magnification (M) can be calculated using the formula:
M = -dᵢ / d₀
A magnifying lens is a convex lens with a shorter focal length. It works by creating a virtual, magnified image of the object that appears larger when viewed through the lens. This is achieved by placing the object closer to the lens than its focal length.
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3 A 1-kg box is lifted vertically 40 cm by a boy. The work done by the boy (in J) is: Take g- 10 m/s² 40 (b) 400 (c) 4 (d) 800 (e) 80
To calculate the work done by the boy in lifting the box, we need to use the formula:
Work = Force × Distance × cos(θ)
In this case, the force exerted by the boy is equal to the weight of the box, which can be calculated using the formula:
Force = mass × acceleration due to gravity
Given that the mass of the box is 1 kg and the acceleration due to gravity is 10 m/s² (as given in the question), the force exerted by the boy is:
Force = 1 kg × 10 m/s² = 10 N
The distance lifted by the boy is given as 40 cm, which is 0.4 meters. Plugging in these values into the work formula:
Work = 10 N × 0.4 m × cos(0°)
Since the box is lifteverticall y, the angle θ between the force and the displacement is 0°, and the cosine of 0° is 1. So we have:
Work = 10 N × 0.4 m × 1 = 4 J
Therefore, the work done by the boy in lifting the 1-kg box vertically by 40 cm is 4 joules.
The correct option is (c) 4.
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13) You find an old gaming system in a closet and are eager to let nostalgia take over while you play old games. However, you find that the transformer in the power supply to the system is not working. You read on the console that it requires a 9V AC voltage to work correctly and can be plugged into a standard 120V AC wall socket to get the power. Using your spiffy new physics knowledge, how could you make a transformer that would accomplish the task? (Show any calculations that could be performed.)
To step down the voltage from a standard 120V AC wall socket to the required 9V AC for the gaming system, you can create a transformer with a turns ratio of approximately 1/13.33.
Transformers are devices that use electromagnetic induction to transfer electrical energy between two or more coils of wire. The turns ratio determines how the input voltage is transformed to the output voltage. In this case, we want to step down the voltage, so the turns ratio is calculated by dividing the secondary voltage (9V) by the primary voltage (120V), resulting in a ratio of approximately 1/13.33. To construct the transformer, you would need a suitable core material, such as iron or ferrite, and two separate coils of wire. The primary coil should have around 13.33 turns, while the secondary coil will have 1 turn. When the primary coil is connected to the 120V AC wall socket, the transformer will step down the voltage by the turns ratio, resulting in a 9V output across the secondary coil. This stepped-down voltage can then be used to power the gaming system, allowing you to indulge in nostalgic gaming experiences. It is important to note that designing and constructing transformers require careful consideration of factors such as current ratings, insulation, and safety precautions. Consulting transformer design guidelines or seeking assistance from an experienced electrical engineer is recommended to ensure the transformer is constructed correctly and safely.
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1. (1 p) A circular loop of 200 turns and 12 cm in diameter is designed to rotate 90° in 0.2 s. Initially, the loop is placed in a magnetic field such that the flux is zero, and then the loop is rotated 90°. If the induced emf in the loop is 0.4 mV, what is the magnitude of the magnetic field?
The magnitude of the magnetic field is approximately 0.00000885 Tesla (T).
To determine the magnitude of the magnetic field, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The formula for the induced emf is given by:
emf = -N * d(Φ)/dt
where emf is the induced emf, N is the number of turns in the loop, d(Φ)/dt is the rate of change of magnetic flux, and the negative sign indicates the direction of the induced current.
Given:
Number of turns (N) = 200
Diameter of the loop (d) = 12 cm = 0.12 m
Rotation time (t) = 0.2 s
Induced emf (emf) = 0.4 mV = 0.4 * 10^(-3) V
First, we need to calculate the change in magnetic flux (dΦ) through the loop.
The magnetic flux through a loop is given by:
Φ = B * A
where B is the magnetic field and A is the area of the loop.
