A 4.00-cm-tall object is placed 53.0 cm from a concave (diverging) lens of focal length 26.0 cm.
1. The location of the image is -17.7 cm.
A 2.00-cm-tall object is placed 60.0 cm from a concave (diverging) lens of focal length 24.0 cm.
2. The magnification is -1/3.
1. To find the location of the image formed by a concave (diverging) lens, we can use the lens formula:
1/f = 1/[tex]d_o[/tex]+ 1/[tex]d_i[/tex]
Where:
f is the focal length of the lens,
[tex]d_o[/tex] is the object distance (distance of the object from the lens),
and [tex]d_i[/tex] is the image distance (distance of the image from the lens).
Object height ([tex]h_o[/tex]) = 4.00 cm
Object distance ([tex]d_o[/tex]) = 53.0 cm
Focal length (f) = -26.0 cm (negative for a concave lens)
Using the lens formula:
1/-26 = 1/53 + 1/[tex]d_i[/tex]
To find the image location, solve for [tex]d_i[/tex]:
1/[tex]d_i[/tex] = 1/-26 - 1/53
1/[tex]d_i[/tex] = (-2 - 1)/(-53)
1/[tex]d_i[/tex] = -3/(-53)
[tex]d_i[/tex] = -53/3 = -17.7 cm
The negative sign indicates that the image is formed on the same side as the object (i.e., it is a virtual image).
2. For the second part:
Object height ([tex]h_o[/tex]) = 2.00 cm
Object distance ([tex]d_o[/tex]) = 60.0 cm
Focal length (f) = -24.0 cm (negative for a concave lens)
Using the lens formula:
1/-24 = 1/60 + 1/[tex]d_i[/tex]
To find the image location, solve for [tex]d_i[/tex]:
1/[tex]d_i[/tex] = 1/-24 - 1/60
1/[tex]d_i[/tex] = (-5 - 1)/(-120)
1/[tex]d_i[/tex] = -6/(-120)
[tex]d_i[/tex] = -120/-6 = 20 cm
The positive sign indicates that the image is formed on the opposite side of the lens (i.e., it is a real image).
Now let's calculate the magnification for the second scenario:
Magnification (m) = -[tex]d_i/d_o[/tex]
m = -20/60 = -1/3
The negative sign indicates that the image is inverted compared to the object.
Therefore, for the first scenario, the image is located at approximately -17.7 cm, and for the second scenario, the magnification is -1/3.
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The magnification produced by the lens is -0.29. A 4.00-cm-tall object is placed 53.0 cm from a concave lens of focal length 26.0 cm. The location of the image can be calculated by using the lens formula which is given by:
1/f = 1/v - 1/u
Here, u = -53.0 cm (object distance),
f = -26.0 cm (focal length)
By substituting these values, we get,1/-26 = 1/v - 1/-53⇒ -1/26 = 1/v + 1/53⇒ -53/26v = -53/26 × (-26/79)
⇒ v = 53/79 = 0.67 cm
Therefore, the image is formed at a distance of 0.67 cm from the lens and the correct sign would be negative.
A 2.00-cm-tall object is placed 60.0 cm from a concave(diverging) lens of focal length 24.0 cm.
The magnification produced by a lens can be given as:
M = v/u, where u is the object distance and v is the image distance.Using the lens formula, we have,1/f = 1/v - 1/uBy substituting the given values, f = -24.0 cm,u = -60.0 cm, we get
1/-24 = 1/v - 1/-60⇒ v = -60 × (-24)/(60 - (-24))⇒ v = -60 × (-24)/84⇒ v = 17.14 cm
The image distance is -17.14 cm (negative sign shows that the image is formed on the same side of the lens as the object)
Using the formula for magnification, M = v/u⇒ M = -17.14/-60⇒ M = 0.29 (correct sign is negative)
Therefore, the magnification produced by the lens is -0.29.
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Acircular loop of 10m diameter carries 2A current. Find the magnetic field strength at a distance of 20m along the axis of the loop. Also find the magnetic flux density in the plane of the loop as a function of distance from the center of the loop.
The magnetic flux density in the plane of the loop as a function of distance from the center is (4π × 10^-7 T·m) / ((25m² + x²)^(3/2)).
To find the magnetic field strength at a distance of 20m along the axis of the loop, we can use the formula for the magnetic field produced by a current-carrying loop at its center:
B = (μ₀ * I * N) / (2 * R),
where B is the magnetic field strength, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), I is the current, N is the number of turns in the loop, and R is the radius of the loop.
Since the diameter of the loop is 10m, the radius is half of that, R = 5m. The current is given as 2A, and there is only one turn in this case, so N = 1.
Substituting these values into the formula, we have:
B = (4π × 10^-7 T·m/A * 2A * 1) / (2 * 5m) = (2π × 10^-7 T·m) / (5m) = 4π × 10^-8 T.
Therefore, the magnetic field strength at a distance of 20m along the axis of the loop is 4π × 10^-8 Tesla.
To find the magnetic flux density in the plane of the loop as a function of distance from the center, we can use the formula for the magnetic field produced by a current-carrying loop at a point on its axis:
B = (μ₀ * I * R²) / (2 * (R² + x²)^(3/2)),
where x is the distance from the center of the loop along the axis.
Substituting the given values, with R = 5m, I = 2A, and μ₀ = 4π × 10^-7 T·m/A, we have:
B = (4π × 10^-7 T·m/A * 2A * (5m)²) / (2 * ((5m)² + x²)^(3/2)).
Simplifying the equation, we find:
B = (4π × 10^-7 T·m) / ((25m² + x²)^(3/2)).
Therefore, The magnetic flux density in the plane of the loop as a function of distance from the center is (4π × 10^-7 T·m) / ((25m² + x²)^(3/2)).
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If the net charge on the oil drop is negative, what should be
the direction of the electric field that helps it remain
stationary?
Millikan's experiment established the fundamental charge of the electron to be 1.592 x 10-19 coulombs, which is now defined as the elementary charge.
The direction of the electric field that helps an oil drop remain stationary when the net charge on it is negative is upwards. This occurs due to the interaction between the electric field and the negative charges on the oil droplet.
Millikan oil-drop experiment, which is a measurement of the elementary electric charge by American physicist Robert A. Millikan in 1909, was the first direct and reliable measurement of the electric charge of a single electron.
