To find the change in gravitational force, we need to calculate the gravitational force at sea level and the gravitational force at an altitude of 1700 m, and then find the difference between the two forces.
Calculation:
Let's denote the gravitational force as F(r), where r is the distance from the center of the Earth.
Calculate the gravitational force at sea level:
F_sea = G * (M_E * m) / (R_E)^2
Calculate the gravitational force at the airplane altitude:
F_airplane = G * (M_E * m) / (R_E + h)^2
Calculate the change in gravitational force:
ΔF = F_airplane - F_sea
Given:
F_sea_level = 770 N
M = 5.972 x 10^24 kg
r_sea_level = Re (Earth's mean radius) = 6.37 x 10^6 m
Now, let's calculate the gravitational force at an altitude of 1700 m above sea level:
r_altitude = r_sea_level + 1700 m
To find the change in gravitational force, we subtract the force at the altitude from the force at sea level:
ΔF = F_sea_level - F_altitude
Let's calculate step by step:
F_sea_level = (G * M * m) / r_sea_level^2
770 N = (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * m) / (6.37 x 10^6 m)^2
Solving the equation above for m (mass of the person), we find:
m = (770 N * (6.37 x 10^6 m)^2) / (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg)
m ≈ 61.14 kg
Now, let's calculate the gravitational force at the altitude:
F_altitude = (G * M * m) / r_altitude^2
F_altitude = (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * 61.14 kg) / (r_sea_level + 1700 m)^2
ΔF = F_sea_level - F_altitude
Finally, let's plug in the values and calculate:
ΔF = 770 N - [(6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * 61.14 kg) / (6.37 x 10^6 m + 1700 m)^2]ΔF ≈ -9.86 N
The change in gravitational force for someone who weighs 770 N at sea level compared to when on an airplane 1700 m above sea level is approximately -9.86 N (decrease in force).
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3. What would happen if you put an object at the focal point of the lens? 4. What would happen if you put an object at the focal point of the mirror? 5. What would happen if you put an object between the focal point and the lens? 6. What would happen if you put an object between the focal point and the mirror?
The specific placement of an object relative to the focal point of a lens or mirror determines the characteristics of the resulting image, such as its nature (real or virtual), size, and orientation.
Let's provide a more detailed explanation for each scenario:
3. Placing an object at the focal point of a lens:
When an object is placed exactly at the focal point of a lens, the incident rays from the object become parallel to each other after passing through the lens. This occurs because the lens refracts (bends) the incoming rays in such a way that they converge at the focal point on the opposite side. However, when the object is positioned precisely at the focal point, the refracted rays become parallel and do not converge to form a real image. Therefore, in this case, no real image is formed on the other side of the lens.
4. Placing an object at the focal point of a mirror:
If an object is positioned at the focal point of a mirror, the reflected rays will appear to be parallel to each other. This happens because the light rays striking the mirror surface are reflected in a way that they diverge as if they were coming from the focal point behind the mirror. Due to this divergence, the rays never converge to form a real image. Instead, the reflected rays appear to originate from a virtual image located at infinity. Consequently, no real image can be projected onto a screen or surface.
5. Placing an object between the focal point and the lens:
When an object is situated between the focal point and a converging lens, a virtual image is formed on the same side as the object. The image appears magnified and upright. The lens refracts the incoming rays in such a way that they diverge after passing through the lens. The diverging rays extend backward to intersect at a point where the virtual image is formed. This image is virtual because the rays do not actually converge at that point. The virtual image is larger in size than the object, making it appear magnified.
6. Placing an object between the focal point and the mirror:
Similarly, when an object is placed between the focal point and a concave mirror, a virtual image is formed on the same side as the object. The virtual image is magnified and upright. The mirror reflects the incoming rays in such a way that they diverge after reflection. The diverging rays appear to originate from a point behind the mirror, where the virtual image is formed. Again, the virtual image is larger than the object and is not a real convergence point of light rays.
In summary, the placement of an object relative to the focal point of a lens or mirror determines the behavior of the light rays and the characteristics of the resulting image. These characteristics include the nature of the image (real or virtual), its size, and its orientation (upright or inverted).
Note: In both cases (5 and 6), the images formed are virtual because the light rays do not actually converge or intersect at a point.
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Pick the correct statement. You can't put a virtual image on a screen. You can't take a picture of a virtual image. Mirrors reflect light, therefore they always make real images. You can't see a virtual image with unaided eyes. Real images are always upright.
The correct statement is, You can't put a virtual image on a screen.
A virtual image is formed when the light rays appear to diverge from a point behind the mirror or lens. Virtual images cannot be projected onto a screen because they do not actually exist at a physical location. They are perceived by the observer as if the light rays are coming from a certain point, but they do not converge to form a real image.
In contrast, real images are formed when the light rays converge to a point, and they can be projected onto a screen. Real images can be captured by a camera or observed directly with the eyes because they are formed by the actual intersection of light rays.
So, the correct statement is that you can't put a virtual image on a screen because virtual images do not have a physical existence at a specific location.
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972 Two bodies of masses ma and my undergo a perfectly elastic collision that is central (head-on). Both are moving in opposite directions along the same straight line before collision with velocities vai and VBI. (Call all v's +) (a) Find the velocity of each body after the collision, in terms of the masses and the velocities given. (b) For the special case in which B is at rest before collision, find the ratio kinetic energy of_B_after_collision K= , in terms of (m/m). kinetic_energy_of_A_before_collision (c) Letr stand for the ratio (m/m). Find the value of that's makes K(r) a maximum. What does me have to be in terms of mx) for the maximum transfer of kinetic energy in the collision? (Would you have guessed this without working it out?). Notice why much more energy is transferred when an electron collides with another electron than when an electron collides with an atom ("Interacts" would be a little more accurate than "collides.") Can you see what a graph of K(T) vs. r looks like?