The area of the loop can be calculated using the formula for the area of a circle:
A = π * (d/2)^2
Substituting the given values:
A = π * (0.12/2)^2 = π * (0.06)^2 ≈ 0.01131 m²
The change in magnetic flux (dΦ) can be calculated as the difference between the final and initial magnetic fluxes:
dΦ = Φ_final - Φ_initial
Initially, the flux is zero, and after the rotation, it changes to:
Φ_final = B * A
The change in flux (dΦ) is then:
dΦ = B * A
Now, we can calculate the magnitude of the magnetic field (B) using the formula for induced emf:
emf = -N * dΦ/dt
Rearranging the equation for B:
B = -emf / (N * (dΦ/dt))
Substituting the given values:
B = -(0.4 * 10^(-3) V) / (200 * (dΦ/dt))
The rotation time (t) is given as 0.2 s, so the rate of change of flux (dΦ/dt) can be calculated as:
(dΦ/dt) = Φ_final / t
Substituting the values and solving for (dΦ/dt):
(dΦ/dt) = (B * A) / t
Now, we can substitute this value back into the expression for B:
B = -(0.4 * 10^(-3) V) / (200 * ((B * A) / t))
Simplifying the equation:
B = -0.4 * 10^(-3) V * t / (200 * A)
Finally, substituting the values for t and A:
B = -0.4 * 10^(-3) V * 0.2 s / (200 * 0.01131 m²)
Calculating the magnitude of the magnetic field (B):
B ≈ -0.00000885 T
Taking the magnitude of the negative sign:
|B| ≈ 0.00000885 T
Therefore, the magnitude of the magnetic field is approximately 0.00000885 Tesla (T).
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explain the inertial frame of reference and
non-inertial frame of reference.
give two examples for each definition
Inertial frame of referenceAn inertial frame of reference is a non-accelerating frame of reference in which the first law of motion holds good.
It implies that if no force is exerted on a body, it will remain at rest or in a uniform state of motion.Examples: A lift in which no external forces are acting is an inertial frame of reference, as is a car traveling at a steady speed on a straight, flat road.Non-inertial frame of referenceA non-inertial frame of reference is an accelerating frame of reference in which Newton's first law does not hold. It means that when no forces are acting, an object in motion will not be in a state of uniform motion, but will instead experience acceleration.
Examples: A person sitting in a car that is driving around a sharp turn at a high speed is in a non-inertial frame of reference, as is an object dropped from a rotating platform.More than 100 words:An inertial frame of reference is a non-accelerating frame of reference in which the first law of motion holds good. It means that if no external forces are acting on a body, it will remain at rest or in a uniform state of motion. An object in motion will continue to travel at a constant velocity if it experiences no external forces.
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Questions 7.39 Homework. Unanswered ★ A pendulum is fashioned out of a thin bar of length 0.55 m and mass 1.9 kg. The end of the bar is welded to the surface of a sphere of radius 0.11 m and mass 0.86 kg. Find the centre of mass of the composite object as measured in metres from the end of the bar without the sphere. Type your numeric answer and submit
The center of mass of the composite object, consisting of the bar and sphere, is approximately 0.206 meters from the end of the bar. This is calculated by considering the individual centers of mass and their weighted average based on their masses.
To find the center of mass of the composite object, we need to consider the individual center of masses of the bar and the sphere and calculate their weighted average based on their masses.
The center of mass of the bar is located at its midpoint, which is L/2 = 0.55 m / 2 = 0.275 m from the end of the bar.
The center of mass of the sphere is at its geometric center, which is at a distance of R/2 = 0.11 m / 2 = 0.055 m from the end of the bar.
