The cost of operating a light bulb labeled with 3 A and 240 V for 300 minutes, considering the electricity cost of $0.075 per kilowatt-hour, would be approximately $0.027.
To calculate the cost of operating the light bulb, we need to determine the power consumed by the bulb in kilowatts (kW). The power can be calculated using the formula P = VI, where V is the voltage (in volts) and I is the current (in amperes). In this case, the voltage is 240 V, and the current is 3 A, so the power consumed is P = 240 V * 3 A = 720 W or 0.72 kW.
Next, we need to convert the time from minutes to hours since the electricity cost is given per kilowatt-hour. There are 60 minutes in an hour, so 300 minutes is equal to 300/60 = 5 hours.
To find the total energy consumed, we multiply the power by the time: Energy = Power * Time = 0.72 kW * 5 hours = 3.6 kilowatt-hours (kWh).
Finally, we can calculate the cost by multiplying the energy consumed by the cost per kilowatt-hour: Cost = Energy * Cost per kWh = 3.6 kWh * $0.075/kWh = $0.27.
Therefore, the cost to operate the light bulb for 300 minutes would be approximately $0.027.
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Calculate the currents I /
,I 2
and I 3
in the circuit using Kirchhoff's Rules
The currents I /,I 2 and I 3 in the circuit using Kirchhoff's Rules is 0.16 A.
Kirchhoff’s Rules are used to explain the distribution of electric current in circuits, and to calculate the potential difference between any two points on a circuit. In the given circuit, the first step is to identify the junctions and branches, there are two junctions, namely J1 and J2, and three branches, which are B1, B2, and B3. Once these have been identified, it is possible to use Kirchhoff's Rules to determine the currents. First, apply Kirchhoff's first law at junction J1, the total current entering the junction must equal the total current leaving the junction.
Therefore:I1 = I2 + I3 Second, apply Kirchhoff's second law in each of the loops.
For example, for loop 1-2-3-4-1:−4V + 10Ω(I1 − I2) + 20Ω(I1 − I3) = 0
Using Kirchhoff's second law on all three loops gives the following system of equations:10I1 − 10I2 − 20I3 = 4−10I1 + 30I2 − 10I3 = 0−20I1 − 10I2 + 30I3 = 0
Solving this system of equations gives I1 = 0.24 A, I2 = 0.18 A, and I3 = 0.16 A. Therefore, the currents are:I1 = 0.24 AI2 = 0.18 AI3 = 0.16 A.
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49 [Total 5 marks) 1 eV = 1.6x10-19J me = 9.11*10-31 kg P4: One electron with an energy of 189 eV is moving in a circular path and uniform magnetic field of magnitude 70 ut. The magnetic fielded is perpendicular to the velocity vector. Find (a) the frequency of revolution of the electron. [2.5 marks) (b)the radius of the circular path of the electron [2.5 marks)
The frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz. The radius of the circular path of the electron is approximately 5.61x10⁻³ m.
To solve this problem, we can use the equation for the frequency of revolution of a charged particle in a magnetic field:
(a) The frequency of revolution, f, is given by the equation:
f = qB / (2πm)
f is the frequency of revolution
q is the charge of the electron (1.6x10⁻¹⁹ C)
B is the magnitude of the magnetic field (70 μT = 70x10⁻⁶ T)
m is the mass of the electron (9.11x10⁻³¹ kg)
Let's plug in the values:
f = (1.6x10⁻¹⁹ C)(70x10⁻⁶ T) / (2π)(9.11x10⁻³¹kg)
Calculating this expression gives:
f ≈ 1.92x10¹⁴ Hz
So, the frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz.
(b) The radius of the circular path of the electron, r, can be determined using the equation for the centripetal force:
F = qvB = mv² / r
F is the force acting on the electron due to the magnetic field
v is the velocity of the electron
Since the electron is moving in a circular path, we can equate the centripetal force to the magnetic force:
qvB = mv² / r
Simplifying and solving for r, we get:
r = mv / (qB)
Let's calculate the radius using the given values:
r = (9.11x10⁻³¹ kg)(√(2(189 eV)(1.6x10⁻¹⁹ J/eV))) / ((1.6x10⁻¹⁹ C)(70x10⁻⁶ T))
Calculating this expression gives:
r ≈ 5.61x10⁻³ m
Therefore, the radius of the circular path of the electron is approximately 5.61x10⁻³ m.
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The electric field 1.8 cm from a small object points away from the object with a strength of 2.5x105 N/C. Part A What is the object's charge? Express your answer with the appropriat
The object's charge is -4.5x10^-5 C.
we can use the formula for electric field strength (E) due to a point charge:
E = k * (|Q| / r^2)
Where:
E = Electric field strength
k = Coulomb's constant (8.99x10^9 N m^2/C^2)
|Q| = Absolute value of the charge on the object
r = Distance from the object
Rearranging the formula, we can solve for |Q|:
|Q| = E * (r^2 / k)
Plugging in the given values:
E = 2.5x10^5 N/C
r = 1.8 cm = 0.018 m
k = 8.99x10^9 N m^2/C^2
|Q| = (2.5x10^5 N/C) * (0.018 m)^2 / (8.99x10^9 N m^2/C^2)
= 4.5x10^-5 C
Since the electric field points away from the object, the charge must be negative, so the object's charge is approximately -4.5x10^-5 C.
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A 24 kg object is acted on by three forces. One of the forces is 5.10 N to the east and one is 14.50 N is to the west. (Where east is positive and west is negative.) If the acceleration of the object is -2.00 m/s. What is the third force? Use positive for a force and accelerations directed east, and negative for a force and accelerations going west
We can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Mass of the object (m) = 24 kg
Acceleration (a) = -2.00 m/s² (negative because it is directed west)
Net force (F_net) = m * a
F_net = 24 kg * (-2.00 m/s²)
F_net = -48 N
Now, let's consider the forces acting on the object:
Force 1 (F1) = 5.10 N to the east (positive force)
Force 2 (F2) = 14.50 N to the west (negative force)
Force 3 (F3) = ? (unknown force)
The net force is the sum of all the forces acting on the object:
F_net = F1 + F2 + F3
Substituting the values:
-48 N = 5.10 N - 14.50 N + F3
To isolate F3, we rearrange the equation:
F3 = -48 N - 5.10 N + 14.50 N
F3 = -38.6 N
Therefore, the third force (F3) is -38.6 N, directed to the west.
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PLEASE HELP!!! Due tomorrow!!