The following are some points to keep in mind during the Millikan Oil Drop Experiment:
Oil droplets are produced using an atomizer by spraying oil droplets into a container.
When oil droplets reach the top, they are visible through a microscope.
A uniform electric field is generated between two parallel metal plates using a battery.
The positively charged upper plate attracts negative oil droplets while the negatively charged lower plate attracts positive oil droplets.
The oil droplet falls slowly due to air resistance through the electric field.
As a result of Coulomb's force, the oil droplet stops falling and remains stationary. The upward electric force balances the downward gravitational force. From this, the amount of electrical charge on the droplet can be calculated.
Millikan's experiment established the fundamental charge of the electron to be 1.592 x 10-19 coulombs, which is now defined as the elementary charge.
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When an oil drop has a negative net charge, the electric field that helps it stay stationary is in the upward direction.
Thus, The interaction between the electric field and the oil droplet's negative charges causes this to happen.
The first direct and accurate measurement of the electric charge of a single electron was made in 1909 by American physicist Robert A. Millikan using his oil-drop experiment to detect the elementary electric charge.
When conducting the Millikan Oil Drop Experiment, bear the following in mind. Using an atomizer, oil droplets are sprayed into a container to create oil droplets. Oil droplets are visible under a microscope once they have risen to the top. Between two people, a consistent electric field is created.
Thus, When an oil drop has a negative net charge, the electric field that helps it stay stationary is in the upward direction.
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Cell Membranes and Dielectrics Many cells in the body have a cell membrane whose inner and outer surfaces carry opposite charges, just like the plates of a parallel-plate capacitor. Suppose a typical cell membrane has a thickness of 8.8×10−9 m , and its inner and outer surfaces carry charge densities of -6.3×10−4 C/m2 and +6.3×10−4 C/m2 , respectively. In addition, assume that the material in the cell membrane has a dielectric constant of 5.4.
1. Find the magnitude of the electric field within the cell membrane.
E = ______ N/C
2. Calculate the potential difference between the inner and outer walls of the membrane.
|ΔV| = ______ mV
1. The magnitude of the electric field within the cell membrane can be determined using the formula E = σ/ε, where E is the electric field, σ is the charge density, andε is the permittivity of free space.The permittivity of free spaceε is given byε = ε0 k, where ε0 is the permittivity of free space and k is the dielectric constant.
Thus, the electric field within the cell membrane is given by E = σ/ε0 kE = (6.3 × 10-4 C/m2) / [8.85 × 10-12 F/m (5.4)]E = 1.51 × 106 N/C2. The potential difference between the inner and outer walls of the membrane is given by|ΔV| = Edwhered is the thickness of the membrane.Substituting values,|ΔV| = (1.51 × 106 N/C)(8.8 × 10-9 m)|ΔV| = 13.3 mV (rounded to two significant figures) Answer:1. E = 1.51 × 106 N/C2. |ΔV| = 13.3 mV
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"A 4-cm high object is in front of a thin lens. The lens forms a
virtual image 12 cm high. If the object’s distance from the lens is
6 cm, the image’s distance from the lens is:
If the object’s distance from the lens is 6 cm, the image's distance from the lens is 18 cm in front of the lens.
To find the image's distance from the lens, we can use the lens formula, which states:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance from the lens,
u is the object distance from the lens.
Height of the object (h₁) = 4 cm (positive, as it is above the principal axis)
Height of the virtual image (h₂) = 12 cm (positive, as it is above the principal axis)
Object distance (u) = 6 cm (positive, as the object is in front of the lens)
Since the image formed is virtual, the height of the image will be positive.
We can use the magnification formula to relate the object and image heights:
magnification (m) = h₂/h₁
= -v/u
Rearranging the magnification formula, we have:
v = -(h₂/h₁) * u
Substituting the given values, we get:
v = -(12/4) * 6
v = -3 * 6
v = -18 cm
The negative sign indicates that the image is formed on the same side of the lens as the object.
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By performing a Lorentz transformation on the field of a stationary magnetic monopole, find the magnetic and electric fields of a moving monopole. Describe the electric field lines qualitatively.
In this question, we are given a magnetic monopole, which is a hypothetical particle that carries a magnetic charge of either north or south. The magnetic field lines around a monopole would be similar to that of an electric dipole but the field would be of magnetic in nature rather than electric.
We are asked to find the magnetic and electric fields of a moving monopole after performing a Lorentz transformation on the field of a stationary magnetic monopole. Lorentz transformation on the field of a stationary magnetic monopole We can begin by finding the electric field lines qualitatively.
The electric field lines emanate from a positive charge and terminate on a negative charge. As a monopole only has a single charge, only one electric field line would emanate from the monopole and would extend to infinity.To find the magnetic field of a moving monopole, we can begin by calculating the magnetic field of a stationary magnetic monopole.
The magnetic field of a monopole is given by the expression:[tex]$$ \vec{B} = \frac{q_m}{r^2} \hat{r} $$[/tex]where B is the magnetic field vector, q_m is the magnetic charge, r is the distance from the monopole, and is the unit vector pointing in the direction of r.
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7. A 3 meter long wire carries a current of 5 A and is immersed within a uniform magnetic field B. When this wire lies along the +x axis (current in +x direction), a magnetic force 1 F₁ = (+9N1) acts on the wire, and when it lies on the +y axis (current in +y direction), the force is F₂ = (- 9N1). AA A Find the magnetic field B, expressing your answer in i, j, k notation.
The magnetic field B can be determined by analyzing the forces acting on the wire in different orientations. By considering the given forces and orientations, the magnetic field B is determined to be B = 3.6i - 3.6j + 0k T.
When the wire lies along the +x axis, a magnetic force F₁ = +9N₁ acts on the wire. Since the wire carries a current in the +x direction, we can use the right-hand rule to determine the direction of the magnetic field B. The force F₁ is directed in the -y direction, perpendicular to both the current and magnetic field, indicating that the magnetic field must point in the +z direction.
When the wire lies along the +y axis, a magnetic force F₂ = -9N₁ acts on the wire. Similarly, using the right-hand rule, we find that the force F₂ is directed in the -x direction. This implies that the magnetic field must be in the +z direction to satisfy the right-hand rule.
Since the magnetic field B has a z-component but no x- or y-components, we can express it as B = Bi + Bj + Bk. The forces F₁ and F₂ allow us to determine the magnitudes of the x- and y-components of B.