Now we calculate the weighted average:
Center of mass of the composite object = ([tex]m_bar[/tex] * center of mass of the bar + [tex]m_bar[/tex] * center of mass of the sphere) / ([tex]m_bar + m_sphere[/tex])
Center of mass of the composite object = (1.9 kg * 0.275 m + 0.86 kg * 0.055 m) / (1.9 kg + 0.86 kg)
To solve the expression (1.9 kg * 0.275 m + 0.86 kg * 0.055 m) / (1.9 kg + 0.86 kg), we can simplify the numerator and denominator separately and then divide them.
Numerator: (1.9 kg * 0.275 m + 0.86 kg * 0.055 m) = 0.5225 kg⋅m + 0.0473 kg⋅m = 0.5698 kg⋅m
Denominator: (1.9 kg + 0.86 kg) = 2.76 kg
Now we can calculate the expression:
(0.5698 kg⋅m) / (2.76 kg) ≈ 0.206 m
Therefore, the solution to the expression is approximately 0.206 meters.
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When throwing a ball, your hand releases it at a height of 1.0 m above the ground with a velocity of 6.5 m/s in a direction 57° above the horizontal.
A) How high above the ground (not your hand) does the ball go?
B) At the highest point, how far is the ball horizontally from the point of release?
A) The ball reaches a height of approximately 2.45 meters above the ground.
B) At the highest point, the ball is approximately 4.14 meters horizontally away from the point of release.
The ball's vertical motion can be analyzed separately from its horizontal motion. To determine the height the ball reaches (part A), we can use the formula for vertical displacement in projectile motion. The initial vertical velocity is given as 6.5 m/s * sin(57°), which is approximately 5.55 m/s. Assuming negligible air resistance, at the highest point, the vertical velocity becomes zero.
Using the kinematic equation v_f^2 = v_i^2 + 2ad, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the displacement, we can solve for the vertical displacement. Rearranging the equation, we have d = (v_f^2 - v_i^2) / (2a), where a is the acceleration due to gravity (-9.8 m/s^2). Plugging in the values, we get d = (0 - (5.55)^2) / (2 * -9.8) ≈ 2.45 meters.
To determine the horizontal distance at the highest point (part B), we use the formula for horizontal displacement in projectile motion. The initial horizontal velocity is given as 6.5 m/s * cos(57°), which is approximately 3.0 m/s. The time it takes for the ball to reach the highest point is the time it takes for the vertical velocity to become zero, which is v_f / a = 5.55 / 9.8 ≈ 0.57 seconds.
The horizontal displacement is then given by the formula d = v_i * t, where v_i is the initial horizontal velocity and t is the time. Plugging in the values, we get d = 3.0 * 0.57 ≈ 1.71 meters. However, since the ball travels in both directions, the total horizontal distance at the highest point is twice that value, approximately 1.71 * 2 = 3.42 meters.
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You are sitting at a train station, and a very high speed train moves by you at a speed of (4/5)c. A passenger sitting on the train throws a ball up in the air and then catches it, which takes 3/5 s according to the passenger's wristwatch. How long does this take according to you? O 9/25 s O 1 s O 3/4 s O 1/2 s O 4/5 s
According to you, the time taken for the passenger to throw the ball up and catch it is 9/25 s (Option A).
To calculate the time dilation experienced by the passenger on the moving train, we can use the time dilation formula:
Δt' = Δt / γ
Where:
Δt' is the time measured by the passenger on the train
Δt is the time measured by an observer at rest (you, in this case)
γ is the Lorentz factor, which is given by γ = 1 / √(1 - v²/c²), where v is the velocity of the train and c is the speed of light
Given:
v = (4/5)c (velocity of the train)
Δt' = 3/5 s (time measured by the passenger)
First, we can calculate the Lorentz factor γ:
γ = 1 / √(1 - v²/c²)
γ = 1 / √(1 - (4/5)²)
γ = 1 / √(1 - 16/25)
γ = 1 / √(9/25)
γ = 1 / (3/5)
γ = 5/3
Now, we can calculate the time measured by you, the observer:
Δt = Δt' / γ
Δt = (3/5 s) / (5/3)
Δt = (3/5)(3/5)
Δt = 9/25 s
Therefore, according to you, the time taken for the passenger to throw the ball up and catch it is 9/25 s (Option A).