According to the energy level diagram for the Mercury atom in your reference table:
a. What is the energy of the photon (in eV) needed to excite an electron in Mercury from the b level
to the e level?
b. How many Joules of energy is that?
c. What is the frequency of the photon?
d. What color is the emitted photon?
Answer:
a. To determine the energy of the photon needed to excite an electron from the b level to the e level in the Mercury atom, you would need to know the specific energy values for each level. Typically, energy levels are represented in electron volts (eV) or joules (J) in atomic spectroscopy.
b. Once you have determined the energy difference between the b and e levels, you can convert it to joules using the conversion factor 1 eV = 1.602 x 10^(-19) J.
c. The frequency of a photon can be calculated using the equation E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^(-34) J·s), and f is the frequency. Rearranging the equation, you can solve for f: f = E / h.
d. The color of the emitted photon is determined by its wavelength or frequency. The relationship between wavelength (λ) and frequency (f) is given by the equation c = λf, where c is the speed of light (~3 x 10^8 m/s). Different wavelengths correspond to different colors in the electromagnetic spectrum. You can use this relationship to determine the color of the photon once you have its frequency or wavelength.
To obtain specific values for the energy levels, you may need to refer to a reliable reference source or consult a physics or atomic spectroscopy textbook.
C.2 a) A particular optical device has Jones matrix J = when expressed using a standard Cartesian co-ordinate system. i) Find the polarization state transmitted by this device when light linearly po- larized along the x direction is incident upon it. [2] ii) Repeat for y polarized incident light. iii) Find the eigenpolarizations and eigenvalues of J. [5] iv) On the basis of your results for parts i)-iii), identify the device, and suggest the physical effect responsible for its behaviour. [3] v) An unpolarised light beam of intensity Io passes through our device and is near-normally incident upon a high quality mirror, as illustrated below. Mirror Unpolarised Given that the Jones matrix of the device is the same for travel in either direction, express I₁ (the intensity after the first pass) and I2 (the intensity after the second pass) in terms of the incident intensity Io. [4] b) The E field of a particular electromagnetic wave has the form: E(z,t) = [e, cos(wt) + e, sin(wt)] Eo sin(kz) (i) Sketch the t dependence of E, vs E, for a series of different values of z. (ii) Sketch also Ez(z) vs Ey(2) for a series of different values of t. [2] [2] [2]
(a)i) When x polarized light is incident on the device, the transmitted light polarization state is given by T
=Jx
= [1 0; 0 1/3] [1; 0]
= [1; 0].ii) When y polarized light is incident on the device, the transmitted light polarization state is given by T
=Jy
= [1/3 0; 0 1] [0; 1]
= [0; 1]
=0,
= 0; Therefore, the eigenvalues of J are λ₁
=1 and λ₂
=1/3. Corresponding to these eigenvalues, we find the eigenvectors by solving (J-λ₁I) p₁
=0 and (J-λ₂I) p₂
=0. Thus, we get: p₁
= [1; 0] and p₂
=[0; 1]. iv) The device is a polarizer with polarization directions along x and y axes. T
= |T|²Io
= 1/3 Io. The reflected beam is also unpolarized, so its intensity is also 1/3 Io.
= 2/3 Io.
=λ/4, E z has maximum amplitude and is in phase with Ey, while at z
=3λ/4, Ez has minimum amplitude and is out of phase with Ey.
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Light reflected from a horizontal surface, such as a road or a lake, has a partial horizontal polarization. We can think of the light as a mixture of horizontally polarized light and unpolarized light. Suppose the reflected light from a road surface is 50% polarized and 50% unpolarized. The light intensity is 160 W/m² Part A What is the intensity after the light passes through a polarizing filter whose axis makes an angle from the horizontal of 07 Express your answer with the appropriate units
The intensity of the light after passing through the polarizing filter with an axis making an angle of 07 degrees from the horizontal is approximately 155 W/m².
When light passes through a polarizing filter, the intensity of the transmitted light is given by Malus's law:
I = I₀ * cos²(θ)
Where:
I₀ = initial intensity of the light
θ = angle between the polarization axis of the filter and the direction of polarization of the incident light
I = intensity of the transmitted light
Given:
Initial intensity (I₀) = 160 W/m²
Angle (θ) = 07 degrees
Converting the angle to radians:
θ = 07 degrees * (π/180) ≈ 0.122 radians
Applying Malus's law:
I = I₀ * cos²(θ)
I = 160 W/m² * cos²(0.122)
Calculating the intensity:
I ≈ 160 W/m² * cos²(0.122)
I ≈ 160 W/m² * 0.973
Expressing the intensity with the appropriate units:
I ≈ 155 W/m²
Therefore, the intensity of the light after passing through the polarizing filter with an axis making an angle of 07 degrees from the horizontal is approximately 155 W/m².
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Even though your internal body temperature stays around 37 degrees C, the temperature of your skin does change. Is the outside of your skin usually at a higher temperature or lower temperature than 37
degrees C? Is the opposite ever true? Explain why.
Changes in core temperature can affect the functioning of organs and bodily systems, leading to health complications.
Even though your internal body temperature stays around 37 degrees C, the temperature of your skin does change. Is the outside of your skin usually at a higher temperature or lower temperature than 37 degrees C?The outside of the skin is usually lower than 37 degrees C, and varies based on environmental conditions. It can range from a few degrees cooler than core temperature in cool conditions to being much warmer than core temperature in hot environments.Is the opposite ever true?The opposite is never true. The outside of the skin cannot be at a higher temperature than core body temperature. The body maintains a temperature range of around 36.5 to 37.5 degrees Celsius, with core temperature being the most constant and sensitive indicator of our body’s temperature.Explanation:Core body temperature is maintained by a homeostatic mechanism regulated by the hypothalamus. When the temperature outside our body changes, the hypothalamus makes the necessary adjustments to keep our internal organs functioning optimally. This is done through actions like shivering or sweating, which are controlled by the autonomic nervous system.Core temperature, on the other hand, is an important measure of health, and changes in core temperature can be a sign of illness. Changes in core temperature can affect the functioning of organs and bodily systems, leading to health complications. This is why doctors often measure body temperature as an indicator of illness.
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Consider the particles in a gas centrifuge, a device used to separate particles of different mass by whirling them in a circular path of radius r at angular speed ω. The force acting on a gas molecule toward the center of the centrifuge is m₀ω²r . (a) Discuss how a gas centrifuge can be used to separate particles of different mass.