(a) The velocity of each body after the collision can be calculated using the conservation of momentum and kinetic energy.
ma * vai + mb * vbi = ma * vaf + mb * vbf
(1/2) * ma * (vai)^2 + (1/2) * mb * (vbi)^2 = (1/2) * ma * (vaf)^2 + (1/2) * mb * (vbf)^2
(b) For the special case where B is at rest before the collision (vbi = 0), we can simplify the expressions:
vaf = vai * (mb / (ma + mb))
vbf = vai * (ma / (ma + mb))
K = (1/2) * mb * (vbf)^2 / ((1/2) * ma * (vai)^2)
K = (mb^2 / (ma + mb)^2) * (ma / ma)
K = mb^2 / (ma + mb)^2
(c) To find the value of r that maximizes K, we need to differentiate K with respect to r and set it to zero:
dK/dr = 0
K = mb^2 / (ma + mb)^2 with respect to r:
dK/dr = -2 * mb^2 / (ma + mb)^3 + 2 * mb^2 * ma / (ma + mb)^4
dK/dr to zero and solving for r:
-2 * mb^2 / (ma + mb)^3 + 2 * mb^2 * ma / (ma + mb)^4 = 0
Therefore, for the maximum transfer of kinetic energy in the collision, the mass of A (me) needs to be equal to the mass of B (mx).
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A 380 kg piano is pushed at constant speed a distance of 3.9 m up a 27° incline by a mover who is pushing parallel to the incline. The coefficient of friction between the piano & ramp is 0.45. (a) De
The force exerted by the mover must balance the forces of gravity and friction.
The work done by the mover would be the force exerted by the mover multiplied by the distance the piano is pushed up the incline.
The piano is being pushed at a constant speed and there is no change in vertical position, the work done by the force of gravity is zero.
(a) To determine the force exerted by the mover, we need to consider the forces acting on the piano. These forces include the force of gravity, the normal force, the force exerted by the mover, and the frictional force. By analyzing the forces, we can find the force exerted by the mover parallel to the incline.
The force exerted by the mover must balance the forces of gravity and friction, as well as provide the necessary force to push the piano up the incline at a constant speed.
(b) The work done by the mover is calculated using the formula
W = F * d, where
W is the work done,
F is the force exerted by the mover
d is the distance moved.
In this case, the work done by the mover would be the force exerted by the mover multiplied by the distance the piano is pushed up the incline.
(c) The work done by the force of gravity can be calculated as the product of the force of gravity and the distance moved vertically. Since the piano is being pushed at a constant speed and there is no change in vertical position, the work done by the force of gravity is zero.
By considering the forces, work formulas, and the given values, we can determine the force exerted by the mover, the work done by the mover, and the work done by the force of gravity in pushing the piano up the incline.
Complete Question-
A 380 kg piano is pushed at constant speed a distance of 3.9 m up a 27° incline by a mover who is pushing parallel to the incline. The coefficient of friction between the piano & ramp is 0.45. (a) Determine the force exerted by the man (include an FBD for the piano): (b) Determine the work done by the man: (c) Determine the work done by the force of gravity
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At what temperature will both the reading in Celsius and Fahrenheit read the same value? Show your calculations.
The temperature at which both the Celsius and Fahrenheit scales read the same value is -40 °C/°F.
The Celsius temperature scale is used by most of the world, while the Fahrenheit scale is used primarily in the United States. The formula to convert Fahrenheit to Celsius is C = (5/9)(F - 32), and the formula to convert Celsius to Fahrenheit is F = (9/5)C + 32.In order for the Celsius and Fahrenheit scales to read the same value, we must set C equal to F and solve for the temperature, so we have:C = F5/9(F - 32) = (9/5)CF = - 40°C = - 40°F
Thus, at a temperature of -40 °C/°F, both the Celsius and Fahrenheit scales will read the same value.Calculations:As per the formula,F = (9/5)C + 32Putting C = F, we get;C = (9/5)C + 32C - (9/5)C = 32-4/5C = 32C = - 40Therefore, both the Celsius and Fahrenheit scales read the same value at -40 °C/°F.
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5.0-C charge experiences a 0.58-N force in the positive y rection Part A If this charge is replaced with a -2.7μC charge, what is the magnitude of the force will it experience? Express your answer u
If the charge is replaced , it will experience a force in the negative y-direction. The magnitude of the force can be calculated using Coulomb's Law.
Coulomb's Law states that the force between two charges is given by the equation:
F = k * |q1 * q2| / r^2where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
Given:
q1 = 0 C (initial charge)
F1 = 0.58 N (force experienced by the initial charge)
To find the magnitude of the force when the charge is replaced with -2.7 μC, we can use the ratio of the charges to calculate the new force:F2 = (q2 / q1) * F1
Converting -2.7 μC to coulombs:
q2 = -2.7 μC * (10^-6 C/1 μC)
q2 = -2.7 * 10^-6 C
Substituting the values into the equation:
F2 = (-2.7 * 10^-6 C / 0 C) * 0.58 N
Calculating the magnitude of the force:
F2 ≈ -1.566 * 10^-6 N
Therefore, if the charge is replaced with a -2.7 μC charge, it will experience a force of approximately 1.566 * 10^-6 N in the negative y-direction.
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An LC circuit consists of a 2.5 mH inductor and a 4.5 μF
capacitor. its impedance Z at 55 Hz in Ω.Find its impedance
Z at 5 kHz in Ω.
The impedance of the LC circuit at 55 Hz is approximately 269.68 Ω and at 5 kHz is approximately 4.43 Ω.