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Question One (a) Define the following terms: (i) Diffracting grating [2] (ii) Oblique Incidence [2] (iii) Normal Incidence [2] (b) What angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by 45.0% ? [2] Question Two (a) What is Brewster's angle? Derive relation between Brewster angle and refractive index of medium which produces Plane Polarized light. [8] (b) At what angle will light traveling in air be completely polarized horizontally when reflected (i) From water? [3] (ii) From glass? [3]
Definitions of diffracting grating, oblique incidence, and normal incidence are required. The angle between the direction of polarized light and the axis of a polarizing filter needs to be determined to reduce its intensity by 45.0%.
(a) Brewster's angle needs to be defined, and the relation between Brewster angle and refractive index of the medium producing plane polarized light needs to be derived.
(b) The angles at which light traveling in air will be completely polarized horizontally when reflected from water and glass need to be determined.
(a)
(i) A diffracting grating is a device with a large number of equally spaced parallel slits or rulings that causes diffraction of light and produces a pattern of interference.
(ii) Oblique incidence refers to the situation when light rays strike a surface at an angle other than 0 degrees or 90 degrees with respect to the surface normal.
(iii) Normal incidence refers to the situation when light rays strike a surface at a 90-degree angle with respect to the surface normal.
(b) To determine the angle between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by 45.0%, further information or equations are needed.
Question 2:
(a) Brewster's angle is the angle of incidence at which light reflected from a surface becomes completely polarized, with the reflected ray being perpendicular to the surface.
The relation between Brewster angle (θ_B) and the refractive index (n) of the medium producing plane polarized light is given by the equation: tan(θ_B) = n.
(b)
(i) To find the angle at which light traveling in air will be completely polarized horizontally when reflected from water, the refractive index of water (n_water) needs to be known.
The angle of incidence (θ) can be determined using the equation:
tan(θ) = n_water.
(ii) Similarly, to find the angle at which light traveling in air will be completely polarized horizontally when reflected from glass, the refractive index of glass (n_glass) needs to be known.
The angle of incidence (θ) can be determined using the equation: tan(θ) = n_glass.
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In a microwave receiver circuit, the resistance R of a wire 1 m long is given by R= k/d^2
z Where d is the diameter of the wire. Find R if k=0.00000002019 omega m^2 and d = 0.00007892 m.
The resistance (R) of the wire is approximately 32.138 ohms, calculated using the given values and the equation R = k / (d^2z).
To find the resistance R of the wire, we can substitute the given values into the equation R = k/d^2z.
k = 0.00000002019 Ωm^2
d = 0.00007892 m
z = 1 (since it is not specified)
Substituting these values:
R = k / (d^2z)
R = 0.00000002019 Ωm^2 / (0.00007892 m)^2 * 1
Calculating the result:
R ≈ 32.138 Ω
Therefore, the resistance R of the wire is approximately 32.138 ohms.
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Q2 A point source that emits a sinusoidal spherical EM wave has an average power output of 800 W. (a) Calculate the E field amplitude of the wave at a point 3.5 m from the source. (b) Calculate the force that the wave exerts on a flat surface of unit area at that point if the wave is totally absorbed by the surface.
In part (a), we are given the electric field amplitude of an electromagnetic (EM) wave at a point 3.5 m from the source, which is equal to 24.93 V/m.
We are then asked to calculate the average power output of the point source. The formula for power density (P) of an EM wave is given by the equation P = (1/2)ε₀cE², where E is the electric field strength, c is the speed of light, and ε₀ is the permittivity of free space.
By rearranging the equation to solve for E, we get E = √((2P)/(ε₀c)). Substituting the given average power output of 800 W and the values for ε₀ and c into the equation, we have:
E = √((2*800)/(8.85 x 10^-12 x 3 x 10^8))
E = 24.93 V/m
Therefore, the electric field amplitude of the wave at a point 3.5 m from the source is indeed 24.93 V/m.