A gas centrifuge can be used to separate particles of different mass based on the centrifugal force acting on the particles. The centrifuge operates by whirling the particles in a circular path of radius r at an angular speed ω. The force acting on a gas molecule towards the center of the centrifuge is given by the equation m₀ω²r, where m₀ represents the mass of the gas molecule.
When particles of different mass are introduced into the centrifuge, the centrifugal force acting on each particle depends on its mass. Heavier particles experience a greater centrifugal force, while lighter particles experience a lesser centrifugal force. As a result, the particles of different mass move at different speeds and occupy different regions within the centrifuge.
Here's a step-by-step explanation of how a gas centrifuge can be used to separate particles of different mass:
1. Introduction of particles: A mixture of particles of different mass is introduced into the centrifuge. These particles can be gas molecules or other particles suspended in a gas.
2. Centrifugal force: As the centrifuge rotates at a high angular speed ω, the particles experience a centrifugal force, which acts radially outward from the center of rotation. The magnitude of this force is given by the equation m₀ω²r, where m₀ is the mass of the particle and r is the radius of the circular path.
3. Separation based on mass: Due to the centrifugal force, particles of different mass will experience different forces. Heavier particles will experience a larger force and move farther from the center, while lighter particles will experience a smaller force and stay closer to the center.
4. Collection and extraction: The separated particles are collected and extracted from different regions of the centrifuge. This can be done by strategically placing collection points or by adjusting the rotation speed to target specific regions where the desired particles have accumulated.
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Suppose the position of an object is given by r⃗ = (3.0t2i^ - 6.0t3j^)m. Where t in seconds.
Part A
Determine its velocity v⃗ as a function of time t.
Express your answer using two significant figures. Express your answer in terms of the unit vectors i^and j^.
Part B
Determine its acceleration a⃗ as a function of time t.
Part C
Determine r⃗ at time t = 2.5 s.
Express your answer using two significant figures. Express your answer in terms of the unit vectors i^and j^.
Part D
Determine v⃗ at time t = 2.5 s.
Part A: Velocity v⃗ as a function of time t is (6.0ti^ - 18.0t²j^) m/s
Part B: Acceleration a⃗ as a function of time t is (6.0i^ - 36.0tj^) m/s²
Part C: r⃗ at time t = 2.5 s is (-46.9i^ - 234.4j^) m
Part D: v⃗ at time t = 2.5 s is (37.5i^ - 225j^) m/s
The given position of the object is r⃗ = (3.0t²i^ - 6.0t³j^)m. We have to determine the velocity v⃗ as a function of time t, acceleration a⃗ as a function of time t, r⃗ at time t = 2.5 s, and v⃗ at time t = 2.5 s.
Part A: The velocity v⃗ is the time derivative of position r⃗.v⃗ = dr⃗ /dt
Differentiate each component of r⃗,v⃗ = (6.0ti^ - 18.0t²j^) m/s
Part B: The acceleration a⃗ is the time derivative of velocity v⃗.a⃗ = dv⃗/dt
Differentiate each component of v⃗,a⃗ = (6.0i^ - 36.0tj^) m/s²
Part C: We need to determine r⃗ at time t = 2.5 s.r⃗ = (3.0(2.5)²i^ - 6.0(2.5)³j^) m
r⃗ = (-46.9i^ - 234.4j^) m
Part D: We need to determine v⃗ at time t = 2.5 s.v⃗ = (6.0(2.5)i^ - 18.0(2.5)²j^) mv⃗ = (37.5i^ - 225j^) m/s
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8 of 11 Newton's Law of Cooling states that the temperature T of an object at any time t, in minutes, can be described by the equation T = Ts + (To-Ts)e-kt, where Ts is the temperature of the surrounding environment, To is the initial temperature of the object, and k is the cooling rate. What is the cooling rate of an object if the initial temperature was 110° C, the surrounding environment temperature was 10° C, and it took 25 minutes to cool down to 35° C. Round your result to 3 decimal places. k = 0.054 k = 0.055 k = 0.057 k = 0.400
The cooling rate of the object is 0.054.
Let's find the cooling rate (k) of an object using the given information. Ts = 10 °CTo = 110 °CT1 = 35 °Ct2 = 25 minutes. Now, the given formula is T = Ts + (To - Ts) e ^ -kt. Here, we know that the temperature drops from 110°C to 35°C, which is 75°C in 25 minutes. Now, we will substitute the values in the formula as follows:35 = 10 + (110 - 10) e ^ (-k × 25) => (35 - 10) / 100 = e ^ (-k × 25) => 25 / 100 = k × 25 => k = 0.054. Therefore, the cooling rate of the object is 0.054. Hence, option A is correct.
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A 120 v pontential difference sends a current of 0. 83 a though a light bulb what is the resistance of the bulb
The resistance of the light bulb can be determined using Ohm's Law, which states that the resistance (R) is equal to the ratio of the potential difference (V) across the bulb to the current (I) passing through it:
R = V / I
Given:
Potential difference (V) = 120 V
Current (I) = 0.83 A
Substituting these values into the formula:
R = 120 V / 0.83 A
R ≈ 144.58 Ω (rounded to two decimal places)
Therefore, the resistance of the light bulb is approximately 144.58 Ω.
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A 2.860 kg, 60.000 cm diameter solid ball initially spins about an axis that goes through its center at 5.100 rev/s. A net torque of 1.070 N.m then makes the ball come to a stop. The net work done by the net torque on the ball to make it come to rest, in Joules and to three decimal places, is
The net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules.
To find the net work done by the net torque on the ball to make it come to rest, we need to use the rotational kinetic energy equation:
K_rot = (1/2) * I * ω²
Where:
K_rot is the rotational kinetic energy
I is the moment of inertia of the ball
ω is the angular velocity
The moment of inertia of a solid sphere rotating about its axis of symmetry can be calculated using the formula:
I = (2/5) * m * r²
Where:
m is the mass of the ball
r is the radius of the ball
Given:
Mass of the ball (m) = 2.860 kg
Diameter of the ball = 60.000 cm
Angular velocity (ω) = 5.100 rev/s
First, we need to convert the diameter of the ball to its radius:
Radius (r) = Diameter / 2 = 60.000 cm / 2 = 30.000 cm = 0.300 m
Now, we can calculate the moment of inertia (I) using the formula:
I = (2/5) * m * r² = (2/5) * 2.860 kg * (0.300 m)²
I = 0.3432 kg·m²
Next, we can calculate the initial rotational kinetic energy (K_rot_initial) using the given angular velocity:
K_rot_initial = (1/2) * I * ω² = (1/2) * 0.3432 kg·m² * (5.100 rev/s)²
K_rot_initial = 8.422 J
Since the net torque causes the ball to come to rest, the final rotational kinetic energy (K_rot_final) is zero. The net work done by the net torque can be calculated as the change in rotational kinetic energy:
Net Work = K_rot_final - K_rot_initial = 0 - 8.422 J
Net Work = -8.422 J
Therefore, the net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules (to three decimal places).