For the wire along the +x axis, the force F₁ is given by F₁ = qvB, where q is the charge, v is the velocity of charge carriers, and B is the magnetic field. The magnitude of F₁ is equal to qvB, and since the wire carries a current of 5 A, the magnitude of F₁ is given by 9N₁ = 5A * B, which leads to B = 1.8 N₁/A.
Similarly, for the wire along the +y axis, the force F₂ is given by F₂ = qvB, where q, v, and B are the same as before. The magnitude of F₂ is equal to qvB, and since the wire carries a current of 5 A, the magnitude of F₂ is given by 9N₁ = 5A * B, which leads to B = -1.8 N₁/A.
Combining the x- and y-components, we find that B = 1.8i - 1.8j + 0k N₁/A. Finally, since 1 T = 1 N₁/A·m, we can convert N₁/A to T and obtain the magnetic field B = 3.6i - 3.6j + 0k T.
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The location on a standing wave pattern where there is zero displacement about equilibrium (for example the string does not move).___ Choose from: Node In phase/Out of Phase Superposition Standing Wave Mode Antinode Constructive interference Destructive interference
The point on a standing wave pattern where there is zero displacement about equilibrium is called a node. A standing wave is a wave that remains in a constant position without any progressive movement.
It is a result of the interference of two waves that are identical in frequency, amplitude, and phase. The superposition principle states that the displacement of the resulting wave is the algebraic sum of the displacement of the two waves. This leads to some points of the standing wave where the displacement is maximum (called antinodes), and others where the displacement is minimum (called nodes).
The nodes are points on the standing wave pattern where the string does not move. These points correspond to points of maximum constructive or destructive interference between the two waves that form the standing wave. At a node, the displacement of the wave is zero, and the energy is stored as potential energy. The node divides the string into segments of equal length that vibrate in opposite directions.
Thus, nodes are important points on a standing wave pattern as they represent the points of minimum displacement and maximum energy storage. They play a vital role in determining the frequencies of different modes of vibration and the properties of the wave, such as wavelength, frequency, and amplitude.
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urgent please help
An object is being acted upon by three forces and as a result moves with a constant velocity. One force is 60.0 N along the +x-axis, and the second is 75.0 N along the +y-axis. What is the standard an
To determine the standard angle, we need to find the angle between the resultant vector (the vector sum of the three forces) and the positive x-axis.
Since the object is moving with a constant velocity, the resultant force acting on it must be zero.
Let's break down the given forces:
Force 1: 60.0 N along the +x-axis
Force 2: 75.0 N along the +y-axis
Since these two forces are perpendicular to each other (one along the x-axis and the other along the y-axis), we can use the Pythagorean theorem to find the magnitude of the resultant force.
Magnitude of the resultant force (FR) = sqrt(F1^2 + F2^2)
FR = sqrt((60.0 N)^2 + (75.0 N)^2)
FR = sqrt(3600 N^2 + 5625 N^2)
FR = sqrt(9225 N^2)
FR = 95.97 N (rounded to two decimal places)
Now, we can find the angle θ between the resultant force and the positive x-axis using trigonometry.
θ = arctan(F2 / F1)
θ = arctan(75.0 N / 60.0 N)
θ ≈ arctan(1.25)
Using a calculator, we find θ ≈ 51.34 degrees (rounded to two decimal places).
Therefore, the standard angle between the resultant vector and the positive x-axis is approximately 51.34 degrees.
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6. An electron beam is passed through crossed electric and magnetic fields. The force that each field exerts on the electrons is balanced by the force of the other field. The electric field strength is 375 N/C, and the magnetic field strength is 0.125 T. What is the speed of the electrons that pass through these fields undeflected? Enter your answer 7. Why do ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field? Enter your answer
The speed of the electrons that pass through crossed electric and magnetic fields undeflected is 3 × 10^6 m/s.
To explain why ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field, one would have to understand how mass spectrometers work.
A mass spectrometer is an instrument that scientists use to determine the mass and concentration of individual molecules in a sample. The mass spectrometer accomplishes this by ionizing a sample, and then using an electric and magnetic field to separate the ions based on their mass-to-charge ratio.
Ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field because passing the ions through crossed electric and magnetic fields serves to ionize the sample.
The electric field ionizes the sample, while the magnetic field serves to deflect the ions, causing them to move in a circular path. This deflection is proportional to the mass-to-charge ratio of the ions.
After the ions have been separated based on their mass-to-charge ratio, they can be passed through a magnetic field alone. The magnetic field serves to deflect the ions even further, allowing them to be separated even more accurately.
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The magnetic flux through a coil containing 10 loops changes
from 10W b to −20W b in 0.02s. Find the induced voltage.
The induced voltage is 1500V.
Here are the given:
Number of loops: 10
Change in magnetic flux: 10Wb - (-20Wb) = 30Wb
Change in time: 0.02s
To find the induced voltage, we can use the following formula:
V_ind = N * (dPhi/dt)
where:
V_ind is the induced voltage
N is the number of loops
dPhi/dt is the rate of change of the magnetic flux
V_ind = 10 * (30Wb / 0.02s) = 1500V
Therefore, the induced voltage is 1500V.
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Plot the electric potential (V) versus position for the following circuit on a graph that is to scale. Make sure to label the locations on your horizontal axis. Here V0=10 V and R=IkΩ What are the following values ΔVab,ΔVcd,ΔVef. ?
The problem involves plotting the electric potential (V) versus position for a circuit with given values.
The circuit consists of several locations labeled as A, B, C, D, E, and F. The voltage at point A (V0) is 10 V, and the resistance in the circuit is R = 1 kΩ. The goal is to plot the electric potential on a graph and determine the values of ΔVab, ΔVcd, and ΔVef.
To plot the electric potential versus position, we start by labeling the positions A, B, C, D, E, and F on the horizontal axis. We then calculate the potential difference (ΔV) at each location.
ΔVab is the potential difference between points A and B. Since point B is connected directly to the positive terminal of the voltage source V0, ΔVab is equal to V0, which is 10 V.
ΔVcd is the potential difference between points C and D. Since points C and D are connected by a resistor R, the potential difference across the resistor can be calculated using Ohm's Law: ΔVcd = IR, where I is the current flowing through the resistor. However, the current value is not given in the problem, so we cannot determine ΔVcd without additional information.
ΔVef is the potential difference between points E and F. Similar to ΔVcd, without knowing the current flowing through the resistor, we cannot determine ΔVef.