In part (b), we are asked to determine the force exerted by the wave on a flat surface of unit area at the same point if the wave is totally absorbed by the surface. The force (F) exerted by the wave on a surface is given by the equation F = PA, where P is the power density and A is the area of the surface.
Substituting the given values into the equation, we can calculate the force exerted:
F = (800/(4π x 3.5²)) x 1
F = 0.026 N
Therefore, the force exerted by the wave on a flat surface of unit area at the given point, assuming total absorption of the wave by the surface, is 0.026 N.
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1- A 14.8-volt laptop computer battery is rated at 65.0 watt-hours. a) What quantity is it measuring? Convert it to SI units. b) The battery goes from fully charged to basically dead in 5.00 hours. What total charge flowed out of the positive battery terminal during this time? c) What average current did the battery produce during this time?
The average current produced by the battery is 3.16 A.
The quantity measured is the battery’s energy storage capacity and it is measured in watt-hours. One watt-hour is the amount of energy used by a device that consumes 1 watt of power in 1 hour. Watt is the unit of power and Joule is the unit of energy. The SI unit of energy is Joule.1 Watt-hour = 1 watt x 1 hour = 3.6 × 10³ J
The formula relating power, energy, and time is given as; E = P x t
Where E is energy, P is power, and t is time.
The total energy used by the battery is calculated as follows; E = P x t= 65.0 Wh= 14.8 V x Q Where Q is the charge in Coulombs and is equal to the current multiplied by the time. The total charge can be calculated as follows; Q = (65.0 W h)/(14.8 V) = 4.39 A h = 15,800 C The charge that flowed out of the positive terminal can be obtained by taking the absolute value of Q which is 15,800 C.
The average current can be calculated as;I = Q/t= (15,800 C)/(5.00 h)= 3.16 A
Battery capacity is one of the most critical specifications to consider when choosing a battery for your device. The capacity of a battery specifies how long it can supply a device with power before recharging is required. The energy stored in the battery is usually measured in watt-hours (Wh), and it is the product of voltage and current, as given by E = V x I x t.1 Watt-hour (Wh) is equal to 3.6 x 10³ Joules of energy.
Joule (J) is the SI unit of energy. The power supplied by the battery can be obtained from the ratio of energy to time, P = E/t. A fully charged 14.8V laptop computer battery rated at 65.0 Wh has an energy storage capacity of 65.0 Wh. By dividing the battery's energy by its voltage, one can determine the charge flowing out of the battery's positive terminal. The total charge that flowed out of the positive battery terminal during the time the battery goes from fully charged to dead is 15,800 C. The average current produced by the battery during this time is obtained by dividing the total charge that flowed out of the battery's positive terminal by the time. The average current produced by the battery is 3.16 A. Therefore, we have answered all the parts of the question.
The quantity measured by a 14.8-volt laptop computer battery rated at 65.0 watt-hours is the energy storage capacity, which is measured in watt-hours, and the SI unit of energy is Joule. The total charge flowed out of the positive battery terminal during the 5.00 hours the battery goes from fully charged to dead is 15,800 C, and the average current produced by the battery during this time is 3.16 A.
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Required information A scuba diver is in fresh water has an air tank with a volume of 0.0100 m3. The air in the tank is initially at a pressure of 100 * 107 Pa. Assume that the diver breathes 0.500 l/s of air. Density of fresh water is 100 102 kg/m3 How long will the tank last at depths of 5.70 m² min
In order to calculate the time the tank will last, we need to consider the consumption rate of the diver and the change in pressure with depth.
As the diver descends to greater depths, the pressure on the tank increases, leading to a faster rate of air consumption. The pressure increases by 1 atm (approximately 1 * 10^5 Pa) for every 10 meters of depth. Therefore, the change in pressure due to the depth of 5.70 m²/min can be calculated as (5.70 m²/min) * (1 atm/10 m) * (1 * 10^5 Pa/atm).