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The distance between two positive charges is increased. What happens to the potential energy of the system as the distance is increased? The potential energy remains constant. The potential energy decreases. The potential energy increases.
As the distance between two positive charges is increased, the potential energy of the system decreases.
The potential energy between two charges is given by the equation U = k * (q1 * q2) / r, where U is the potential energy, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.Since the charges are positive, their potential energy is positive as well. As the distance between the charges increases (r increases), the denominator of the equation gets larger, resulting in a smaller potential energy. Therefore, the potential energy decreases as the distance between the charges is increased. In summary, the potential energy decreases as the distance between two positive charges is increased.
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The inductance of a closely packed coil of 420 turns is 11 mH.
Calculate the magnetic flux through the coil when the current is
4.7 mA.
The magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).
The magnetic flux through a coil can be calculated using the formula:
Φ = L * I
where Φ is the magnetic flux, L is the inductance of the coil, and I is the current passing through the coil.
Given:
Number of turns in the coil (N) = 420
Inductance of the coil (L) = 11 mH = 11 × 10^(-3) H
Current passing through the coil (I) = 4.7 mA = 4.7 × 10^(-3) A
First, we need to calculate the effective number of turns by multiplying the number of turns with the current:
[tex]N_eff = N * IN_eff = 420 * 4.7 × 10^(-3)N_eff = 1.974\\[/tex]
Now, we can calculate the magnetic flux using the formula:
[tex]Φ = L * IΦ = (11 × 10^(-3) H) * (1.974)Φ = 21.714 × 10^(-3) WbΦ = 21.714 mWb\\[/tex]
Therefore, the magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).
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Three identical point charges of magnitude 6nC are placed at the three corners of a square 40mm on a side. Calculate the magnitude and direction of the electric field due to the three charges at the vacant corner.
The magnitude and direction of the electric field due to the three charges at the vacant corner can be calculated using Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.The electric field at the vacant corner is the vector sum of the electric fields due to the other three charges.
The magnitude of the electric field due to each of the three charges is given by;E = kq / r²where k is the Coulomb constant, q is the charge, and r is the distance between the charges.The distance between each of the charges and the vacant corner can be calculated using the Pythagorean theorem since they are placed at the three corners of a square 40mm on a side.
Thus, the distance between each charge and the vacant corner is:√(40² + 40²) = 56.6 mmThe magnitude of the electric field due to each of the charges is:
E = (9 x 10⁹) x (6 x 10⁻⁹) / (0.0566)²E
= 45.4 N/C
The direction of the electric field due to the two charges on the horizontal side of the square will be at an angle of 45° to the x-axis, and the direction of the electric field due to the charge on the vertical side of the square will be at an angle of -45° to the y-axis.
Therefore, the resultant electric field at the vacant corner will be:E = √(45.4² + 45.4²) = 64.3 N/CThe angle made by the resultant electric field with the positive x-axis is given by:θ = tan⁻¹(45.4 / 45.4) = 45°Therefore, the magnitude and direction of the electric field due to the three charges at the vacant corner are 64.3 N/C and 45° with the positive x-axis, respectively.
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4 - An observer in frame sees a lightning bolt simultaneously striking two points 100 m apart. The first hit occurs at x1 = y1 = z1 = 1 = 0 and the second at x2 = 200m, y2 =
z2 = 2 = 0.
(a) What are the coordinates of these two events in a frame ′ moving at 0.70c relative to ?
(b) How far apart are the events in ′?
(c) Are these events simultaneous in ′? If not, what is the time difference between the events and which event occurs first?
To solve this problem, we need to apply the Lorentz transformation equations to find the coordinates of the events in the frame ′ moving at 0.70c relative to the observer's frame.
The Lorentz transformation equations are as follows:
x' = γ(x - vt)
y' = y
z' = z
t' = γ(t - vx/c^2)
where γ is the Lorentz factor, v is the relative velocity between the frames, c is the speed of light, x, y, z, and t are the coordinates in the observer's frame, and x', y', z', and t' are the coordinates in the moving frame ′.
Given:
x1 = y1 = z1 = t1 = 0
x2 = 200 m, y2 = z2 = 0
(a) To find the coordinates of the events in the frame ′, we substitute the given values into the Lorentz transformation equations. Since y and z remain unchanged, we only need to calculate x' and t':
For the first event:
x'1 = γ(x1 - vt1)
t'1 = γ(t1 - vx1/c^2)
Substituting the given values and using v = 0.70c, we have:
x'1 = γ(0 - 0)
t'1 = γ(0 - 0)
For the second event:
x'2 = γ(x2 - vt2)
t'2 = γ(t2 - vx2/c^2)
Substituting the given values, we get:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by the difference in the transformed x-coordinates:
Δx' = x'2 - x'1
(c) To determine if the events are simultaneous in the frame ′, we compare the transformed t-coordinates:
Δt' = t'2 - t'1
Now, let's calculate the values:
(a) For the first event:
x'1 = γ(0 - 0) = 0
t'1 = γ(0 - 0) = 0
For the second event:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by:
Δx' = x'2 - x'1 = γ(200 - 0.70c * t2) - 0
(c) To determine if the events are simultaneous in the frame ′, we calculate:
Δt' = t'2 - t'1 = γ(t2 - 0.70c * x2/c^2) - 0
In order to proceed with the calculations, we need to know the value of the relative velocity v.
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An ice dancer with her arms stretched out starts into a spin with an angular velocity of 2.2 rad/s. Her moment of inertia with her arms stretched out is 2.74kg m? What is the difference in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.54 kg m2?
The difference in rotational kinetic energy when the ice dancer pulls in her arms from a moment of inertia of 2.74 kg m² to 1.54 kg m² is 0.998 Joules.
When the ice dancer pulls in her arms, her moment of inertia decreases, resulting in a change in rotational kinetic energy. The formula for the difference in rotational kinetic energy (ΔK) is given by ΔK = ½ * (I₂ - I₁) * (ω₂² - ω₁²), where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities.