To find the impedance (Z) of the LC circuit at 55 Hz and 5 kHz, we can use the formula for the impedance of an LC circuit:
Z = √((R^2 + (ωL - 1/(ωC))^2))
Given:
L = 2.5 mH = 2.5 × 10^(-3) H
C = 4.5 μF = 4.5 × 10^(-6) F
1. For 55 Hz:
ω = 2πf = 2π × 55 = 110π rad/s
Z = √((0 + (110π × 2.5 × 10^(-3) - 1/(110π × 4.5 × 10^(-6)))^2))
≈ √((110π × 2.5 × 10^(-3))^2 + (1/(110π × 4.5 × 10^(-6)))^2)
≈ √(0.3025 + 72708.49)
≈ √72708.79
≈ 269.68 Ω (approximately)
2. For 5 kHz:
ω = 2πf = 2π × 5000 = 10000π rad/s
Z = √((0 + (10000π × 2.5 × 10^(-3) - 1/(10000π × 4.5 × 10^(-6)))^2))
≈ √((10000π × 2.5 × 10^(-3))^2 + (1/(10000π × 4.5 × 10^(-6)))^2)
≈ √(19.635 + 0.00001234568)
≈ √19.63501234568
≈ 4.43 Ω (approximately)
Therefore, the impedance of the LC circuit at 55 Hz is approximately 269.68 Ω and at 5 kHz is approximately 4.43 Ω.
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Battery 2 Resistor A Added wire M Resistor B Battery 1 -) () Starting with the original circuit from part (a) above, how can a wire be ac cause a short circuit? Give your answer by drawing a diagram of the circuit with th ded wire in your solutions. Explain why this additional wire shorts the circuit.
To cause a short circuit in the original circuit, an additional wire can be connected between the two ends of Resistor B. This wire creates a direct path for the current to flow, bypassing the resistance of Resistor B.
By connecting an additional wire between the two ends of Resistor B in the circuit, we create a short circuit. In this configuration, the current will follow the path of least resistance, which is the wire with negligible resistance.
Since the wire provides a direct connection between the positive and negative terminals of the battery, it bypasses Resistor B, effectively shorting it. As a result, the current will flow through the wire instead of going through Resistor B, causing a significant increase in the current flow and potentially damaging the circuit or components.
The short circuit occurs because the added wire creates a low-resistance path that diverts the current away from its intended path through Resistor B.
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Question 16 An element, X has an atomic number 45 and a atomic mass of 133.559 u. This element is unstable and decays by decay, with a half life of 68d. The beta particle is emitted with a kinetic energy of 11.71 MeV. Initially there are 9.41×10¹² atoms present in a sample. Determine the activity of the sample after 107 days (in μCi). 1 pts
The activity of the sample after 107 days is 0.2777 μCi.
Atomic number of an element, X = 45
Atomic mass of an element, X = 133.559 u
Half-life = 68 d
Initial number of atoms in the sample = 9.41 x 10¹²
Beta particle emitted with kinetic energy = 11.71 MeV
To determine the activity of the sample after 107 days (in μCi), we use the formula given below:
Activity = λN
Where,
λ is the decay constant
N is the number of radioactive nuclei.
We know that the decay constant (λ) of an element is related to the half-life (t1/2) of an element as follows:
λ = 0.693/t1/2
Hence, the decay constant (λ) of the element can be calculated as follows:
λ = 0.693/68 = 0.01019 per day
Thus, the activity of the sample can be calculated using the formula as shown below:
Activity = λN = (0.01019 per day) x (9.41 x 10¹² atoms) = 9.604 x 10¹⁰ decays per day
Now, the activity is calculated for one day. To find the activity for 107 days, we multiply it by 107.
Activity after 107 days = 9.604 x 10¹⁰ decays/day x 107 days = 1.0275 x 10¹³ decays
Thus, the activity of the sample after 107 days is 1.0275 x 10¹³ decays.
The activity is measured in Becquerel (Bq) and microcurie (μCi) units.
1 Bq = 27 nCi (nano Curies)
1 μCi = 37 MBq
Hence, the activity of the sample after 107 days (in μCi) is calculated as shown below:
Activity in μCi = 1.0275 x 10¹³ decays x (1 Bq/decays) x (27 nCi/1 Bq) x (1 μCi/10⁶ nCi) = 0.2777 μCi
Therefore, the activity of the sample after 107 days is 0.2777 μCi (rounded to four significant figures).
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Victor is a Civil Engineer and goes to rural cities throughout California to provide environmentally sustainable ways of supplying water. In one community he builds a water tower consisting of a 15 m tall tub of water that is elevated 20 m off the ground, with a pipe tube that descends to ground level to provide water to the community. How fast will water flow out of the tube of Victor's water tower?
[the density of water is 1,000 kg/m^3]
Group of answer choices
A. 26.2 m/s
B. 21.7 m/s
C. 13.5 m/s
D. 8.9 m/s
The water will flow out of the tube at a speed of 8.9 m/s.
To determine the speed at which water will flow out of the tube, we can apply the principles of fluid dynamics. The speed of fluid flow is determined by the height of the fluid above the point of discharge, and it is independent of the shape of the container. In this case, the water tower has a height of 15 m, which provides the potential energy for the flow of water.
The potential energy of the water can be calculated using the formula: Potential Energy = mass × gravity × height. Since the density of water is given as 1,000 kg/m³ and the height is 15 m, we can calculate the mass of the water in the tower as follows: mass = density × volume. The volume of the water in the tower is equal to the cross-sectional area of the tub multiplied by the height of the water column.