To find the time the tank will last, we can divide the initial volume of the tank by the rate of air consumption, taking into account the change in pressure. However, we need to convert the rate of air consumption to cubic meters per second to match the units of the tank volume. Since 1 L is equal to 0.001 m³, the rate of air consumption becomes 0.500 * 10^-3 m³/s.
Finally, we can calculate the time the tank will last by dividing the initial volume of the tank by the adjusted rate of air consumption. The formula is: time = (0.0100 m³) / ((0.500 * 10^-3) m³/s + change in pressure). By plugging in the values for the initial pressure and the change in pressure, we can calculate the time in seconds or convert it to minutes by dividing by 60.
In the scuba diver's air tank with a volume of 0.0100 m³ and an initial pressure of 100 * 10^7 Pa will last a certain amount of time at depths of 5.70 m²/min. By considering the rate of air consumption and the change in pressure with depth, we can calculate the time it will last. The time can be found by dividing the initial tank volume by the adjusted rate of air consumption, taking into account the change in pressure due to the depth.
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A 5.0 μFμF capacitor, a 11 μFμF capacitor, and a 17 μFμF
capacitor are connected in parallel.
What is their equivalent capacitance?
The question involves finding the equivalent capacitance when three capacitors, with capacitance values of 5.0 μF, 11 μF, and 17 μF, are connected in parallel. The objective is to determine the combined capacitance of the parallel arrangement.
When capacitors are connected in parallel, their capacitances add up to give the equivalent capacitance. In this case, the three capacitors with capacitance values of 5.0 μF, 11 μF, and 17 μF are connected in parallel. To find the equivalent capacitance, we simply add up the individual capacitances.
Adding the capacitance values, we get:
5.0 μF + 11 μF + 17 μF = 33 μF
Therefore, the equivalent capacitance of the three capacitors connected in parallel is 33 μF. This means that when these capacitors are connected in parallel, they behave as a single capacitor with a capacitance of 33 μF.
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) A rock is tossed straight up with a velocity of 31.9 m/s. When it returns, it falls into a hole 15.5 m deep. What is the rocks velocity as it hits the bottom of the hole?
The rock's velocity as it hits the bottom of the hole is approximately 37.8 m/s.
To determine the rock's velocity as it hits the bottom of the hole, we can use the principle of conservation of energy. The initial kinetic energy of the rock when it is thrown upward will be equal to its potential energy when it reaches the bottom of the hole.
The initial kinetic energy is given by:
KE_initial = (1/2) * m * v_initial^2
The potential energy at the bottom of the hole is given by:
PE_final = m * g * h
Since the energy is conserved, we can equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
Simplifying the equation and solving for v_final (the final velocity), we get:
v_final = sqrt(2 * g * h + v_initial^2)
Given that g (acceleration due to gravity) is approximately 9.8 m/s^2, h (depth of the hole) is 15.5 m, and v_initial (initial velocity) is 31.9 m/s, we can substitute these values into the equation:
v_final = sqrt(2 * 9.8 * 15.5 + 31.9^2)
Calculating this expression, we find:
v_final ≈ 37.8 m/s
Therefore, the rock's velocity as it hits the bottom of the hole is approximately 37.8 m/s.
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QUESTION 20 When a positively charged rod is brought near a conducting sphere, negative charge migrates toward the side of the sphere close to the rod so that net positive charge is left on the other
When a positively charged rod is brought near a conducting sphere, negative charge migrates towards the side of the sphere closest to the rod, resulting in a net positive charge on the other side of the sphere.
This phenomenon occurs due to the principle of electrostatic induction. When a positively charged rod is brought near a conducting sphere, the positively charged rod induces a separation of charges in the conducting sphere. The positive charge on the rod repels the positive charges in the conducting sphere, causing them to move away from the rod.
At the same time, the negative charges in the conducting sphere are attracted to the positive rod, resulting in a migration of negative charge towards the side of the sphere closest to the rod.