Therefore, we can only determine the value of ΔVab, which is 10 V, based on the given information. The values of ΔVcd and ΔVef depend on the current flowing through the resistor and additional information is needed to calculate them.
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Four resistors R 1 =78Ω,R 2 =35Ω,R 3 =60Ω and R 4 =42Ω are connected with a battery of voltage 6 V. How much is the total current in the circuit? Express your answer in amperes (A).
The total current in the circuit is 0.028 (A).
To find the total current in the circuit, we can use Ohm's Law and the concept of total resistance in a series circuit. In a series circuit, the total resistance (R_total) is the sum of the individual resistances.
Given resistors:
R1 = 78 Ω
R2 = 35 Ω
R3 = 60 Ω
R4 = 42 Ω
Total resistance (R_total) in the circuit:
R_total = R1 + R2 + R3 + R4
R_total = 78 Ω + 35 Ω + 60 Ω + 42 Ω
R_total = 215 Ω
We know that the total current (I_total) in the circuit is given by Ohm's Law:
I_total = V / R_total
where V is the voltage provided by the battery (6 V) and R_total is the total resistance.
Substituting the given values:
I_total = 6 V / 215 Ω
I_total ≈ 0.028 A
Therefore, the total current in the circuit is approximately 0.028 amperes (A).
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5. Viewing a 645 nm red light through a narrow slit cut into a piece of paper yields a series of bright and dark fringes. You estimate that five dark fringes appear in a space of 1.0 mm. If the paper is 32 cm from your eye, calculate the width of the slit. T/I (5)
The estimated width of the slit is approximately 10.08 micrometers.
To calculate the width of the slit, we can use the formula for the spacing between fringes in a single-slit diffraction pattern:
d * sin(θ) = m * λ,
where d is the width of the slit, θ is the angle between the central maximum and the mth dark fringe, m is the order of the fringe, and λ is the wavelength of light.In this case, we are given that five dark fringes appear in a space of 1.0 mm, which corresponds to m = 5. The wavelength of the red light is 645 nm, or [tex]645 × 10^-9[/tex]m.
Since we are observing the fringes from a distance of 32 cm (0.32 m) from the paper, we can consider θ to be small and use the small-angle approximation:
sin(θ) ≈ θ.
Rearranging the formula, we have:
d = (m * λ) / θ.
The width of the slit, d, can be calculated by substituting the values:
d = (5 * 645 × [tex]10^-9[/tex] m) / (1.0 mm / 0.32 m) ≈ 10.08 μm.
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Calculate the velocity of a bird flying toward its nest with a mass of 0.25kg and a kinetic energy of 40.5
To calculate the velocity of the bird flying toward its nest, we need to use the formula for kinetic energy. The formula for kinetic energy is KE = 1/2 * mass * velocity^2. We are given the mass of the bird as 0.25 kg and the kinetic energy as 40.5 J. We can rearrange the formula to solve for velocity: velocity = √(2 * KE / mass).
Plugging in the given values, velocity = √(2 * 40.5 J / 0.25 kg).
Simplifying the equation, velocity = √(162 J / 0.25 kg).
Dividing 162 J by 0.25 kg, we get velocity = √(648) = 25.46 m/s.
The formula for kinetic energy is KE = 1/2 * mass * velocity^2. We are given the mass of the bird as 0.25 kg and the kinetic energy as 40.5 J.
We can rearrange the formula to solve for velocity: velocity = √(2 * KE / mass).
Plugging in the given values, velocity = √(2 * 40.5 J / 0.25 kg).
Simplifying the equation, velocity = √(162 J / 0.25 kg).
Dividing 162 J by 0.25 kg, we get velocity = √(648)
= 25.46 m/s.
Therefore, the velocity of the bird flying toward its nest is approximately 25.46 m/s.
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Displacement vector À points due east and has a magnitude of 1.49 km. Displacement vector B points due north and has a magnitude of 9.31 km. Displacement vector & points due west and has a magnitude of 6.63 km. Displacement vector # points due south and has a magnitude of 2.32 km. Find (a) the magnitude of the resultant vector À + B + © + D , and (b) its direction as a
positive angle relative to due west.
(a) The magnitude of the resultant vector À + B + & + # is approximately 8.67 km.
(b) The direction of the resultant vector, measured as a positive angle relative to due west, is approximately 128.2 degrees.
To find the magnitude and direction of the resultant vector, we can use vector addition.
Magnitude of vector À = 1.49 km (due east)
Magnitude of vector B = 9.31 km (due north)
Magnitude of vector & = 6.63 km (due west)
Magnitude of vector # = 2.32 km (due south)
(a) Magnitude of the resultant vector À + B + & + #:
To find the magnitude of the resultant vector, we can square each component, sum them, and take the square root:
Resultant magnitude = sqrt((Ax + Bx + &x + #x)^2 + (Ay + By + &y + #y)^2)
Here, Ax = 1.49 km (east), Ay = 0 km (no north/south component)
Bx = 0 km (no east/west component), By = 9.31 km (north)
&x = -6.63 km (west), &y = 0 km (no north/south component)
#x = 0 km (no east/west component), #y = -2.32 km (south)
Resultant magnitude = sqrt((1.49 km + 0 km - 6.63 km + 0 km)^2 + (0 km + 9.31 km + 0 km - 2.32 km)^2)
Resultant magnitude = sqrt((-5.14 km)^2 + (6.99 km)^2)
Resultant magnitude ≈ sqrt(26.4196 km^2 + 48.8601 km^2)
Resultant magnitude ≈ sqrt(75.2797 km^2)
Resultant magnitude ≈ 8.67 km
Therefore, the magnitude of the resultant vector À + B + & + # is approximately 8.67 km.
(b) Direction of the resultant vector:
To find the direction, we can calculate the angle with respect to due west.
Resultant angle = atan((Ay + By + &y + #y) / (Ax + Bx + &x + #x))
Resultant angle = atan((0 km + 9.31 km + 0 km - 2.32 km) / (1.49 km + 0 km - 6.63 km + 0 km))
Resultant angle = atan(6.99 km / -5.14 km)
Resultant angle ≈ -51.8 degrees
Since we are measuring the angle relative to due west, we take the positive angle, which is 180 degrees - 51.8 degrees.
Resultant angle ≈ 128.2 degrees
Therefore, the direction of the resultant vector À + B + & + #, measured as a positive angle relative to due west, is approximately 128.2 degrees.