The cross-sectional area of the tub can be calculated using the formula: area = π × radius². Assuming the tub has a uniform circular cross-section, we need to determine the radius. The radius can be calculated as the square root of the ratio of the cross-sectional area to π. With the given information, we can find the radius and subsequently calculate the mass of the water in the tower.
Once we have the mass of the water, we can use the formula for potential energy to calculate the potential energy of the water. The potential energy is given by the equation: Potential Energy = mass × gravity × height. The potential energy is then converted to kinetic energy as the water flows out of the tube. The kinetic energy is given by the equation: Kinetic Energy = (1/2) × mass × velocity².
By equating the potential energy to the kinetic energy, we can solve for the velocity. Rearranging the equation, we get: velocity = √(2 × gravity × height). Plugging in the values of gravity (9.8 m/s²) and height (20 m), we can calculate the velocity to be approximately 8.9 m/s.
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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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The position of a 51 gg oscillating mass is given by
x(t)=(1.5cm)cos11t, where t is in seconds. Determine the
amplitude.
The given position equation is x(t) = (1.5 cm)cos(11t). In this equation, the coefficient of the cosine function represents the amplitude of the oscillation.
To determine the amplitude of the oscillating mass, we can observe that the equation for position, x(t), is given by:
x(t) = (1.5 cm) * cos(11t)
The amplitude of an oscillating mass is the maximum displacement from the equilibrium position. In this case, the maximum displacement is the maximum value of the cosine function.
The maximum value of the cosine function is 1, so the amplitude of the oscillating mass is equal to the coefficient in front of the cosine function, which is 1.5 cm.
Therefore, the amplitude of the oscillating mass is 1.5 cm.
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In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 100Hz. Find 1.The amplitude 2.The maximum blade speed 3. The magnitude of the maximum blade acceleration
The amplitude of the blade's simple harmonic motion is 1.0 mm (0.001 m). The maximum blade speed is approximately 0.628 m/s. The magnitude of the maximum blade acceleration is approximately 1256.64 m/s².
The amplitude, maximum blade speed, and magnitude of maximum blade acceleration in the electric shaver:
1. Amplitude (A): The amplitude of simple harmonic motion is equal to half of the total distance covered by the blade. In this case, the blade moves back and forth over a distance of 2.0 mm, so the amplitude is 1.0 mm (or 0.001 m).
2. Maximum blade speed (V_max): The maximum blade speed occurs at the equilibrium position, where the displacement is zero. The maximum speed is given by the product of the amplitude and the angular frequency (ω).
V_max = A * ω
The angular frequency (ω) can be calculated using the formula ω = 2πf, where f is the frequency. In this case, the frequency is 100 Hz.
ω = 2π * 100 rad/s = 200π rad/s
V_max = (0.001 m) * (200π rad/s) ≈ 0.628 m/s
3. Magnitude of maximum blade acceleration (a_max): The maximum acceleration occurs at the extreme positions of the motion, where the displacement is maximum. The magnitude of maximum acceleration is given by the product of the square of the angular frequency (ω^2) and the amplitude (A).
a_max = ω² * A
a_max = (200π rad/s)² * 0.001 m ≈ 1256.64 m/s²
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A car moving at 38 km/h negotiates a 160 m -radius banked turn
designed for 60 km/h. What coefficient of friction is needed to
keep the car on the road?
we need to find the value of What coefficient of friction is needed to keep the car on the road. The concepts we can use are centripetal force, gravity etc.
Given data:
The speed of the car v = 38 km/h
Radius of the turn r = 160 m
The turn is designed for the speed of the car v' = 60 km/h
The coefficient of friction between the tires and the road = μ
First, we convert the speed of the car into m/s.1 km/h = 0.27778 m/s
Therefore, 38 km/h = 38 × 0.27778 m/s = 10.56 m/s
Similarly, we convert the speed designed for the turn into m/s
60 km/h = 60 × 0.27778 m/s
60 km/h = 16.67 m/s
To keep the car on the road, the required centripetal force must be provided by the frictional force acting on the car. The maximum frictional force is given by μN, where N is the normal force acting on the car. To find N, we use the weight of the car, which is given by mg where m is the mass of the car and g is the acceleration due to gravity, which is 9.81 m/s². We assume that the car is traveling on a level road. So, N = mg. We can find the mass of the car from the centripetal force equation. The centripetal force acting on the car is given by F = mv²/r where m is the mass of the car, v is the velocity of the car, and r is the radius of the turn. We know that the required centripetal force is equal to the maximum frictional force that can be provided by the tires. Therefore,
F = μN
F = μmg
So,
mv²/r = μmg
m = μgr/v²
Now we can substitute the values in the above formula to calculate the required coefficient of friction.
μ = mv²/(gr)
μ = v²/(gr) × m = (10.56)²/(160 × 9.81)
μ = 0.205
So, the required coefficient of friction to keep the car on the road is μ = 0.205.