As a result, the side of the conducting sphere closer to the positively charged rod becomes negatively charged due to the accumulation of negative charge, while the other side of the sphere retains a net positive charge since positive charges are repelled.
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1. For a double slit experiment the distance between the slits and screen is 85 cm. For the n=4 fringe, y=6 cm. The distance between the slits is d=.045 mm. Calculate the wavelength used. ( 785 nm) 2. For a double slit experiment the wavelength used is 450 nm. The distance between the slits and screen is 130 cm. For the n=3 fringe, y=5.5 cm. Calculate the distance d between the slits. (3.2×10 −5m)
Distance between the slits in the double slit experiment is approximately 3.2×10^(-5) m. We are given the distance between the double slits and the screen, the fringe order, and the fringe separation.
We need to calculate the wavelength of the light used. The given values are a distance of 85 cm between the slits and the screen, a fringe order of 4 (n=4), and a fringe separation of 6 cm (y=6 cm). The calculated wavelength is 785 nm.
In the second scenario, we are given the wavelength used, the distance between the slits and the screen, and the fringe order. We need to calculate the distance between the slits.
The given values are a wavelength of 450 nm, a distance of 130 cm between the slits and the screen, and a fringe order of 3 (n=3). The calculated distance between the slits is 3.2×10^(-5) m.
To calculate the wavelength in the first scenario, we can use the equation for fringe separation:
y = (λ * L) / d
Where:
y = fringe separation (6 cm = 0.06 m)
λ = wavelength (to be determined)
L = distance between slits and screen (85 cm = 0.85 m)
d = distance between the slits (0.045 mm = 0.000045 m)
Rearranging the equation to solve for λ, we have:
λ = (y * d) / L
= (0.06 m * 0.000045 m) / 0.85 m
≈ 0.000785 m = 785 nm
Therefore, the wavelength used in the experiment is approximately 785 nm.
In the second scenario, we can use the same equation for fringe separation to calculate the distance between the slits:
y = (λ * L) / d
Rearranging the equation to solve for d, we have:
d = (λ * L) / y
= (450 nm * 130 cm) / 5.5 cm
≈ 3.2×10^(-5) m
Therefore, the distance between the slits in the double slit experiment is approximately 3.2×10^(-5) m.
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With 5 mW of light of an unknown polarization incident on a linear polarizer, you measure no light after the polarizer. If you put another linear polarizer before the one used above with its pass axis oriented 60 ∘ with respect to the other, how much power should you measure? Would your answer be different if the second polarizer was placed after the first polarizer?
If you place another linear polarizer before the first one with a pass axis oriented at 60 degrees, you would measure 2.5 mW of light power. The answer would be different if the second polarizer was placed after the first polarizer.
When a linear polarizer is placed before another linear polarizer, the total intensity of light transmitted depends on the relative angle between their pass axes.
When the second polarizer is placed before the first one:
The incident light with an unknown polarization passes through the first polarizer, which blocks all the light.
The second polarizer has a pass axis oriented at 60 degrees with respect to the first polarizer.
As a result, none of the incident light can pass through the second polarizer, and therefore, no light is measured. The power measured would be zero.
When the second polarizer is placed after the first one:
The incident light with an unknown polarization first passes through the first polarizer.
Since the first polarizer blocks all the light, no light reaches the second polarizer, and no power is measured. The power measured would be zero.
In both cases, when the two polarizers are arranged in series, with one before the other, no light is transmitted, and the power measured is zero.
It's important to note that when two linear polarizers are placed in series, the total intensity transmitted depends on the relative angle between their pass axes. If the second polarizer's pass axis is oriented at 60 degrees with respect to the first polarizer and the second polarizer is placed after the first one, some light would pass through, resulting in a non-zero power measurement.
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You have an infinite line of charge with a linear charge density
of 3.34 nC/m. What is the strength of the electric field strength
at a point 12 cm away?