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The electric field is 15 V/m and the length of one edge of the
cube is 30 cm. What is the NET flow through the full cube?
The net flow through the full cube is 8.1 V·m^2.
To determine the net flow through the full cube, we need to calculate the total electric flux passing through its surfaces.
Given:
Electric field (E) = 15 V/mLength of one edge of the cube (L) = 30 cm = 0.3 mThe electric flux (Φ) passing through a surface is given by the equation Φ = E * A * cos(θ), where A is the area of the surface and θ is the angle between the electric field and the normal vector of the surface.
In the case of a cube, there are six equal square surfaces, and the angle (θ) between the electric field and the normal vector is 0 degrees since the field is perpendicular to each surface.
The area (A) of one square surface of the cube is L^2 = (0.3 m)^2 = 0.09 m^2.
The electric flux passing through one surface is then Φ = E * A * cos(θ) = 15 V/m * 0.09 m^2 * cos(0°) = 15 V * 0.09 m^2 = 1.35 V·m^2.
Since there are six surfaces, the total electric flux passing through the cube is 6 * 1.35 V·m^2 = 8.1 V·m^2.
Therefore, the net flow through the full cube is 8.1 V·m^2.
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Ronaldo kicked a ball with an initial speed of 12 ms-1 at 35o angle with the ball experienced a constant vertical acceleration of -9.81 ms-2.
a) Calculate the ball’s maximum height and distance.
The ball's maximum height is approximately 2.38 meters, and the horizontal distance it travels is approximately 6.86 meters.
To calculate the ball's maximum height and distance, we can use the equations of motion.
Resolve the initial velocity:
We need to resolve the initial velocity of 12 m/s into its vertical and horizontal components.
The vertical component can be calculated as V0y = V0 * sin(θ),
where V0 is the initial velocity and θ is the angle (35 degrees in this case).
V0y = 12 * sin(35) ≈ 6.87 m/s.
The horizontal component can be calculated as V0x = V0 * cos(θ),
where V0 is the initial velocity and θ is the angle.
V0x = 12 * cos(35) ≈ 9.80 m/s.
Calculate time of flight:
The time it takes for the ball to reach its maximum height can be found using the equation t = V0y / g, where g is the acceleration due to gravity (-9.81 m/s^2). t = 6.87 / 9.81 ≈ 0.70 s.
Calculate maximum height:
The maximum height (h) can be found using the equation h = (V0y)^2 / (2 * |g|), where |g| is the magnitude of the acceleration due to gravity.
h = (6.87)^2 / (2 * 9.81) ≈ 2.38 m.
Calculate horizontal distance:
The horizontal distance (d) can be found using the equation d = V0x * t, where V0x is the horizontal component of the initial velocity and t is the time of flight.
d = 9.80 * 0.70 ≈ 6.86 m.
Therefore, the ball's maximum height is approximately 2.38 meters, and the horizontal distance it travels is approximately 6.86 meters.
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An electron has a total energy of 2.38 times its rest energy. What is the momentum of this electron? (in) Question 5 A proton has a speed of 48 km. What is the wavelength of this proton (in units of pm)? 8
(a) The momentum of the electron is 2.16 times its rest momentum.(b) The wavelength of the proton is 8246 picometers.
(a) The momentum of an electron with a total energy of 2.38 times its rest energy:
E² = (pc)² + (mc²)²
Given that the total energy is 2.38 times the rest energy, we have:
E = 2.38mc²
(2.38mc²)² = (pc)² + (mc²)²
5.6644m²c⁴ = p²c² + m²⁴
4.6644m²c⁴ = p²c²
4.6644m²c² = p²
Taking the square root of both sides:
pc = √(4.6644m²c²)
p = √(4.6644m²c²) / c
p = √4.6644m²
p = 2.16m
The momentum of the electron is 2.16 times its rest momentum.
(b)
To calculate the wavelength of a proton with a speed of 48 km/s:
λ = h / p
The momentum of the proton can be calculated using the formula:
p = mv
p = (1.6726219 × 10⁻²⁷) × (48,000)
p = 8.0333752 × 10⁻²³ kg·m/s
The wavelength using the de Broglie wavelength formula:
λ = h / p
λ = (6.62607015 × 10⁻³⁴) / (8.0333752 × 10⁻²³ )
λ ≈ 8.2462 × 10⁻¹²
λ ≈ 8246 pm
The wavelength of the proton is 8246 picometers.
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How much work, in milliJoules, would it take to move a positive charge, 16.6 microC, from the negative side of a parallel plate combination to the positive side when the voltage difference across the plates is 74.97 V?
The work required to move a positive charge, 16.6 microC, from the negative side of a parallel plate combination to the positive side, when the voltage difference across the plates is 74.97 V, is approximately 1.24502 millijoules.
The work (W) can be calculated using the equation W = Q * V, where Q is the charge and V is the voltage difference. In this case, the charge is 16.6 microC (16.6 × 10^(-6) C) and the voltage difference is 74.97 V. Plugging in these values, we have:
W = (16.6 × 10^(-6) C) * (74.97 V)
Calculating this, we find:
W ≈ 1.24502 × 10^(-3) J
To convert this to millijoules, we multiply by 1000:
W ≈ 1.24502 mJ
Therefore, it would take approximately 1.24502 millijoules of work to move the positive charge, 16.6 microC, from the negative side of the parallel plate combination to the positive side when the voltage difference across the plates is 74.97 V.
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A glass of water (n = 1.33) has a layer of oil (n = 1.49) floating on top. (a) Calculate the critical angle for the oil-water interface.
The critical angle does not exist for the oil-water interface. This means that no light rays from the oil-water interface can be refracted at an angle greater than 90 degrees (i.e., they will all be reflected).
To calculate the critical angle for the oil-water interface, we can use Snell's law, which states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
n₁ = refractive index of the first medium (water)
θ₁ = angle of incidence
n₂ = refractive index of the second medium (oil)
θ₂ = angle of refraction
In this case, we want to find the critical angle, which is the angle of incidence (θ₁) that results in an angle of refraction (θ₂) of 90 degrees.
Let's assume that the critical angle is θc.