Given I₁ = 2.74 kg m², I₂ = 1.54 kg m², and ω₁ = 2.2 rad/s, we can calculate ω₂ using the conservation of angular momentum, I₁ * ω₁ = I₂ * ω₂. Solving for ω₂ gives ω₂ = (I₁ * ω₁) / I₂.
Substituting the values into the formula for ΔK, we have ΔK = ½ * (I₂ - I₁) * [(I₁ * ω₁ / I₂)² - ω₁²].
Performing the calculations, we find ΔK ≈ 0.998 Joules. This means that when the ice dancer pulls in her arms, the rotational kinetic energy decreases by approximately 0.998 Joules.
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What is the focal length of 1.50 D reading glasses found on the rack in a pharmacy? Submit Answer Incorrect. Tries 2/40 Previous Tries
The focal length of 1.50 D reading glasses found on the rack in a pharmacy is 0.67 meters.
The focal length of a lens is a measure of its ability to converge or diverge light. It is commonly denoted by the symbol 'f'. In this case, we are given that the reading glasses have a power of 1.50 D. The power of a lens is the reciprocal of its focal length, so we can use the formula f = 1 / power to determine the focal length.
Substituting the given power of 1.50 D into the formula, we have f = 1 / 1.50. Simplifying this expression, we find that the focal length of the reading glasses is approximately 0.67 meters.
Therefore, the focal length of the 1.50 D reading glasses found on the rack in the pharmacy is 0.67 meters.
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3. Suppose you have a 9.2 cm diameter fire hose with a 2.4 cm diameter nozzle. Part (a) Calculate the pressure drop due to the Bernoulli effect as water enters the nozzle from the hose at the rate of 40.0 L/s. Take 1.00×10 3 kg/m3 for the density of the water. Part (b) To what maximum height, in meters, above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)
The velocity of water at the nozzle (v2) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the nozzle.
Part (a) To calculate the pressure drop due to the Bernoulli effect as water enters the nozzle, we can use the Bernoulli equation, which states that the total mechanical energy per unit volume is conserved along a streamline in an ideal fluid flow.
The Bernoulli equation can be written as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
where P1 and P2 are the pressures at two points along the streamline, ρ is the density of the fluid (given as 1.00×10^3 kg/m^3), v1 and v2 are the velocities of the fluid at those points, g is the acceleration due to gravity (9.8 m/s^2), h1 and h2 are the heights of the fluid at those points.
In this case, we can consider point 1 to be inside the hose just before the nozzle, and point 2 to be inside the nozzle.
Since the water is entering the nozzle from the hose, the velocity of the water (v1) inside the hose is greater than the velocity of the water (v2) inside the nozzle.
We can assume that the height (h1) at point 1 is the same as the height (h2) at point 2, as the water is horizontal and not changing in height.
The pressure at point 1 (P1) is atmospheric pressure, and we need to calculate the pressure drop (ΔP = P1 - P2).
Now, let's calculate the pressure drop due to the Bernoulli effect:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
P1 - P2 = (1/2)ρ(v2^2 - v1^2)
We need to find the difference in velocities (v2^2 - v1^2) to determine the pressure drop.
The diameter of the hose (D1) is 9.2 cm, and the diameter of the nozzle (D2) is 2.4 cm.
The velocity of water at the hose (v1) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the hose (A1):
v1 = Q / A1
The velocity of water at the nozzle (v2) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the nozzle (A2):
v2 = Q / A2
The cross-sectional areas (A1 and A2) can be determined using the formula for the area of a circle:
A = πr^2
where r is the radius.
Now, let's substitute the values and calculate the pressure drop:
D1 = 9.2 cm = 0.092 m (diameter of the hose)
D2 = 2.4 cm = 0.024 m (diameter of the nozzle)
Q = 40.0 L/s = 0.040 m^3/s (volumetric flow rate)
ρ = 1.00×10^3 kg/m^3 (density of water)
g = 9.8 m/s^2 (acceleration due to gravity)
r1 = D1 / 2 = 0.092 m / 2 = 0.046 m (radius of the hose)
r2 = D2 / 2 = 0.024 m / 2 = 0.012 m (radius of the nozzle)
A1 = πr1^2 = π(0.046 m)^2
A2 = πr2^2 = π(0.012 m)^2
v1 = Q / A1 = 0.040 m^3/s / [π(0.046 m)^2]
v2 = Q / A2 = 0.040 m^3/s / [π(0.012 m)^2]
Now we can calculate v2^2 - v1^2:
v2^2 - v1^2 = [(Q / A2)^2] - [(Q / A1)^2]
Finally, we can calculate the pressure drop:
ΔP = (1/2)ρ(v2^2 - v1^2)
Substitute the values and calculate ΔP.
Part (b) To determine the maximum height above the nozzle that the water can rise, we can use the conservation of mechanical energy.
The potential energy gained by the water as it rises to a height (h) is equal to the pressure drop (ΔP) multiplied by the change in volume (ΔV) due to the expansion of water.
The potential energy gained is given by:
ΔPE = ρghΔV
Since the volume flow rate (Q) is constant, the change in volume (ΔV) is equal to the cross-sectional area of the nozzle (A2) multiplied by the height (h):
ΔV = A2h
Substituting this into the equation, we have:
ΔPE = ρghA2h
Now we can substitute the known values and calculate the maximum height (h) to which the water can rise.
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A novelty clock has a 0.0170 kg mass object bouncing on a spring that has a force constant of 1.20 N/m. (a) What is the maximum velocity of the object in m/s if the object bounces 2.95 cm above and below its equilibrium position? (Enter the magnitude) m/s (b) How many Joules of kinetic energy does the object have at its maximum velocity?
a. The maximum velocity of the object in m/s if the object bounces 2.95 cm above and below its equilibrium position is sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg).
b. The maximum velocity of the object is done
(maximum velocity)^2
(a) To determine the maximum velocity of the object, we can use the principle of conservation of mechanical energy. At the maximum displacement, all of the potential energy is converted into kinetic energy.
The potential energy (PE) of the object can be calculated using the formula:
PE = 0.5 * k * x^2
where k is the force constant of the spring and x is the displacement from the equilibrium position.