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Louis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ to every particle possessing some momentum p by the relationship λ=ph, where h is Planck's constant (h=6.626×10−34 J⋅S). To help you develop some number sense for what this relationship means, try below calculations. You may find these two constants useful: Planck's constant h=6.626×10−34 J⋅s and electron mass 9.109×10−31 kg. a. The de Broglie wavelength of an electron moving at speed 4870 m/s is nm. (This speed corresponds to thermal speed of an electron that has been cooled down to about 1 kelvin.) b. The de Broglie wavelength of an electron moving at speed 610000 m/s is nm. (This speed corresponds to the speed of an electron with kinetic energy of about 1eV.) c. The de Broglie wavelength of an electron moving at speed 17000000 m/s is nm. (At speeds higher than this, we will need to start accounting for effects of specialurelativity to avoid significant (greater than a few percents) errors in calculation.) Question Help: buis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ every particle possessing some momentum p by the relationship λ=ph, where h Planck's constant (h=6.626×1034 J⋅s). This applies not only to subatomic articles like electrons, but every particle and object that has a momentum. To help ou develop some number sense for de Broglie wavelengths of common, everyday bjects, try below calculations. Use Planck's constant h=6.626×10−34 J⋅s; other necessary constants will be given below. To enter answers in scientific notation below, use the exponential notation. For example, 3.14×10−14 would be entered as "3.14E-14". a. Air molecules (mostly oxygen and nitrogen) move at speeds of about 270 m/s. If mass of air molecules are about 5×10−26 kg, their de Broglie wavelength is m. b. Consider a baseball thrown at speed 50 m/s. If mass of the baseball is 0.14 kg, its de Broglie wavelength is c. The Earth orbits the Sun at a speed of 29800 m/s. Given that the mass of the Earth is about 6.0×1024 kg, its de Broglie wavelength is Yes, many of these numbers are absurdly small, which is why I think you should enter the powers of 10. Question Help: □ Message instructor
a. The de Broglie wavelength of an electron moving at a speed of 4870 m/s is approximately 2.72 nanometers (2.72 nm).
b. The de Broglie wavelength of an electron moving at a speed of 610,000 m/s is approximately 0.022 nanometers (0.022 nm).
c. The de Broglie wavelength of an electron moving at a speed of 17,000,000 m/s is approximately 0.00077 nanometers (0.00077 nm).
To calculate the de Broglie wavelength using Louis de Broglie's hypothesis, we can use the formula λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.
a. For an electron moving at a speed of 4870 m/s:
Given:
Speed of the electron (v) = 4870 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Given:
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (4870 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (4870 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 2.72 × 10^−9 m ≈ 2.72 nm
b. For an electron moving at a speed of 610,000 m/s:
Given:
Speed of the electron (v) = 610,000 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Given:
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (610,000 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (610,000 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 2.2 × 10^−11 m ≈ 0.022 nm
c. For an electron moving at a speed of 17,000,000 m/s:
Given:
Speed of the electron (v) = 17,000,000 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (17,000,000 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (17,000,000 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 7.7 × 10^−13 m ≈ 0.00077 nm
The de Broglie wavelength of an electron moving at
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The ideal gas in a container is under a pressure of 17.0 atm at a temperature of 25.0°C. If half of the gas is released from the container and the temperature is increased by 42.0°C, what is the final pressure of the gas?
The final pressure of the gas is 22.5 atm.
To solve this problem, we can use the combined gas law, which relates the initial and final states of a gas sample.
The combined gas law is given by:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes (assuming the volume remains constant in this case), and T1 and T2 are the initial and final temperatures.
Given:
P1 = 17.0 atm (initial pressure)
T1 = 25.0°C (initial temperature)
ΔT = 42.0°C (change in temperature)
P2 = ? (final pressure)
First, let's convert the temperatures to Kelvin:
T1 = 25.0°C + 273.15 = 298.15 K
ΔT = 42.0°C = 42.0 K
Next, we can rearrange the combined gas law equation to solve for P2:
P2 = (P1 * V1 * T2) / (V2 * T1)
Since the volume remains constant, V1 = V2, and we can simplify the equation to:
P2 = (P1 * T2) / T1
Substituting the given values, we have:
P2 = (17.0 atm * (298.15 K + 42.0 K)) / 298.15 K = 22.5 atm
Therefore, the final pressure of the gas is 22.5 atm.
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As has focal length 44 cm Part A Find the height of the image produced when a 22 cas high obard is placed at stance +10 cm Express your answer in centimeters
The height of the image is 58.74 cm.
Given data:
Focal length = 44 cm
Height of object = 22 cm
Object distance (u) = -10 cm
Image distance (v) =?
Formula: Using the lens formula `1/f = 1/v - 1/u`,
Find the image distance (v).
Using the magnification formula m = -v/u`,
Find the magnification (m).
Using the magnification formula m = h₂/h₁`,
Find the height of the image (h₂).
As per the formula, `
1/f = 1/v - 1/u`
1/44 = 1/v - 1/(-10)
1/v =1/44 + 1/10
v = 26.7 cm.
The image distance (v) is 26.7 cm.
As per the formula, `m = -v/u`
m = -26.7/-10
m = 2.67.
The magnification is 2.67.
As per the formula, `m = h₂/h₁`
2.67 = h₂/22
h₂ = 58.74 cm.
Therefore The height of the image is 58.74 cm.
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An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 10.6 m/s in 3.40 s. (a) What is the magnitude and direction of the bird's acceleration? (b) Assuming that the acceleration remains the same, what is the bird's velocity after an additional 2.70 s has elapsed?
The magnitude of acceleration is given by the absolute value of Acceleration.
Given:
Initial Velocity,
u = 13.0 m/s
Final Velocity,
v = 10.6 m/s
Time Taken,
t = 3.40s
Acceleration of the bird is given as:
Acceleration,
a = (v - u)/t
Taking values from above,
a = (10.6 - 13)/3.40s = -0.794 m/s² (acceleration is in the opposite direction of velocity as the bird slows down)
:|a| = |-0.794| = 0.794 m/s²
The direction of the bird's acceleration is in the opposite direction of velocity,
South.
To calculate the velocity after an additional 2.70 s has elapsed,
we use the formula:
Final Velocity,
v = u + at Taking values from the problem,
u = 13.0 m/s
a = -0.794 m/s² (same as part a)
v = ?
t = 2.70 s
Substituting these values in the above formula,
v = 13.0 - 0.794 × 2.70s = 10.832 m/s
The final velocity of the bird after 2.70s has elapsed is 10.832 m/s.