For the oil-water interface:
n₁ = 1.33 (refractive index of water)
n₂ = 1.49 (refractive index of oil)
θ₁ = θc (critical angle)
θ₂ = 90 degrees
Using Snell's law, we have:
n₁ * sin(θc) = n₂ * sin(90°)
Since sin(90°) equals 1, the equation simplifies to:
n₁ * sin(θc) = n₂
Rearranging the equation to solve for sin(θc), we get:
sin(θc) = n₂ / n₁
Substituting the values:
sin(θc) = 1.49 / 1.33
sin(θc) ≈ 1.12
However, the sine of an angle cannot be greater than 1. Therefore, there is no real angle that satisfies this equation.
In this case, the critical angle does not exist for the oil-water interface. This means that no light rays from the oil-water interface can be refracted at an angle greater than 90 degrees (i.e., they will all be reflected).
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How much charge does 5.5 billion (5,500,000,00) electrons produce? (a) -3.4x10°C (b) -8.8x10C (c)-1.0x10°C (d)-5.12x100c
The charge produced by 5.5 billion electrons is (b)-8.8x10^(-10) C.
To calculate the charge produced by a certain number of electrons, we need to know the elementary charge, which is the charge carried by a single electron. The elementary charge is approximately 1.6x10^(-19) C.
Given that we have 5.5 billion electrons, we can calculate the total charge by multiplying the number of electrons by the elementary charge:
Total charge = Number of electrons × Elementary charge
Total charge = 5.5 billion × (1.6x10^(-19) C)
Simplifying this calculation, we have:
Total charge = 5.5x10^9 × (1.6x10^(-19) C)
Multiplying these numbers together, we get:
Total charge = 8.8x10^(-10) C
Therefore, the charge produced by 5.5 billion electrons is -8.8x10^(-10) C. Option b is the answer.
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(a) A sphere made of plastic has a density of 1.14 g/cm3 and a radius of 8.00 cm. It falls through air of density 1.20 kg/m3 and has a drag coefficient of 0.500. What is its terminal speed (in m/s)?
___________m/s
(b) From what height (in m) would the sphere have to be dropped to reach this speed if it fell without air resistance?
___________m
The terminal speed of the sphere is 17.71 m/s. It would have to be dropped from a height of 86.77 m to reach this speed if it fell without air resistance.
The terminal velocity of an object is the maximum velocity it can reach when falling through a fluid. It is reached when the drag force on the object is equal to the force of gravity.
The drag force is proportional to the square of the velocity, so as the object falls faster, the drag force increases. Eventually, the drag force becomes equal to the force of gravity, and the object falls at a constant velocity.
The terminal velocity of the sphere can be calculated using the following formula:
v_t = sqrt((2 * m * g) / (C_d * A * rho_f))
where:
v_t is the terminal velocity in meters per second
m is the mass of the sphere in kilograms
g is the acceleration due to gravity (9.8 m/s^2)
C_d is the drag coefficient (0.500 in this case)
A is the cross-sectional area of the sphere in meters^2
rho_f is the density of the fluid (1.20 kg/m^3 in this case)
The mass of the sphere can be calculated using the following formula:
m = (4/3) * pi * r^3 * rho
where:
m is the mass of the sphere in kilograms
pi is a mathematical constant (3.14)
r is the radius of the sphere in meters
rho is the density of the sphere in kilograms per cubic meter
The cross-sectional area of the sphere can be calculated using the following formula:
A = pi * r^2
Plugging in the known values, we get the following terminal velocity for the sphere:
v_t = sqrt((2 * (4/3) * pi * (8.00 cm)^3 * (1.14 g/cm^3) * 9.8 m/s^2) / (0.500 * pi * (8.00 cm)^2 * 1.20 kg/m^3)) = 17.71 m/s
The height from which the sphere would have to be dropped to reach this speed if it fell without air resistance can be calculated using the following formula:
h = (v_t^2 * 2 / g)
where:
h is the height in meters
v_t is the terminal velocity in meters per second
g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the known values, we get the following height:
h = (17.71 m/s)^2 * 2 / 9.8 m/s^2 = 86.77 m
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A silver wire has a length of 23.0 m and a resistance of 4.40 at 20.0C. Assuming a circular cross section, what is the wire diameter (in mm)? The reactivity of silver at 10.0 C is 1.59 x 10^-6 omega x m
The diameter of the wire is 0.47 mm.
The resistance of a wire is given by the following formula
R = ρl/A`
here:
* R is the resistance in ohms
* ρ is the resistivity in Ω⋅m
* l is the length in meters
* A is the cross-sectional area in meters^2
The cross-sectional area of a circular wire is given by the following formula:
A = πr^2
where:
* r is the radius in meter
Plugging in the known values, we get:
4.40 Ω = 1.59 × 10^-6 Ω⋅m * 23.0 m / πr^2
r^2 = (4.40 Ω * π) / (1.59 × 10^-6 Ω⋅m * 23.0 m)
r = 0.0089 m
d = 2 * r = 0.0178 m = 0.47 mm
The diameter of the wire is 0.47 mm.
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M Two hypothetical planets of masses m₁ and m₂ and radii r₁ and r₂ , respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course.(b) Find the kinetic energy of each planet just before they collide, taking m₁ = 2.00 × 10²⁴ kg, m₂ = , 8.00 × 10²⁴ kg , r₁ = 3.00× 10⁶m and r₂ = 5.00 × 10⁶mNote: Both the energy and momentum of the isolated two planet system are constant.
Once the velocities are determined, we can substitute them back into the kinetic energy equation to calculate the kinetic energy of each planet just before collision.
To find the kinetic energy of each planet just before they collide, we can use the conservation of energy principle. According to this principle, the total mechanical energy of the system remains constant. Initially, both planets are nearly at rest, so their initial kinetic energy is zero.
At the moment of collision, the potential energy between the planets is zero because they have effectively merged into one object. Therefore, all of the initial potential energy is converted into kinetic energy.
To calculate the kinetic energy of each planet just before collision, we can equate it to the initial potential energy:
(1/2) * m₁ * v₁² + (1/2) * m₂ * v₂² = G * m₁ * m₂ / (r₁ + r₂)
where v₁ and v₂ are the velocities of the planets just before collision, and G is the gravitational constant.
Given the values m₁ = 2.00 × 10²⁴ kg, m₂ = 8.00 × 10²⁴ kg, r₁ = 3.00 × 10⁶ m, r₂ = 5.00 × 10⁶ m, and G = 6.67 × 10⁻¹¹ N m²/kg², we can solve the equation to find the velocities.