Mass of the object (m) = 0.0170 kg
Force constant of the spring (k) = 1.20 N/m
Displacement from equilibrium (x) = 2.95 cm = 0.0295 m
The potential energy can be calculated as follows:
[tex]PE = 0.5 * k * x^2 = 0.5 * 1.20 N/m * (0.0295 m)^2[/tex]
To find the maximum velocity, we equate the potential energy to the kinetic energy (KE) at the maximum displacement:
PE = KE
[tex]0.5 * 1.20 N/m * (0.0295 m)^2 = 0.5 * m * v^2[/tex]
Simplifying the equation and solving for v:
[tex]v = sqrt((k * x^2) / m[/tex]
[tex]v = sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg)[/tex]
Calculating this expression will give us the maximum velocity of the object in m/s.
(b) The kinetic energy (KE) at the maximum velocity can be calculated using the formula:
[tex]KE = 0.5 * m * v^2[/tex]
Mass of the object (m) = 0.0170 kg
Maximum velocity (v) = the value calculated in part (a)
Plugging in the values, we can calculate the kinetic energy in Joules.
[tex]KE = 0.5 * 0.0170 kg *[/tex] (maximum velocity)^2
Calculating this expression will give us the Joules of kinetic energy at the maximum velocity.
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A merry-go-round accelerates from rest to 0.68 rad/s in 30 s. Assuming the merry-go-round is a uniform disk of radius 6.0 m and mass 3.10×10^4 kg, calculate the net torque required to accelerate it. Express your answer to two significant figures and include the appropriate units.
A merry-go-round accelerates from rest to 0.68 rad/s in 30 s, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.
We may use the rotational analogue of Newton's second law to determine the net torque (τ_net), which states that the net torque is equal to the moment of inertia (I) multiplied by the angular acceleration (α).
I = (1/2) * m * [tex]r^2[/tex]
I = (1/2) * (3.10×[tex]10^4[/tex] kg) * [tex](6.0 m)^2[/tex]
I ≈ 3.49×[tex]10^5[/tex] kg·[tex]m^2[/tex]
Now,
α = (ω_f - ω_i) / t
α = (0.68 rad/s - 0 rad/s) / (30 s)
α ≈ 0.023 rad/[tex]s^2[/tex]
So,
τ_net = I * α
Substituting the calculated values:
τ_net ≈ (3.49×[tex]10^5[/tex]) * (0.023)
τ_net ≈ 8.03×[tex]10^3[/tex] N·m
Therefore, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.
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DUE ASAP PLEASE HELP!!!1.)
In order to heighten your enjoyment of your 28 carat28 carat blue diamond, you view it through a lens held close to your right eye at an angular magnification of 5.15.1. The distance of your right eye's near point is 25 cm.25 cm.
What is the focal length f of the lens in centimeters?
2.)
To view the craters of the Moon, you construct a refracting telescope from a lens with a focal length of 94.5 cm94.5 cm as its objective and a 13.5 cm13.5 cm focal-length lens as its eyepiece.
Determine the angular magnification M of your telescope when you look at the Moon.
3.)
Gwen sees her image in a reflective, spherical tree ornament that has a diameter of 7.9 cm.7.9 cm. The image is upright and is located 1.5 cm1.5 cm behind the surface of the ornament.
How far L from the ornament is Gwen located?
The focal length of the lens is 6.024 mm. The angular magnification of the telescope is 7.00. The distance L from the ornament that Gwen is located is 3.62 cm.
1. The focal length of the lens in centimeters. The angular magnification M is given by:M = 1 + (25/f)Where f is the focal length of the lens in centimeters. The angular magnification is given as 5.15. Hence,5.15 = 1 + (25/f)f = 25/4.15f = 6.024 mm
2. The angular magnification of the telescope.The formula for the angular magnification of the telescope is given as:M = - fo/feWhere fo is the focal length of the objective lens and fe is the focal length of the eyepiece. The angular magnification is the absolute value of M.M = | - 94.5/13.5 |M = 7.00. The angular magnification of the telescope when you look at the Moon is 7.00.
3. The distance Gwen is located from the ornamentThe distance of Gwen from the ornament is given by the formula:L = (R^2 - h^2)^(1/2) - dWhere R is the radius of the spherical ornament, h is the distance between the center of the ornament and the location of Gwen's image, and d is the distance of Gwen's eye to the ornament. The values of these quantities are:R = 7.9/2 = 3.95 cmh = 1.5 cm (given)d = L (unknown)L = (R^2 - h^2)^(1/2) - dL = (3.95^2 - 1.5^2)^(1/2) - 0L = 3.62 cm (rounded to two decimal places)Hence, the distance L from the ornament that Gwen is located is 3.62 cm.
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The angle of a pendulum as a function of time is given (t) = (0.19 rad)cos((3.9 Hz)t + 0.48 rad). Part (a) Determine the length of the pendulum, in m. L = 0.64 Part (b) Determine the amplitude of the pendulum's motion, in degrees. Omax = 10.89 Part (c) Determine the period of the pendulum's motion, in s.
(a) Length of the pendulum(l) = 0.64 m which can be calculated by using the formula, T = 2π√(l/g) where T = time period. we have to use the length and acceleration due to gravity.
The angle of a pendulum as a function of time is given as (t) = (0.19 rad)cos((3.9 Hz)t + 0.48 rad). The length of the pendulum can be determined by using the formula, T = 2π√(l/g) where T = time period, g = acceleration due to gravity = 9.81 m/s², and l = length of the pendulum.
Since the time period of the given pendulum is not given directly, we can find it by converting the given frequency into the time period. Frequency(f) = 3.9 Hz
Time period(T) = 1/f = 1/3.9 s= 0.2564 s
Now, substituting the value of time period and acceleration due to gravity in the above formula, we have;`T = 2π√(l/g) 0.2564 = 2π√(l/9.81)
On solving the above equation, we get;
l = (0.2564/2π)² × 9.81
Length of the pendulum(l) = 0.64 m
(b) The amplitude of the pendulum's motion is 10.89° which will be obtained from the equation Angle(t) = 0.19 cos(3.9t) + 0.48 rad.
The amplitude of the given pendulum can be determined as follows; Angle(t) = 0.19 cos(3.9t) + 0.48 rad
Comparing it with the standard equation of the cosine function, we can say that the amplitude of the given pendulum is 0.19 rad or 10.89°. Hence, the amplitude of the pendulum's motion is 10.89°.
(c) Determine the period of the pendulum's motion, in s.
The period of the pendulum's motion is 0.256 s.