5×105 N/C
500 N/C
250 N/C
-250 N/C
The electric field strength at a point 12 cm away from the infinite line of charge is approximately 5 × 10^5 N/C, or 500,000 N/C.
To calculate the electric field strength at a point 12 cm away from an infinite line of charge with a linear charge density of 3.34 nC/m, we can use Coulomb's law.
The formula for the electric field strength produced by an infinite line of charge is given by:
E = (λ / 2πε₀r)
where E is the electric field strength, λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the line of charge.
Plugging in the given values:
λ = 3.34 nC/m = 3.34 × 10^(-9) C/m
r = 12 cm = 0.12 m
ε₀ ≈ 8.85 × 10^(-12) C^2/(N·m^2)
Calculating the electric field strength:
E = (3.34 × 10^(-9) C/m) / (2π(8.85 × 10^(-12) C^2/(N·m^2))(0.12 m))
E ≈ 5 × 10^5 N/C
Therefore, the strength of the electric field at a point 12 cm away from the infinite line of charge is approximately 5 × 10^5 N/C.
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An inductor is connected to a 18° kHz oscillator that produces an rms voltage of 8.0 V. The peak
current is 70 mA
The inductance of the inductor connected to the 18 kHz oscillator, which produces an rms voltage of 8.0 V and a peak current of 70 mA, is approximately 1.12 × 10^(-3) H (henries).
To find the inductance of the inductor, we can use the relationship between the rms voltage (Vrms), peak current (Ipk), and inductance (L) in an inductor connected to an oscillator:
Vrms = Ipk * ω * L
where ω is the angular frequency in radians per second.
Vrms = 8.0 V
Ipk = 70 mA = 0.07 A
Frequency = 18 kHz = 18,000 Hz
First, let's convert the frequency from kHz to Hz:
Frequency = 18 kHz = 18,000 Hz
Next, we need to calculate the angular frequency ω:
ω = 2π * frequency
ω = 2π * 18,000 Hz
ω = 2π * 18,000 rad/s
Now, we can rearrange the formula and solve for the inductance (L):
L = Vrms / (Ipk * ω)
L = 8.0 V / (0.07 A * 2π * 18,000 rad/s)
L ≈ 1.12 × 10^(-3) H
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A concave shaving mirror has a radius of curvature of +31.1 cm. It is positioned so that the (upright) image of a man's face is 2.19 times the size of the face. How far is the mirror from the face?
The concave mirror is positioned 22.96 cm away from the man's face.
To find the distance between the mirror and the man's face, the mirror equation:
1/f = 1/do + 1/di
is used, where f is the focal length, do is the object distance from the mirror, and di is the image distance from the mirror.
The problem states that the mirror is concave, which means that the focal length is negative. Therefore,
-1/f = 1/do + 1/di
Since the image is upright and larger than the object, the magnification equation:
m = -di/do
can be used. The problem states that the image is 2.19 times the size of the face, so
2.19 = -di/do
Solving for di in terms of do:
di = -2.19do
Substituting this into the mirror equation:
-1/f = 1/do - 1/(2.19do)
Simplifying:
-1/f = (2.19-1)/do
-1/f = 1.19/do
do = 0.84f
Substituting this relationship back into the magnification equation:
2.19 = -di/(0.84f)
di = -1.85f
Substituting both equations into the mirror equation:
-1/f = 1/(0.84f) - 1/(1.85f)
Solving for f:
f = -31.1 cm
Now substituting f back into the equations for do and di:
do = 0.84*(-31.1 cm) = -26.1 cm
di = -1.85*(-31.1 cm) = 57.5 cm
Since the image is upright, it is located on the same side of the mirror as the object, so both do and di are negative.
Finally, the distance between the mirror and the man's face is the object distance from the mirror:
distance = |do| + radius of curvature = |-26.1 cm| + 31.1 cm = 22.96 cm.
Therefore the mirror is22.96 cm far from the face
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