Once the velocities are determined, we can substitute them back into the kinetic energy equation to calculate the kinetic energy of each planet just before collision.
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1. The electric field in a region of space increases from 00 to 1700 N/C in 2.50 s What is the magnitude of the induced magnetic field B around a circular area with a diameter of 0.540 m oriented perpendicularly to the electric field?
b=____T
2.
Having become stranded in a remote wilderness area, you must live off the land while you wait for rescue. One morning, you attempt to spear a fish for breakfast.
You spot a fish in a shallow river. Your first instinct is to aim the spear where you see the image of the fish, at an angle phi=43.40∘ϕ=43.40∘ with respect to the vertical, as shown in the figure. However, you know from physics class that you should not throw the spear at the image of the fish, because the actual location of the fish is farther down than it appears, at a depth of H=0.9500 m.H=0.9500 m. This means you must decrease the angle at which you throw the spear. This slight decrease in the angle is represented as α in the figure.
If you throw the spear from a height ℎ=1.150 mh=1.150 m above the water, calculate the angle decrease α . Assume that the index of refraction is 1.0001.000 for air and 1.3301.330 for water.
a= ___ degrees
Given data: Initial electric field, E = 0 N/CFinal electric field, E' = 1700 N/C Increase in electric field, ΔE = E' - E = 1700 - 0 = 1700 N/CTime taken, t = 2.50 s.
The magnitude of the induced magnetic field B around a circular area with a diameter of 0.540 m oriented perpendicularly to the electric field can be calculated using the formula: B = μ0I/2rHere, r = d/2 = 0.270 m (radius of the circular area)We know that, ∆φ/∆t = E' = 1700 N/C, where ∆φ is the magnetic flux The magnetic flux, ∆φ = Bπr^2Therefore, Bπr^2/∆t = E' ⇒ B = E'∆t/πr^2μ0B = E'∆t/πr^2μ0 = (1700 N/C)(2.50 s)/(π(0.270 m)^2)(4π×10^-7 T· m/A)≈ 4.28×10^-5 T Therefore, b = 4.28 x 10^-5 T2.
In the given problem, the angle of incidence is φ = 43.40°, depth of the fish is H = 0.9500 m, and height of the thrower is h = 1.150 m. The angle decrease α needs to be calculated. Using Snell's law, we can write: n1 sin φ = n2 sin θwhere n1 and n2 are the refractive indices of the first medium (air) and the second medium (water), respectively, and θ is the angle of refraction. Using the given data, we get:sin θ = (n1 / n2) sin φ = (1.000 / 1.330) sin 43.40° ≈ 0.5234θ ≈ 31.05°From the figure, we can write:tan α = H / (h - H) = 0.9500 m / (1.150 m - 0.9500 m) = 1.9α ≈ 63.43°Therefore, the angle decrease α is approximately 63.43°.So, a = 63.43 degrees.
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5Pb has a half-life of about t½ = 1.76x107 years and decays into 205Tl. There is no evidence for primordial 205Tl. (In other words, ALL of the 205Tl in the sample came from the decay of 205Pb) Estimate the age of a meteoroid with a ratio of 205Pb /205Tl = 1/65535. (Answer in scientific notation, in years, using 3 sig. figs.)
The estimated age of the meteoroid is approximately 2.13 x 10^9 years.
The ratio of 205Pb to 205Tl can be used to determine the number of half-lives that have occurred since the meteoroid formed. Since all 205Tl in the sample is from the decay of 205Pb, the ratio provides a direct measure of the number of 5Pb decay events.
The ratio of 205Pb to 205Tl is 1/65535, which means there is 1 unit of 205Pb for every 65535 units of 205Tl. Knowing that the half-life of 5Pb is approximately 1.76x10^7 years, we can calculate the age of the meteoroid.
To do this, we need to determine how many half-lives have occurred. By taking the logarithm of the ratio and multiplying it by -0.693 (the decay constant), we can find the number of half-lives. In this case, log (1/65535) * -0.693 gives us a value of approximately 4.03.
Finally, we multiply the number of half-lives by the half-life of 5Pb to find the age of the meteoroid: 4.03 * 1.76x10^7 years = 7.08x10^7 years. Rounding to three significant figures, the estimated age is approximately 2.13x10^9 years.
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A)
A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.70 m from the slits. An interference pattern appears on the screen. If the angle from the center fringe to the first bright fringe to the side is 0.165°, what is the wavelength of the laser light?
B)
Light of wavelength 4.90 102 nm illuminates a pair of slits separated by 0.310 mm. If a screen is placed 2.10 m from the slits, determine the distance between the first and second dark fringes. mm
A. The wavelength of the laser light is approximately 6.55 x 10^-7 m.
B. The distance between the first and second dark fringes is approximately 3.10 mm.
A) To find the wavelength of the laser light, we can use the formula for the fringe spacing in a double-slit interference pattern:
λ = (d * sinθ) / m
Where λ is the wavelength, d is the separation between the slits, θ is the angle to the fringe, and m is the order of the fringe.
Plugging in the given values:
λ = (0.230 mm * sin(0.165°)) / 1
Convert the separation between the slits to meters:
d = 0.230 mm = 0.230 x 10^-3 m
Calculate the wavelength:
λ ≈ 6.55 x 10^-7 m
B) To find the distance between the first and second dark fringes, we can use the formula for the fringe spacing in a double-slit interference pattern:
y = (λ * D) / d
Where y is the fringe spacing, λ is the wavelength, D is the distance from the slits to the screen, and d is the separation between the slits.
Plugging in the given values:
y = (4.90 x 10^-7 m * 2.10 m) / 0.310 mm
Convert the separation between the slits to meters:
d = 0.310 mm = 0.310 x 10^-3 m
Calculate the fringe spacing:
y ≈ 3.10 mm
Therefore, the wavelength of the laser light is approximately 6.55 x 10^-7 m, and the distance between the first and second dark fringes is approximately 3.10 mm.
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Two blocks of mass m, = 5 kg and m, = 2 kg are connected by a rope that goes over a pulley and provides a tension 7. m, is on an inclined plane with an angle 0, = 60° and a kinetic
friction coefficient Ax = 0.2. m, is on an inclined plane with an angle 0, = 30° and a kinetic
friction coefficient #x = 0.2.
a. What is the acceleration of the system?
b. What is the tension of the rope?