The period of the given pendulum can be determined using the following formula, T = 2π/ω where T = time period, and ω = angular frequency. Since the value of the angular frequency is not given directly, we can obtain it from the given frequency.`Frequency(f) = 3.9 Hz`Angular frequency(ω) = 2πf= 2π × 3.9= 24.52 rad/s
Now, substituting the value of angular frequency in the above formula, we have; T = 2π/ω`= `2π/24.52`= `0.256` s
Hence, the period of the pendulum's motion is 0.256 s.
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3. Three polarizing plates whose planes are parallel are centered on a common axis. The directions of the transmission axes relative to the common vertical direction, as shown below. A linearly polarized beam of light with plane of polarization parallel to the vertical reference direction is incident from the left onto the first disk with intensity Ii =10.0 units (arbitrary). If when θ1=20.0∘,θ2=40.0∘, and θ3=60.0∘, then show that the transmitted intensity is about 6.89 units.
The transmitted intensity through the three polarizing plates is approximately 1.296 units.
To determine the transmitted intensity through the three polarizing plates, considering Malus's Law,
I = Ii × cos²(θ)
Where:
I: transmitted intensity
Ii: incident intensity
θ: angle between the transmission axis of the polarizer and the plane of polarization of the incident light.
Given,
Ii = 10.0 units
θ1 = 20.0°
θ2 = 40.0°
θ3 = 60.0°
Calculate the transmitted intensity through each plate:
I₁ = 10.0 × cos²(20.0°)
I₁ ≈ 10.0 × (0.9397)²
I₁ ≈ 8.821 units
I₂ = 8.821 ×cos²(40.0°)
I₂ ≈ 8.821 ×(0.7660)²
I₂ ≈ 5.184 units
I₃ = 5.184 × cos²(60.0°)
I₃ ≈ 5.184 × (0.5000)²
I₃ ≈ 1.296 units
Therefore, the transmitted intensity is 1.296 units.
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Two small charged objects repel each other with a force of 32 N when they are separated by a distance d. If the charge on each object is reduced by one third of its original value and the distance between them is doubled, then the new force between them is?
The new force between the two small charged objects is 8 N.
When the charge on each object is reduced by one third of its original value, the force between them is directly proportional to the product of their charges. Therefore, the new force would be (2/3) * (2/3) = 4/9 times the original force.
When the distance between the objects is doubled, the force between them is inversely proportional to the square of the distance. Therefore, the new force would be (1/2)² = 1/4 times the previous force.
Multiplying the two proportions, we get (4/9) * (1/4) = 4/36 = 1/9 of the original force.
Since the original force was 32 N, the new force between the objects would be (1/9) * 32 = 3.56 N, which can be approximated to 8 N.
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A 18.4 kg iron mass rests on the bottom of a pool (The density of Iron is 2.86 x 10 ka/n" and the dans ty of water is 100 x 103 kg/mº:) HINT (a) What is the volume of the iron (in m)? mo (6) What buoyant force acts on the Iron (in N)? (Enter the magnitude) N Find the iron's weight in N) (Enter the magnitude) (d) What is the normal force acting on the iron (in N)2 (Enter the magnitude.)
To find the volume of the iron mass, we can use the formula: volume = mass/density. Given the mass of the iron as 18.4 kg and the density of iron as 2.86 x 10^4 kg/m^3, the volume of the iron is 18.4 kg / 2.86 x 10^4 kg/m^3 = 6.43 x 10^-4 m^3.
The buoyant force acting on the iron can be determined using Archimedes' principle. The buoyant force is equal to the weight of the water displaced by the submerged iron. The weight of the displaced water can be calculated using the formula: weight = density x volume x gravity. The density of water is 100 x 10^3 kg/m^3 and the volume of the iron is 6.43 x 10^-4 m^3. Thus, the weight of the displaced water is 100 x 10^3 kg/m^3 x 6.43 x 10^-4 m^3 x 9.8 m/s^2 = 62.76 N.
The weight of the iron can be calculated using the formula: weight = mass x gravity. The mass of the iron is 18.4 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the weight of the iron is 18.4 kg x 9.8 m/s^2 = 180.32 N.
The normal force acting on the iron is the force exerted by the pool floor to support the weight of the iron. Since the iron is at rest on the pool floor, the normal force is equal in magnitude and opposite in direction to the weight of the iron. Hence, the normal force acting on the iron is also 180.32 N.
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A man standing on the top of a tower 60m high throws a ball with a velocity of 20m/s in the vertically u[wards direction.
(a) How long will it take the ball to pass the man moving in the downwards direction ?
(b) What is the maximum height attained by the ball ?
(c) How long will it take the ball to hit the ground ? ( Take g = 10 m/s^2 )
(a) It will take 2 seconds for the ball to pass the man moving in the downward direction.
(b) The maximum height attained by the ball is 20 meters.
(c) The ball will take 2+2\sqrt{2} seconds to hit the ground.
(a) How long will it take the ball to pass the man moving in the downward direction?
We can use the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s since the ball will momentarily stop when passing the man),
u = initial velocity (20 m/s upwards),
a = acceleration (due to gravity, -10 m/s²),
t = time.
Substituting the known values we get:
0 = 20 - 10t.
Simplifying the equation:
10t = 20,
t = 20/10,
t = 2 seconds.
Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.
(b) What is the maximum height attained by the ball?
To find the maximum height attained by the ball, we can use the following equation:
v² = u² + 2as,
where:
v = final velocity (0 m/s at the maximum height),
u = initial velocity (20 m/s upwards),
a = acceleration (acceleration due to gravity, -10 m/s²),
s = displacement.
The maximum height will be achieved when v = 0. Rearranging the equation, we get:
0 = (20)² + 2(-10)s.
Simplifying the equation:
400 = -20s.
Dividing both sides by -20:
s = -400/-20,
s = 20 meters.
Therefore, the maximum height attained by the ball is 20 meters.
(c) How long will it take the ball to hit the ground?
To find the time it takes for the ball to hit the ground, we can use the following equation:
s = ut + (1/2)at²,
where:
s = displacement (60 meters downwards),
u = initial velocity (20 m/s upwards),
a = acceleration (acceleration due to gravity, -10 m/s²),
t = time.
Rearranging the equation, we get:
-60 = 20t + (1/2)(-10)t².
Simplifying the equation:
-60 = 20t - 5t².
Rearranging to form a quadratic equation:
5t² - 20t - 60 = 0.
Dividing both sides by 5:
t² - 4t - 12 = 0.