The direction is still North.
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The Fermi Energy, Ep, for a free electron gas at T = 0 K is given as: Ef = h^2/2me (3pi^2 ne)^(2/3
where me is the free electron mass and ne is the number of electrons per unit volume. Zinc is a metal with Ep = 9.4 eV, a relative atomic mass of 65.4, and a mass density of p= 7.13 x 10^3 kgm-3. Estimate how many electrons each zinc atom contributes to the free electron gas.
Zinc is a metal with a Fermi Energy (Ef) of 9.4 eV. Each zinc atom contributes approximately 2.77 electrons to the free electron gas
The equation for Ef is given as Ef = (h^2/2me) * (3π^2ne)^(2/3), where h is Planck's constant, me is the free electron mass, and ne is the number of electrons per unit volume.
To calculate the number of electrons contributed by each zinc atom, we need to rearrange the equation to solve for ne. Taking the cube of both sides and rearranging, we have ne = (Ef / [(h^2/2me) * (3π^2)])^(3/2).
Given the value of Ef for zinc (9.4 eV), we can substitute the known constants (h, me) and solve for ne. Substituting the values and performing the calculations, we find that each zinc atom contributes approximately 2.77 electrons to the free electron gas.
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An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce X-rays. a. How much potential energy does a single electron loose due to being accelerated through the 40 kV potential? Hint: what is the charge of a single electron? b. What would be the maximum speed of these electrons? Hint: Potential energy is converted into another form of energy and the mass of an electron is 9.11x10" kg.
a. A single electron loses 6.408 × 10⁻¹⁵ J of potential energy.
b. The maximum speed of the electrons is 8.9 × 10⁶ m/s.
a. The potential energy lost by a single electron can be calculated using the equation for electric potential energy:
ΔPE = qΔV, where ΔPE is the change in potential energy, q is the charge of the electron (1.6 × 10⁻¹⁹ C), and ΔV is the change in voltage (40,000 V). Plugging in the values,
we get ΔPE = (1.6 × 10⁻¹⁹ C) × (40,000 V)
= 6.4 × 10⁻¹⁵ J.
b. To determine the maximum speed of the electrons, we can equate the loss in potential energy to the gain in kinetic energy.
The kinetic energy of an electron is given by KE = ½mv²,
where m is the mass of the electron (9.1 × 10⁻³¹ kg) and v is the velocity. Equating ΔPE to KE, we have ΔPE = KE.
Rearranging the equation, we get
(1.6 × 10⁻¹⁹ C) × (40,000 V) = ½ × (9.1 × 10⁻³¹ kg) × v².
Solving for v, we find
v = √((2 × (1.6 × 10⁻¹⁹ C) × (40,000 V)) / (9.1 × 10⁻³¹ kg))
= 8.9 × 10⁶ m/s.
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In the partial wave analysis of low-energy scattering, we often find that S-wave scattering phase shift is all we need. Why do the higher partial waves tend not to contribute to scattering at this limit?
In partial wave analysis, the S-wave scattering phase shift is all we need to analyze low-energy scattering. At low energies, the wavelength is large, which makes the effect of higher partial waves to be minimal.
In partial wave analysis, the S-wave scattering phase shift is all we need to analyze low-energy scattering. The reason why the higher partial waves tend not to contribute to scattering at this limit is due to the following reasons:
The partial wave expansion of a scattering wavefunction involves the summation of different angular momentum components. In scattering problems, the energy is proportional to the inverse square of the wavelength of the incoming particles.
Hence, at low energies, the wavelength is large, which makes the effect of higher partial waves to be minimal. Moreover, when the incident particle is scattered through small angles, the dominant contribution to the cross-section comes from the S-wave. This is because the higher partial waves are increasingly suppressed by the centrifugal barrier, which is proportional to the square of the distance from the nucleus.
In summary, the contribution of higher partial waves tends to be negligible in the analysis of low-energy scattering. In such cases, we can get an accurate description of the scattering process by just considering the S-wave phase shift. This reduces the complexity of the analysis and simplifies the interpretation of the results.
This phase shift contains all the relevant information about the interaction potential and the scattering properties. The phase shift can be obtained by solving the Schrödinger equation for the potential and extracting the S-matrix element. The S-matrix element relates the incident and scattered waves and encodes all the scattering information. A simple way to extract the phase shift is to analyze the behavior of the wavefunction as it approaches the interaction region.
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Inside a 138 mm x 346 mm rectangular duct, air at 17 N/s, 20 deg
C, and 112 kPa flows. Solve for the volume flux if R = 28.5 m/K.
Express your answer in 3 decimal places.
The volume flux inside the rectangular duct is 0.028 m³/s.
Volume flux, also known as volumetric flow rate, is a measure of the volume of fluid passing through a given area per unit time. It is commonly expressed in cubic meters per second (m³/s). To calculate the volume flux in the given scenario, we can use the formula:
Volume Flux = (Air flow rate) / (Cross-sectional area)
First, we need to calculate the cross-sectional area of the rectangular duct. The area can be determined by multiplying the length and width of the duct:
Area = (138 mm) * (346 mm)
To maintain consistent units, we convert the dimensions to meters:
Area = (138 mm * 10⁻³ m/mm) * (346 mm * 10⁻³ m/mm)
Next, we can calculate the air flow rate using the given information. The air flow rate is given as 17 N/s, which represents the mass flow rate. We can convert the mass flow rate to volume flow rate using the ideal gas law:
Volume Flow Rate = (Mass Flow Rate) / (Density)
The density of air can be determined using the ideal gas law:
Density = (Pressure) / (Gas constant * Temperature)
where the gas constant (R) is given as 28.5 m/K, the pressure is 112 kPa, and the temperature is 20 degrees Celsius.
With the density calculated, we can now determine the volume flow rate. Finally, we can divide the volume flow rate by the cross-sectional area to obtain the volume flux.
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A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate, and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)
a. How many revolutions does the plate rotate in 5.5 min? How many radians is it?
b. What is the linear distance traveled by a pea which is placed 2/3 the radius from the center of the plate?
c. What is the linear speed of the pea?
d. What is the angular speed of the pea?
a. The plate rotates 33 revolutions (66π radians) in 5.5 minutes.
b. The pea placed 2/3 the radius from the center travels 6.6π meters.
c. The linear speed of the pea is 3.3π meters per minute.
d. The angular speed of the pea is 33π radians per minute.
a. To find the number of revolutions the plate rotates in 5.5 minutes, we can use the formula:
Number of revolutions = (time / period) = (5.5 min / 1 min/6 rev) = 5.5 * 6 / 1 = 33 revolutions.
To find the number of radians, we use the formula: Number of radians = (number of revolutions) * (2π radians/revolution) = 33 * 2π = 66π radians.
b. The linear distance traveled by the pea placed 2/3 the radius from the center of the plate can be calculated using the formula:
Linear distance = (angular distance) * (radius) = (θ) * (r).
Since the pea is placed 2/3 the radius from the center of the plate, the radius would be (2/3 * 0.15 m) = 0.1 m.
The angular distance can be calculated using the formula:
Angular distance = (number of revolutions) * (2π radians/revolution) = 33 * 2π = 66π radians.
Therefore, the linear distance traveled by the pea would be:
Linear distance = (66π radians) * (0.1 m) = 6.6π meters.
c. The linear speed of the pea can be calculated using the formula:
Linear speed = (linear distance) / (time) = (6.6π meters) / (2.0 min) = 3.3π meters per minute.
d. The angular speed of the pea can be calculated using the formula:
Angular speed = (angular distance) / (time) = (66π radians) / (2.0 min) = 33π radians per minute.
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A dry cell having internal resistance r = 0.5 Q has an electromotive force & = 6 V. What is the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q?
I. 4.5 II. 5.5 III.3.5 IV. 2.5 V. 6.5
The power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is 4.5 W. Hence, the correct option is I. 4.5.
The expression for the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is as follows:
Given :The internal resistance of a dry cell is `r = 0.5Ω`.
The electromotive force of a dry cell is `ε = 6 V`.The external resistance is `R = 1.5Ω`.Power is given by the expression P = I²R. We can use Ohm's law to find current I flowing through the circuit.I = ε / (r + R) Substituting the values of ε, r and R in the above equation, we getI = 6 / (0.5 + 1.5)I = 6 / 2I = 3 A Therefore, the power dissipated through the internal resistance isP = I²r = 3² × 0.5P = 4.5 W Therefore, the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is 4.5 W. Hence, the correct option is I. 4.5.
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A m= 5,400 kg trailer with two axles separated by a distance L = 9.4 m has the center of gravity at d = 4.5 m from the front axle. How far from the rear axle should the center of gravity of a M = 2,000 kg load be placed so that the same normal force acts on the front and rear axles?
The center of gravity of the load should be placed at a distance of 5.8 m from the rear axle.
In the case of a vehicle with a trailer, the distribution of the load is critical for stability. In general, it is recommended that the heaviest items be placed in the center of the trailer, as this will help to maintain stability.The normal force is the weight force, which is represented by the force that the load applies to the axles, and is equal to the product of the mass and the acceleration due to gravity. Thus, to maintain stability, the center of gravity of the load must be placed at a certain distance from the rear axle.Let the distance from the rear axle to the center of gravity of the load be x. Then, the weight of the load will be given by:
Mg = F1 + F2
Here, F1 is the normal force acting on the front axle, and F2 is the normal force acting on the rear axle. Since the same normal force acts on both axles, F1 = F2.
Therefore, Mg = 2F1or F1 = Mg/2
Now, let us calculate the weight that acts on the front axle:
W1 = mF1g
where W1 is the weight of the trailer that acts on the front axle, and m is the mass of the trailer. Similarly, the weight that acts on the rear axle is:
W2 = mF2g = mF1g
Thus, to maintain balance, the center of gravity of the load must be placed at a distance of x from the rear axle, such that: W2x = W1(d - x)
where d is the distance between the axles. Substituting the values given, we get:
W2x = W1(d - x)2000*9.81*x
= (5400+2000)*9.81(9.4 - x + 4.5)x = 5.8 m
Therefore, the center of gravity of the load should be placed at a distance of 5.8 m from the rear axle.
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(A=4, C=2) \) Use the principle of superposition to determine the resulting waveform when the waves in figure interfere with each other."
The resulting waveform will have a displacement equal to the sum of their individual displacements at each point.
When waves interfere with each other,
The principle of superposition states that the displacement of the resulting waveform at any point is equal to the algebraic sum of the individual displacements caused by each wave at that point.
In this case, we have two waves, one represented by Figure A and the other by Figure C.
Assuming these waves are traveling in the same medium and have the same frequency, we can determine the resulting waveform by adding the individual displacements at each point.
Let's consider a point in space and time where both waves overlap.
If the amplitude of the wave in Figure A is 4 and the amplitude of the wave in Figure C is 2,
The resulting waveform at that point will have a displacement equal to the sum of the individual displacements, which is
4 + 2 = 6.
The resulting waveform will have a shape and wavelength determined by the characteristics of the individual waves.
The exact form of the resulting waveform will depend on the phase relationship between the waves, which is not specified in the given information.
When the waves in Figure A and Figure C interfere, the resulting waveform will have a displacement equal to the sum of their individual displacements at each point.
The specific shape and wavelength of the resulting waveform will depend on the characteristics and phase relationship of the individual waves.
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1. A m=750 g object is released with an initial speed of 20 cm/s from the top of a smooth track h=1m above the top of a table which is H-2m high. (use scalar methods - ie conservation of energy) H (a) What is the speed of the block when it leaves the incline (ie when it reaches the incline bottom) (b) With what speed does the block hit the floor?
The speed of the block, when it leaves the incline, is approximately 4.43 m/s. With this speed of 7.675 m/s, the block hit the floor.
a) The initial potential energy of the object at the top of the track is given by:
PE(initial) = m × g × h
KE(final) = (1/2) × m × v(final)²
According to the law of conservation of energy,
PE(initial) = KE(final)
m × g × h = (1/2) × m × v(final)²
v(final) = √(2 × g × h)
v_final = √(2 × 9.8 × 1) = 4.43 m/s
Hence, the speed of the block when it leaves the incline is approximately 4.43 m/s.
b) Gravity work done = Change in kinetic energy,
mg(h +H) = (1/2) × m × v(final)² - 1/2 × m × v(20/100)²
9.8 (2+1) = v(final)²/2 - 0.02
v(final) = 7.675 m/s
Hence, with this speed of 7.675 m/s, the block hit the floor.
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If 3.04 m 3 of a gas initially at STP is placed under a pressure of 2.68 atm, the temperature of the gas rises to 33.3 ∘ C. Part A What is the volume?
The volume of the gas at the given condition is 6.5 m³ given that 3.04 m 3 of a gas initially at STP is placed under a pressure of 2.68 atm and the temperature of the gas rises to 33.3° C.
Given: Initial volume of gas = 3.04 m³
Pressure of the gas = 2.68 ATM
Temperature of the gas = 33.3°C= 33.3 + 273= 306.3 K
As per Gay Lussac's law: Pressure of a gas is directly proportional to its temperature, if the volume remains constant. At constant volume, P ∝ T ⟹ P1/T1 = P2/T2 [Where P1, T1 are initial pressure and temperature, P2, T2 are final pressure and temperature]
At STP, pressure = 1 atm and temperature = 273 K
So, P1 = 1 atm and T1 = 273 K
Now, P2 = 2.68 atm and T2 = 306.3 K
V1 = V2 [Volume remains constant]1 atm/273 K = 2.68 atm/306.3 K
V2 = V1 × (P2/P1) × (T1/T2)
V2 = 3.04 m³ × (2.68 atm/1 atm) × (273 K/306.3 K)
V2 = 6.5 m³
Therefore, the volume of the gas at the given condition is 6.5 m³.
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An RLC circuit has a capacitance of 0.47μF. a) What inductance will produce a resonance frequency of 96MHz ? It is desired that the impedance at resonance be one-third the impedance at 27kHz. What value of R should be used to obtain this result?
An RLC circuit has a capacitance of 0.47 μF. We need to find the inductance and value of R.
The solution to it is explained below: Given data:
Capacitance (C) = 0.47 μF
Resonance frequency (f) = 96 MHz
Impedance at resonance (Z) = Impedance at 27 kHz/3
The resonance frequency can be found using the formula:
f = 1 / 2π√(LC)
The above formula is known as the answer and is used to find out the value of inductance (L). So, rearranging the formula we get:
L = (1/4π²f²C)
L = (1/4π²×96×10⁶ ×0.47 ×10⁻⁶)
L = 41.49 μH
So, the inductance value is 41.49 μH.
Impedance at resonance can be determined as:
Z = √(R²+(Xl - Xc)²)
Here, Xl is the inductive reactance and Xc is the capacitive reactance at the resonant frequency. At resonance,
Xl = Xc,
so Xl - Xc = 0
Therefore, Z = R
We know that impedance at resonance (Z) should be one-third the impedance at 27 kHz.
Hence: Z = RZ₁
Z = R/3
Where, Z₁ is the impedance at 27 kHz So, R = 3 Z₁
Now, the conclusion is the formula of L and the value of R that satisfies the given conditions.
L = 41.49 μH
R = 3 Z₁.
The answer to the question is as follows inductance value is 41.49 μH and R = 3 Z₁.
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A ball of mass m= 75.0 grams is dropped from a height of 2.00 m. The ball stays in contact with the ground 25.0 ms. How high did it bounce back up if the ground exerts a force of 30.0 N on it
The ball of mass m=75.0 g is dropped from a height of 2.00 m. It bounces back with a height of 0.5 m.
To determine the height to which the ball bounced back up, use the conservation of energy principle. The total mechanical energy of a system remains constant if no non-conservative forces do any work on the system. The kinetic energy and the potential energy of the ball at the top and bottom of the bounce need to be calculated. The force of the ground is considered a non-conservative force, and it does work on the ball during the impact. Therefore, its work is equal to the loss of mechanical energy of the ball.
The potential energy of the ball before the impact is equal to its kinetic energy after the impact because the ball comes to a halt at the top of its trajectory.
Hence, mgh = 1/2mv²v = sqrt(2gh) v = sqrt(2 x 9.81 m/s² x 2.00 m) v = 6.26 m/s.
The force applied by the ground on the ball is given by the equation
F = m x a where F = 30 N and m = 75.0 g = 0.075 kg.
So, a = F/m a = 30 N / 0.075 kg a = 400 m/s²
Finally, h = v²/2a h = (6.26 m/s)² / (2 x 400 m/s²) h = 0.5 m.
Thus, the ball bounced back to a height of 0.5 meters.
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