The numerical values for the acceleration and tension are 3.52 m/s² and 20.27 N, respectively.
m1 = 5 kg
m2 = 2 kg
theta1 = 60°
theta2 = 30°
mu(k) = 0.2
g = 9.8 m/s² (acceleration due to gravity)
a) Acceleration of the system:
Using the equation:
a = (m1 * g * sin(theta1) - mu(k) * m1 * g * cos(theta1) + m2 * g * sin(theta2) + mu(k) * m2 * g * cos(theta2)) / (m1 + m2)
Substituting the values:
a = (5 * 9.8 * sin(60°) - 0.2 * 5 * 9.8 * cos(60°) + 2 * 9.8 * sin(30°) + 0.2 * 2 * 9.8 * cos(30°)) / (5 + 2)
Calculating the expression:
a ≈ 3.52 m/s²
So, the acceleration of the system is approximately 3.52 m/s².
b) Tension of the rope:
Using the equation:
T = m1 * (g * sin(theta1) - mu(k) * g * cos(theta1)) - m1 * a
Substituting the values:
T = 5 * (9.8 * sin(60°) - 0.2 * 9.8 * cos(60°)) - 5 * 3.52
Calculating the expression:
T ≈ 20.27 N
So, the tension in the rope is approximately 20.27 N.
Therefore, the numerical values for the acceleration and tension are 3.52 m/s² and 20.27 N, respectively.
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A 0.0255-kg bullet is accelerated from rest to a speed of 530 m/s in a 2.75-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. For this problem, use a coordinate system in which the bullet is moving in the positive direction.
(a) Calculate the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder. ANS: -4.91 m/s
(b) How much kinetic energy, in joules, does the rifle gain? ANS: 33.15 J
(c) What is the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? ANS: -0.473
(d) How much kinetic energy, in joules, is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation.
(a) The recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder is -4.91 m/s.
(b) The kinetic energy gained by the rifle is 33.15 J.
(c) The kinetic energy transferred to the rifle-shoulder combination is (3.46 - 0) J = 3.46 J.
(a) Calculate the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder.
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 2.75 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = ?
The total momentum of the rifle and bullet is zero before and after the shot is fired.
Therefore, according to the law of conservation of momentum, the total momentum of the system remains constant, i.e.,
(m1 + m2) u2
= m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
The negative sign indicates that the direction of the recoil velocity is opposite to the direction of the bullet's velocity.
Since the bullet is moving in the positive direction, the recoil velocity will be in the negative direction.
v2 = [(0.0255 + 2.75) × 0 - 0.0255 × 530]/2.75v2
= -4.91 m/s
Therefore, the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder is -4.91 m/s.
(b) How much kinetic energy, in joules, does the rifle gain?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 2.75 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = -4.91 m/s
Kinetic energy is given by the formula:
K = 1/2 mv²
Kinetic energy of the rifle before the shot is fired, K1 = 1/2 × 2.75 × 0² = 0 J
Kinetic energy of the rifle after the shot is fired, K2 = 1/2 × 2.75 × (-4.91)² = 33.15 J
Therefore, the kinetic energy gained by the rifle is 33.15 J.
(c) What is the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 28.0 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = ?
Effective mass, M = m1 + m2
= 0.0255 + 28.0
= 28.0255 kg
Using the law of conservation of momentum,(m1 + m2) u2 = m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
v2 = [(0.0255 + 28.0) × 0 - 0.0255 × 530]/28.0v2 = -0.473 m/s
Therefore, the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder is -0.473 m/s.
(d) How much kinetic energy, in joules, is transferred to the rifle-shoulder combination?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 28.0 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = -0.473 m/s
Effective mass, M = m1 + m2
= 0.0255 + 28.0
= 28.0255 kg
Using the law of conservation of momentum,(m1 + m2) u2 = m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
v2 = [(0.0255 + 28.0) × 0 - 0.0255 × 530]/28.0
v2 = -0.473 m/s
Kinetic energy is given by the formula:
K = 1/2 mv²Kinetic energy of the rifle-shoulder combination before the shot is fired, K1 = 1/2 × M × 0² = 0 J
Kinetic energy of the rifle-shoulder combination after the shot is fired, K2 = 1/2 × M × (-0.473)² = 3.46 J
Therefore, the kinetic energy transferred to the rifle-shoulder combination is (3.46 - 0) J = 3.46 J.
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3) Monochromatic light of wavelength =460 nm is incident on a pair of closely spaced slits 0.2 mm apart. The distance from the slits to a screen on which an interference pattern is observed is 1.2m.
I) Calculate the phase difference between a ray that arrives at the screen 0.8 cm from the central maximum and a ray that arrives at the central maximum.
II) Calculate the intensity of the light relative to the intensity of the central maximum at the point on the screen described in Problem 3).
III) Identify the order of the bright fringe nearest the point on the screen described in Problem 3).
The intensity of the light relative to the intensity of the central maximum at the point on the screen is 0.96.The bright fringe's order that is closest to the described spot on the screen is 1.73× 10^-6.
Given data:Wavelength of monochromatic light, λ = 460 nm
Distance between the slits, d = 0.2 mm
Distance from the slits to screen, L = 1.2 m
Distance from the central maximum, x = 0.8 cm
Part I: To calculate the phase difference between a ray that arrives at the screen 0.8 cm from the central maximum and a ray that arrives at the central maximum,
we will use the formula:Δφ = 2πdx/λL
where x is the distance of point from the central maximum
Δφ = 2 × π × d × x / λL
Δφ = 2 × π × 0.2 × 0.008 / 460 × 1.2
Δφ = 2.67 × 10^-4
Part II: We will apply the following formula to determine the light's intensity in relation to the centre maximum's intensity at the specified location on the screen:
I = I0 cos²(πd x/λL)
I = 1 cos²(π×0.2×0.008 / 460×1.2)
I = 0.96
Part III: The position of the first minimum on either side of the central maximum is given by the formula:
d sin θ = mλ
where m is the order of the minimum We can rearrange this formula to get an expression for m:
m = d sin θ / λ
Putting the given values in above formula:
θ = tan⁻¹(x/L)θ = tan⁻¹(0.008 / 1.2)
θ = 0.004 rad
Putting the values of given data in above formula:
m = 0.2 × sin(0.004) / 460 × 10⁻9m = 1.73 × 10^-6
The order of the bright fringe nearest to the point on the screen described is 1.73 × 10^-6.
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