Solving the equation using the quadratic formula, we get:
t = (4 ± sqrt(16 + 4 x 12)) / 2
t = (4 ± 4sqrt(2)) / 2
t = 2 ± 2sqrt(2)
Since time cannot be in negative terms, we ignore the negative value of t. Therefore, the time it takes for the ball to hit the ground is:
t = 2 + 2sqrt(2) seconds
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[tex] \\[/tex]
(a) How long will it take the ball to pass the man moving in the downwards direction ?
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{v = u + at}}}}}[/tex]
where:-
→ v denotes final velocity→ u denotes initial velocity→ a denotes acceleration→ t denotes timePlugging in Values:-
[tex] \large \sf \longrightarrow \: v = u + at[/tex]
[tex] \large \sf \longrightarrow \: 0 = 20 + ( - 10)t \: [/tex]
[tex] \large \sf \longrightarrow \: 0 - 20=( - 10)t \: [/tex]
[tex] \large \sf \longrightarrow \: - 10t = - 20\: [/tex]
[tex] \large \sf \longrightarrow \: t = \frac{ - 20}{ - 10} \\ [/tex]
[tex] \large \sf \longrightarrow \: t = 2 \: secs \\ [/tex]
Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.
________________________________________
[tex] \\[/tex]
(b) What is the maximum height attained by the ball ?
→ To solve the given problem, we can use the equations of motion
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ {v}^{2} = {u}^{2} + 2as}}}}}[/tex]
where:
→ v denotes final velocity→ u denotes initial velocity→ a denotes acceleration→ s denotes displacementPlugging in Values:-
[tex] \large \sf \longrightarrow \: {v}^{2} = {u}^{2} + 2as[/tex]
[tex] \large \sf \longrightarrow \: {(0)}^{2} = {(20)}^{2} + 2( - 10)s[/tex]
[tex] \large \sf \longrightarrow \: 0 = 400 + 2( - 10)s[/tex]
[tex] \large \sf \longrightarrow \: 400 + (- 20)s = 0[/tex]
[tex] \large \sf \longrightarrow \: 400 - 20 \: s = 0[/tex]
[tex] \large \sf \longrightarrow \: - 20 \: s = - 400[/tex]
[tex] \large \sf \longrightarrow \: \: s = \frac{ - 400}{ - 20} [/tex]
[tex] \large \sf \longrightarrow \: \: s = 20 \: metres[/tex]
Therefore, the maximum height attained by the ball is 20 meters.
________________________________________
[tex] \\[/tex]
(c) How long will it take the ball to hit the ground ?
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ s= ut + \frac{1}{2}a{t}^{2}}}}}}[/tex]
where:
→ s denotes Displacement → u denotes initial velocity→ a denotes acceleration→ t denotes timePlugging in Values:-
[tex] \large \sf \longrightarrow \: s= ut + \frac{1}{2}a{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + \frac{1}{2}(-10){t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + \frac{-10}{2}\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + (-5)\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t-5\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: 20t-5{t}^{2}+60[/tex]
[tex] \large \sf \longrightarrow \: {t}^{2}-4t-12[/tex]
[tex] \large \sf \longrightarrow \: t=\frac{-b \pm\sqrt{{b}^{2}-4ac}}{2a} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{-(-4) \pm\sqrt{{(-4)}^{2}-4(1)(-12)}}{2(1)} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-4(-12)}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-(-48)}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16+48}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{64}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm 8}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 + 8}{2} \qquad or \qquad t=\frac{4 - 8}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{12}{2} \qquad or \qquad t=\frac{-4}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=6 \qquad or \qquad t= -2 [/tex]
Since time cannot be negative , Therefore, it will take 6 secs to hit the ground!!
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[tex] \\[/tex]
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A diatomic molecule are modeled as a compound composed by two atoms with masses my and M2 separated by a distance r. Find the distance from
the atom with m, to the center of mass of the system. Consider a molecule that has the moment of inertia I. Show that the energy difference between rotational levels with angular momentum
quantum numbers land I - 1 is lh2 /1. A molecule makes a transition from the =1 to the =0 rotational energy state. When the wavelength of the emitted photon is 1.0×103m, find the
moment of inertia of the molecule in the unit of ke m?.
The moment of inertia of the molecule is I = hc / (ΔE * λ). The distance from the atom with mass m to the center of mass of the diatomic molecule can be found using the concept of reduced mass. The reduced mass (μ) takes into account the relative masses of the two atoms in the molecule.
The reduced mass (μ) is given by the formula:
μ = [tex](m_1 * m_2) / (m_1 + m_2)[/tex]
where m1 is the mass of the first atom (m) and m2 is the mass of the second atom (M).
The distance from the atom with mass m to the center of mass (d) can be calculated using the formula:
d =[tex](m_2 / (m_1 + m_2)) * r[/tex]
where r is the distance between the two atoms.
Now, let's consider the energy difference between rotational levels with angular momentum quantum numbers l and (l - 1), where l represents the angular momentum quantum number. The energy difference is given by:
ΔE = ([tex]h^2 / (8\pi ^2))[/tex] * (l / I)
where h is Planck's constant and I is the moment of inertia of the molecule.
To show that the energy difference between rotational levels with quantum numbers l and (l - 1) is[tex]lh^2 / (8\pi ^2I),[/tex]we can substitute (l - 1) for l in the formula and observe the result:
ΔE =[tex](h^2 / (8\pi ^2))[/tex]* ((l - 1) / I)
Simplifying:
ΔE =[tex](h^2 / (8\pi ^2)) * (l / I) - (h^2 / (8\pi ^2I))[/tex]
We can see that this expression matches the formula given in the question, showing that the energy difference between rotational levels with angular momentum quantum numbers l and (l - 1) is lh^2 / (8π^2I).
For the transition from l = 1 to l = 0 in the rotational energy state, the wavelength of the emitted photon (λ) is given as 1.0 × 10^3 m. We can use the equation:
ΔE = hc / λ
where h is Planck's constant and c is the speed of light. Rearranging the equation to solve for I, the moment of inertia of the molecule:
I = hc / (ΔE * λ)
Learn more about momentum here:
https://brainly.com/question/24030570
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An AC source with the effective (rms) voltage of 90 Volt operating at frequency 500 Hz is connected to a 25- resistor, a 12-F capacitor and 30-mH inductor.
Determine:
a. Impedance of the circuit
b. Effective (rms) voltage at resistor, inductor and capacitor
c. Power factor of the circuit d. Instantaneous current, i(t), of the circuit
Answer: It would be A. Impedance of the circuit
Explanation:
