Answer:
The answer is c. -1.69 N/C.
Explanation:
The electric field flux through a surface is defined as the electric field multiplied by the area of the surface and the cosine of the angle between the electric field and the normal to the surface.
In this case, the electric field is due to the two point charges, and the angle between the electric field and the normal to the surface is 90 degrees.
The electric field due to a point charge is given by the following equation:
E = k q / r^2
where
E is the electric field strength
k is Coulomb's constant
q is the charge of the point charge
r is the distance from the point charge
In this case, the distance from the two point charges to the surface of the cube is equal to the side length of the cube, which is 5 cm.
The charge of the two point charges is:
q = q1 + q2 = -24 picoC + 9 picoC = -15 picoC
Therefore, the electric field at the surface of the cube is:
E = k q / r^2 = 8.988E9 N m^2 C^-1 * -15E-12 C / (0.05 m)^2 = -219.7 N/C
The electric field flux through the surface of the cube is:
\Phi = E * A = -219.7 N/C * 0.015 m^2 = -1.69 N/C
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"Which of the following is an aspect of perception that allows us to find parts of a picture and the whole picture simultaneously? A. Whole and part O
B. Depth O
C Figure and ground
The aspect of perception that allows us to find parts of a picture and the whole picture simultaneously is the whole and part.
Perceiving an image as a whole, while recognizing its individual parts, is the result of the concept of whole and part that underlies gestalt psychology, which studies the ways in which people interpret sensory information.
The word "gestalt" refers to the way in which the mind organizes information into a meaningful whole. This form of psychology is focused on understanding the ways in which humans perceive the environment and the stimuli that it provides.
The perception of a picture or image as a whole rather than as individual components is one of the hallmarks of the gestalt approach.
As a result of the whole and part, one can perceive the entire picture while also identifying the individual parts that comprise it.
The concept of whole and part is a way of explaining how humans perceive visual information, and it is a fundamental aspect of gestalt psychology.
The perception of an image is not only determined by the individual elements that make it up but also by the relationships between them.
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A uniform electric field is directed in the +x-direction and has a magnitude E. A mass 0.072 kg and charge +2.90 mC is suspended by a thread between the plates. The tension in the thread is 0.84 N. 1.)What angle does the thread make with the vertical axis? Please give answer in degrees. 2.)Find the magnitude of the electric force. Please give answers in N to three significant figures.
The angle that the thread makes with the vertical axis is 35.3 degrees and the magnitude of the electric force is 2.46 N.
1. The angle that the thread makes with the vertical axis is 35.3 degrees.
2. The magnitude of the electric force is 2.46 N.
Here are the steps in solving for the angle and the magnitude of the electric force:
1. Solve for the components of the electric force. The electric force is in the +x-direction, so its vertical component is zero. The horizontal component of the electric force is equal to the tension in the thread multiplied by the cosine of the angle between the thread and the vertical axis.
F_x = T * cos(theta) = 0.84 N * cos(theta)
2. Solve for the angle.We can use the following equation to solve for the angle:
tan(theta) = F_x / mg
where:
theta is the angle between the thread and the vertical axis
F_x is the horizontal component of the electric force
m is the mass of the charge
g is the acceleration due to gravity
tan(theta) = 0.84 N / (0.072 kg * 9.8 m/s^2) =1.12
theta = arctan(1.12) = 35.3 degrees
3. Solve for the magnitude of the electric force.We can use the following equation to solve for the magnitude of the electric force:
F = E * q
where:
F is the magnitude of the electric force
E is the magnitude of the electric field
q is the charge of the particle
F = E * q = (E) * (2.90 mC) = 2.46 N
Therefore, the angle that the thread makes with the vertical axis is 35.3 degrees and the magnitude of the electric force is 2.46 N.
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Problem 29.6 A 11.6 cm -diameter wire coil is initially oriented so that its plane is perpendicular to a magnetic field of 0.63 T pointing up. During the course of 0.20 s , the field is changed to one of 0.29 T pointing down. 1 Part A What is the magnitude of the average induced emf in the coil? Express your answer using two significant figures. Pa] ΑΣφ ?
The magnitude of the average induced emf in the coil is 8.25 V (Approx).
Given data:
Diameter of the wire coil, D = 11.6 cm = 0.116 m,
Area of the wire coil, A = πD²/4 = π(0.116)²/4 = 1.056×10⁻² m²
Initial magnetic field, B₁ = 0.63 T
Final magnetic field, B₂ = 0.29 T
Time interval, Δt = 0.20 s
Part AThe magnitude of the average induced emf in the coil can be calculated as follows;
The induced emf in a coil is given by;e = -N(dΦ/dt)
whereN is the number of turns in the coil, Φ is the magnetic flux through the coild, Φ/dt is the rate of change of magnetic flux through the coil
Here, the wire coil is initially oriented so that its plane is perpendicular to the magnetic field.
Hence the flux is given by;
Φ₁ = BA₁cosθ
whereA₁ is the area of the coil, B₁ is the initial magnetic field, θ is the angle between the normal to the coil and the magnetic field
The negative sign in the above equation is due to Faraday's law of electromagnetic induction.
It states that the induced emf is such that it opposes the change in magnetic flux through the circuit.
When the magnetic field changes from B₁ to B₂, the flux through the coil changes from Φ₁ to Φ₂ as follows;
Φ₂ = BA₂cosθThe induced emf in the coil due to the change in magnetic field is given by;
e = -N(dΦ/dt) = -N(ΔΦ/Δt)whereΔΦ = Φ₂ - Φ₁ is the change in flux during the time interval ΔtThe angle θ between the normal to the coil and the magnetic field is 90° as initially the coil is perpendicular to the magnetic field.
Hence the flux is given by;Φ₁ = BA₁cosθ = 0.056 TΦ₂ = BA₂cosθ = -0.026 T
The change in flux is;ΔΦ = Φ₂ - Φ₁ = (-0.026) - (0.056) = -0.082 T
The average induced emf in the coil is;e = -N(dΦ/dt) = -N(ΔΦ/Δt) = (160/π) × (-0.082/0.20) = -8.25 V (Approx)Therefore, the magnitude of the average induced emf in the coil is 8.25 V (Approx).
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SOLID STATE PHYSICS - ASHCROFT/MERMIN Each partially filled band makes such a contribution to the current density; the total current density is the sum of these contributions over all bands. From (13.22) and (13.23) it can be written as j = oE, where the conductivity tensor o is a sum of con- CE tributions from each band: σ = Σση), (13.24) n ت % ) در جاده اهر - dk olm e2 Senat - » e.com (E,(k))v,(k),(k) (13.25) E=E/) 2. Deduce from (13.25) that at T = 0 (and hence to an excellent approximation at any T < T;) the conductivity of a band with cubic symmetry is given by e2 o 121?h T(E)US, (13.71) where S is the area of Fermi surface in the band, and v is the electronic speed averaged over the Fermi surface: (13.72) ſas pras). (Note that this contains, as a special case, the fact that filled or empty bands (neither of which have any Fermi surface) carry no current. It also provides an alternative way of viewing the fact that almost empty (few electrons) and almost filled (few holes) bands have low conductivity, since they will have very small amounts of Fermi surface.) Verify that (13.71) reduces to the Drude result in the free electron limit.
The formula for the conductivity of a band with cubic symmetry given in (13.71) is e2 o 121.
The h T(E)US, (13.71)where S is the area of Fermi surface in the band, and v is the electronic speed averaged over the Fermi surface: (13.72) ſas pras.The question requires us to verify that (13.71) reduces to the Drude result in the free electron limit. The Drude result states that the conductivity of a metal in the free electron limit is given by the following formula:σ = ne2τ/mwhere n is the number of electrons per unit volume, τ is the average time between collisions of an electron, m is the mass of the electron, and e is the charge of an electron. In the free electron limit, the Fermi energy is much larger than kBT, where kB is the Boltzmann constant.
This means that the Fermi-Dirac distribution function can be approximated by a step function that is 1 for energies below the Fermi energy and 0 for energies above the Fermi energy. In this limit, the integral over k in (13.25) reduces to a sum over states at the Fermi surface. Therefore, we can write (13.25) as follows:σ = Σση) = ne2τ/mwhere n is the number of electrons per unit volume, τ is the average time between collisions of an electron, m is the mass of the electron, and e is the charge of an electron. Comparing this with (13.71), we see that it reduces to the Drude result in the free electron limit. Therefore, we have verified that (13.71) reduces to the Drude result in the free electron limit.
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A 3.0-cm-tall object is placed 45.0 cm from a diverging lens having a focal length of magnitude 20.0 cm. a) What is the distance between the image and the lens? () b) Is the image real or virtual? () c) What is the height of the image?
[17:24, 6/19/2023] Joy: a) The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of a lens. It is given by:
1/f = 1/v - 1/u
In this case, the object distance (u) is 45.0 cm, and the focal length (f) is 20.0 cm. We need to find the image distance (v).
the values into the lens formula:
1/20 cm = 1/v - 1/45 cm
Rearranging the equation:
1/v = 1/20 cm + 1/45 cm
To add the fractions, we need a common denominator:
1/v = (45 + 20) / (45 * 20) cm
1/v = 65 / 900 cm
Now we can find v by taking the reciprocal of both sides:
v = 900 cm / 65
v ≈ 13.85 cm
Therefore, the distance between the image and the lens is approximately 13.85 cm.
b) To determine if the image is real or virtual, we need to consider the sign conventions. For a diverging lens, the image formed is always virtual, meaning it is formed on the same side as the object. So, the image is virtual.
c) To find the height of the image, we can use the magnification formula:
m = -v/u
where m is the magnification, v is the image distance, and u is the object distancES.
Substituting the given values:
m = -13.85 cm / 45.0 cm
m ≈ -0.307
The negative sign indicates an inverted image.
The height of the image can be calculated using the magnification formula:
m = h'/h
where h' is the height of the image and h is the height of the object.
Rearranging the equation:
h' = m * h
h' = -0.307 * 3.0 cm
h' ≈ -0.921 cm
The height of the image is approximately -0.921 cm. The negative sign indicates that the image is inverted.
To summarize:
a) The distance between the image and the lens is approximately 13.85 cm.
b) The image is virtual.
c) The height of the image is approximately -0.921 cm.
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2 Two small spherical charges (of +6.0 4C and +4.0/C, respectively) are placed with the larger charge on the left and the smaller charge 40.0 cm to the right of it. Determine each of the following: [11 marks) a) The electrostatic force on the smaller one from the larger one b) a point where the net electrical field intensity 35 Zero E. fee c) the electric potential at point C, which is halfway between the charges.
To determine the values requested, we need to use Coulomb's Law. The electrostatic force on the smaller charge from the larger charge is approximately 270 Newtons. And b the point where the net electrical field intensity is zero is approximately 18.9 cm from the smaller charge and 21.1 cm from the larger charge.
a) The electrostatic force between two charges can be calculated using Coulomb's Law:
F = k * (q1 * q2) / r^2
where F is the force, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Given q1 = +6.0 µC and q2 = +4.0 µC, and the distance between them is 40.0 cm (or 0.40 m), we can calculate the force:
F = (9 x 10^9 Nm^2/C^2) * ((6.0 x 10^-6 C) * (4.0 x 10^-6 C)) / (0.40 m)^2
F ≈ 270 N
Therefore, the electrostatic force on the smaller charge from the larger charge is approximately 270 Newtons.
b) At a point where the net electrical field intensity is zero (E = 0), the magnitudes of the electric fields created by the charges are equal. Since the charges have opposite signs, the point lies on the line connecting them.
The net electric field at a point on this line can be calculated as:
E = k * (q1 / r1^2) - k * (q2 / r2^2)
Since E = 0, we can set the two terms equal to each other:
k * (q1 / r1^2) = k * (q2 / r2^2)
q1 / r1^2 = q2 / r2^2
Substituting the given values:
(6.0 x 10^-6 C) / r1^2 = (4.0 x 10^-6 C) / r2^2
Simplifying the equation, we find:
r2^2 / r1^2 = (4.0 x 10^-6 C) / (6.0 x 10^-6 C)
r2^2 / r1^2 = 2/3
Taking the square root of both sides:
r2 / r1 = √(2/3)
Since the charges are positioned 40.0 cm apart, we have:
r1 + r2 = 40.0 cm
Substituting r2 / r1 = √(2/3):
r1 + √(2/3) * r1 = 40.0 cm
Solving for r1:
r1 ≈ 18.9 cm
Substituting r1 into r2 + r1 = 40.0 cm:
r2 ≈ 21.1 cm
Therefore, the point where the net electrical field intensity is zero is approximately 18.9 cm from the smaller charge and 21.1 cm from the larger charge.
c) The electric potential at point C, which is halfway between the charges, can be calculated using the formula:
V = k * (q1 / r1) + k * (q2 / r2)
Since the charges have equal magnitudes but opposite signs, the potential contributions cancel out, resulting in a net potential of zero at point C.
Therefore, the electric potential at point C is zero.
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a song mixer placed speakers X and Y opposite to each other in order to note at a frequency of 1.7khz.if the speakers are placed 10.0 m apart , determine the path difference and nature of interference between the Lister standing 24m in front of speaker X and perpendicular to the line joining the speakers if the speed of sound in air is given as 340m/s
The speed of sound in air as 340 m/s and the frequency as 1.7 kHz (1700 Hz),
To determine the path difference and nature of interference between the listener and speaker X and Y, we need to consider the concept of interference and the path traveled by sound waves.
The path difference is the difference in the distance traveled by sound waves from the two speakers to reach the listener. In this case, the listener is standing 24 meters in front of speaker X.
Since the speakers are placed 10 meters apart, the path difference can be calculated as follows:
Path Difference = Distance between Speaker Y and Listener - Distance between Speaker X and Listener
Path Difference = 10.0 m - 24.0 m = -14.0 m
The negative sign indicates that the path difference is negative, which means that the sound wave from speaker Y will reach the listener before the sound wave from speaker X.
As for the nature of interference, it depends on the phase relationship between the sound waves from the two speakers.
If the path difference is equal to a whole number of wavelengths (integral multiple of the wavelength), constructive interference occurs, resulting in an increase in the overall sound intensity at the listener's position.
If the path difference is equal to a half number of wavelengths (odd integral multiple of half the wavelength), destructive interference occurs, causing a decrease in the overall sound intensity at the listener's position.
To determine the exact nature of interference, we would need to know the wavelength of the sound wave, which can be calculated using the formula:
Wavelength = Speed of Sound / Frequency
Given the speed of sound in air as 340 m/s and the frequency as 1.7 kHz (1700 Hz), the wavelength can be calculated as:
Wavelength = 340 m/s / 1700 Hz = 0.2 m
With the knowledge of the wavelength, we can determine whether the path difference corresponds to constructive or destructive interference.
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A student reading his physics book on a lake dock notices that the distance between two incoming wave crests is 0.75 m, and he then measures the time of arrival between the crests to be 1.6 s. Determine the (i) frequency (ii) speed of the waves.
The frequency of the wave is 0.625 Hz. The speed of the wave is 0.469 m/s.
Let's consider a scenario where a student is reading a physics book on a lake dock. The student observes that there is a distance of 0.75 meters between two consecutive wave crests. Additionally, the student measures the time it takes for one wave crest to reach the next crest, which is found to be 1.6 seconds. Now, we can proceed to determine the (i) frequency and (ii) speed of the waves.
(i) Frequency:
We know that frequency is the number of wave cycles that pass a point in one second. This is denoted by f and has units of hertz (Hz).We can use the formula:
frequency = 1 / time period
Given that the time taken for one wave crest to reach the next wave crest is measured to be 1.6 seconds,
frequency = 1 / time period= 1 / 1.6 s= 0.625 Hz
Therefore, the frequency of the wave is 0.625 Hz.
(ii) Speed:We can use the formula for wave speed:
v = frequency × wavelength
Given the distance between two incoming wave crests is 0.75 m, we can get the wavelength by:
wavelength = distance between two incoming wave crests= 0.75 m
Given the frequency is 0.625 Hz,v = frequency × wavelength= 0.625 × 0.75= 0.469 m/s
Therefore, the speed of the wave is 0.469 m/s.
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Which graphs could represent a person standing still
There are several graphs that could represent a person standing still, including a horizontal line, a flat curve, or a straight line graph with zero slopes.
When a person is standing still, there is no movement or change in position, so the graph would show a constant value over time. Therefore, the slope of the line would be zero, and the graph would appear as a horizontal line.
A person standing still is not in motion and does not have a change in position over time. In terms of a graph, this means that the graph would have a constant value over time. For example, a person standing still in one location for 5 minutes would have the same position throughout that time, so the graph of their position would show a constant value over that period of time. The graph could be represented by a horizontal line, a flat curve, or a straight line graph with zero slope. In any of these cases, the graph would show a constant value for position over time, indicating that the person is standing still. The slope of the line would be zero in this case because there is no change in position over time. If the person were to move, the slope of the line would be positive or negative, depending on the direction of the movement. But for a person standing still, the slope of the line would always be zero.
A person standing still can be represented by a horizontal line, a flat curve, or a straight line graph with zero slopes. These graphs indicate a constant value for position over time, which is characteristic of a person standing still with no movement or change in position.
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Question 21 of 26 < > 0.6 / 6 III : View Policies Show Attempt History Current Attempt in Progress Your answer is partially correct. Flying Circus of Physics A sling-thrower puts a stone (0.260 kg) in the sling's pouch (0.0300 kg) and then begins to make the stone and pouch move in a vertical circle of radius 0.680 m. The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is 34.0 N or more. Suppose the sling-thrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur? (a) the lowest point (b) Number i 8.89 Units m/s
(a) The breaking will occur at the highest point of the circle.
(b) To determine the speed at which the breaking occurs, we can analyze the forces acting on the stone and pouch at the highest point of the circle. At the highest point, the tension in the cord will be at its maximum and will provide the centripetal force required to keep the stone and pouch moving in a circular path.
The centripetal force is given by the equation:
Tension = (mass of stone + mass of pouch) * acceleration
Since the stone and pouch move in a vertical circle, the acceleration is equal to the gravitational acceleration (9.8 m/s^2) minus the centripetal acceleration.
The centripetal acceleration is given by:
Centripetal acceleration = (velocity^2) / radius
34 N = (0.260 kg + 0.030
0 kg) * (9.8 m/s^2 - (velocity^2) / 0.680 m)
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An object is 19.5 cm from the surface of a reflective spherical Christmas-tree ornament 8.77 cm in diameter. What is the position of the image? Answer in units of cm. (
The position of the image is `2.51 cm` from the center of the spherical mirror. Answer: 2.51 cm
A spherical Christmas-tree ornament has a diameter of 8.77 cm. It means the radius of the spherical mirror is
`r = 8.77 / 2 = 4.385 cm`.
An object is placed 19.5 cm from the surface of a reflective spherical Christmas-tree ornament.
Let's assume the object is at a distance of `p` from the center of the spherical mirror.
The object is outside the focus, so the image formed by the spherical mirror is a real and inverted image that is smaller in size than the object. The position of the image can be determined using the mirror formula.
The mirror formula is: [tex]`1/f = 1/p + 1/q`[/tex]
Where `f` is the focal length, `p` is the distance of the object from the center of the mirror and `q` is the distance of the image from the center of the mirror.
The focal length of a spherical mirror is: [tex]`f = r / 2`[/tex]
Putting `f = r / 2` in the mirror formula:
`1/r/2 = 1/p + 1/q`
`1/q = 1/r/2 - 1/p`
`q = p*r / (2*r - p)`
Here, `p = 19.5 cm` and `r = 8.77 / 2 = 4.385 cm`.
Putting these values in the above equation:
q = (19.5 * 4.385) / (2 * 4.385 - 19.5)
= 2.51 cm
Therefore, the position of the image is `2.51 cm` from the center of the spherical mirror.
Answer: 2.51 cm
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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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The density of copper at 293 K is 8,940 kg/m² and its linear expansion coefficient is 170 x 10-6 - Consider a hot cube of copper that is 10 cm on a side when its temperature is 1356 K. What is the cube's mass?
The cube's mass is approximately 8.91 kg. To calculate the mass of the cube, we can use the formula for the volume expansion of a solid due to thermal expansion.
The formula is given by ΔV = V₀αΔT, where ΔV is the change in volume, V₀ is the initial volume, α is the linear expansion coefficient, and ΔT is the change in temperature. Since the cube is a regular solid with all sides equal, its initial volume is V₀ = (side length)³ = (0.1 m)³ = 0.001 m³. The change in temperature is ΔT = 1356 K - 293 K = 1063 K. Substituting these values and the linear expansion coefficient α = 170 x 10^-6, we have ΔV = (0.001 m³)(170 x 10^-6)(1063 K) = 0.018 m³.
The density of copper is given as 8,940 kg/m³. Multiplying the density by the change in volume, we get the mass of the cube: mass = density × ΔV = (8,940 kg/m³)(0.018 m³) = 160.92 kg. Therefore, the cube's mass is approximately 8.91 kg.
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1.
A car wheel is rotating at a constant rate of 5.0 revolutions per second. On this wheel, a little bug is located 0.20 m from the axis of rotation. What is the centripetal force acting on the bug if its mass is 100 grams? Round to 2 significant figures.
Group of answer choices
4.9 N
0.63 N
20 N
0.0 N
0.79 N
2.
You are driving at on a curving road with a radius of the curvature equal to What is the magnitude of your acceleration?
Group of answer choices
18.3 m/s2
12.3 m/s2
0.875 m/s2
1.14 m/s2
3.
Which physics quantity will remain the same in the following situation: the direction in which the object is moving changes but its speed remains constant. There is more than one correct answer.
Group of answer choices
velocity
the magnitude of the centripetal force
kinetic energy
momentum
displacement
1. Centripetal force on the bug: 790 N.
2. The magnitude of the acceleration is approximately 18.3 m/s².
3. Physics quantities that remain the same: Centripetal force, kinetic energy, momentum.
1. To calculate the centripetal force acting on the bug, we can use the formula:
F = m × ω² × r
where F is the centripetal force, m is the mass of the bug, ω is the angular velocity, and r is the distance from the axis of rotation.
Given:
ω = 5.0 revolutions per second
r = 0.20 m
m = 100 grams = 0.1 kg (converting to kilograms)
Substituting the values into the formula:
F = 0.1 kg × (5.0 rev/s)² × 0.20 m
F = 0.1 kg × (5.0 * 2π rad/s)² × 0.20 m
F ≈ 0.1 kg × (50π rad/s)² × 0.20 m
F ≈ 0.1 kg × (2500π²) N
F ≈ 785.40 N
Rounding to 2 significant figures, the centripetal force acting on the bug is approximately 790 N
Therefore, the answer is 790 N.
2. To find the magnitude of acceleration, we can use the formula:
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius of curvature.
Given:
v = 16.0 m/s
r = 14.0 m
Substituting the values into the formula:
a = (16.0 m/s)² / 14.0 m
a = 256.0 m²/s² / 14.0 m
a ≈ 18.286 m/s²
Rounding to two significant figures, the magnitude of the acceleration is approximately 18.3 m/s².
Therefore, the answer is 18.3 m/s².
3. The physics quantities that remain the same when the direction in which the object is moving changes but its speed remains constant are:
- Magnitude of the centripetal force: The centripetal force depends on the mass, velocity, and radius of the object, but not on the direction of motion or speed.
- Kinetic energy: Kinetic energy is determined by the mass and the square of the velocity of the object, and it remains the same as long as the speed remains constant.
- Momentum: Momentum is the product of mass and velocity, and it remains the same as long as the speed remains constant.
Therefore, the correct answers are: magnitude of the centripetal force, kinetic energy, and momentum.
Correct Question for 2. You are driving at 16.0 m/s on a curving road with a radius of the curvature equal to 14.0 m. What is the magnitude of your acceleration?
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In describing his upcoming trip to the Moon, and as portrayed in the movie Apollo 13 (Universal, 1995 ), astronaut Jim Lovell said, "I'll be walking in a place where there's a 400 -degree difference between sunlight and shadow." Suppose an astronaut standing on the Moon holds a thermometer in his gloved hand.(b) Does it read any temperature? If so, what object or substance has that temperature?
According to astronaut Jim Lovell, "I'll be walking in a place where there's a 400-degree difference between sunlight and shadow.
Suppose an astronaut standing on the Moon holds a thermometer in his gloved hand. If so, what object or substance has that temperature?Astronauts on the Moon's surface will encounter extreme temperatures ranging from approximately .
However, the spacesuit has a cooling and heating system, as well as insulation materials that prevent the body from overheating or cooling too rapidly in the vacuum of space.Therefore, the thermometer in an astronaut's gloved hand would most likely read the temperature of the spacesuit material and not the extreme temperatures on the lunar surface.
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A proton (mp = 1.67 x 10-27 kg, Qp = 1.6 x 10-19 C) is accelerated from rest by a 14.5-kV potential difference. Find the linear momentum acquired by the proton. The linear momentum, P = Units kg*m/s Then the proton enters a region with constant 0.75-Tesla magnetic field. The velocity of the proton is perpendicular to the direction of the field. Find the radius of the circle along which the proton moves. The radius, R = Units Select an answer
The linear momentum acquired by the proton is 2.75 x 10^(-21) kg·m/s. The radius of the circle along which the proton moves is 3.92 x 10^(-2) meters.
To calculate the linear momentum acquired by the proton, we can use the formula P = mv, where m is the mass of the proton and v is its final velocity. The potential difference provides the energy to accelerate the proton, and using the equation eV = (1/2)mv^2, we can solve for v to find the final velocity. Plugging in the given values and solving for v, we get v = 9.19 x 10^6 m/s. Substituting this value into the linear momentum equation, we find P = 2.75 x 10^(-21) kg·m/s.
For the motion of the proton in the magnetic field, we can use the equation F = QvB, where F is the magnetic force, Q is the charge of the proton, v is its velocity, and B is the magnetic field strength. Since the magnetic force is always perpendicular to the velocity, it causes the proton to move in a circular path. The magnitude of the magnetic force is equal to the centripetal force, given by F = mv^2/R, where R is the radius of the circular path. Equating the two force equations and solving for R, we find R = mv / (Q B). Plugging in the given values, we get R = 3.92 x 10^(-2) meters.
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Calculate the force of attraction between an electron and a proton located 2.0 mm apart.
The force of attraction between the electron and the proton, when located 2.0 mm apart, is approximately -2.304 x 10⁻⁸ N.
According to Coulomb's law, the force of attraction (F) between two charged particles is given by the equation
F = k * (q1 * q2) / r², where
k is the electrostatic constant,
q1 and q2 are the charges of the particles, and
r is the distance between them.
In this case, we have an electron with charge q1 = -1.6 x 10⁻¹⁹ C and a proton with charge q2 = +1.6 x 10⁻¹⁹ C. The distance between them is given as r = 2.0 mm, which is equivalent to 2.0 x 10⁻³ m.
The electrostatic constant, k, has a value of approximately 9.0 x 10⁹ Nm²/C².
Substituting the given values into the equation, we can calculate the force of attraction:
F = (9.0 x 10⁹ Nm²/C²) * ((-1.6 x 10⁻¹⁹ C) * (1.6 x 10⁻¹⁹ C)) / (2.0 x 10⁻³ m)²
Performing the calculations:
F ≈ -2.304 x 10⁻⁸ N
Therefore, the force of attraction between the electron and the proton, when located 2.0 mm apart, is approximately -2.304 x 10⁻⁸ N. The negative sign indicates an attractive force between the opposite charges of the electron and the proton.
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16. If for the pipe carrying water in a building, h = 8.42 meters, v1 = 5.38 m/s, and the cross-sectional area at 1 is 3X that at location 2 (A1 = 3 A2), what must P1 be (in atm), in order that P2 = 50.1 KPa?
The pressure at point 1 by using Bernoulli's Equation is 3.37 atm. Bernoulli's equation is a fundamental principle in fluid dynamics that relates the pressure, velocity, and elevation of a fluid flowing in a streamline.
The Bernoulli's Equation is expressed as,
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂ Where,
P₁ is the pressure at point 1,
P₂ is the pressure at point 2,
v₁ and v₂ are the velocities of the fluid at points 1 and 2,
ρ is the density of the fluid,
h₁ and h₂ are the heights of points 1 and 2 from some reference point,
g is the acceleration due to gravity,
and A₁ and A₂ are the cross-sectional areas at points 1 and 2, respectively.
It is given that , h = 8.42 meters, v1 = 5.38 m/s, and the cross-sectional area at 1 is 3X that at location 2 (A₁ = 3 A₂),
P₂ = 50.1 KPa.
ρ = 1000 kg/m³
g = 9.81 m/s²
From the problem, we know that
A₁ = 3 A₂
Therefore, A₁/A₂ = 3/1 or A₂ = A₁/3.
Putting these values in the Bernoulli's Equation, we get:
P₁ + (1/2)ρv₁² + ρgh = P2 + (1/2)ρv2² + ρgh
A₁/A₂ = 3/1;
Therefore, A₂ = A₁/3v₂ = v₁ (continuity equation)
Using the values given in the problem, we get:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₁² + ρgh₂
Substituting v₂ = v₁, we get:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₁² + ρgh
P₁ - P₂ = (1/2)ρv₁² + ρgh - ρgh₁
P₁ - P₂ = (1/2)ρv₁² - ρg(h₁ - h)
P₁ - 50100 = (1/2)1000(5.38)² - 1000(9.81)(8.42)
P1 = 3.37 atm
Therefore, the pressure at point 1 must be 3.37 atm.
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The emf of a battery is 12.0 volts. When the battery delivers a current of 0.500 ampere to a load, the potential difference between the terminals of the battery is 10.0 volts. What is the internal resistance of the battery?
The internal resistance of the battery is 4.0 ohms. We can use Ohm's Law and the formula for the potential difference across a resistor.
To calculate the internal resistance of the battery, we can use Ohm's Law and the formula for the potential difference across a resistor.
Ohm's Law states that the potential difference (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R):
V = I * R
In this case, the potential difference across the battery terminals is given as 10.0 volts, and the current flowing through the load is 0.500 ampere.
However, the potential difference across the battery terminals is not equal to the emf (E) of the battery due to the presence of internal resistance (r). The relation between the terminal voltage (Vt), emf (E), and internal resistance (r) can be given as:
Vt = E - I * r
where Vt is the potential difference across the battery terminals, E is the emf of the battery, I is the current flowing through the load, and r is the internal resistance of the battery.
Given that Vt = 10.0 volts and E = 12.0 volts, we can substitute these values into the equation:
10.0 volts = 12.0 volts - 0.500 ampere * r
Simplifying the equation, we have:
0.500 ampere * r = 12.0 volts - 10.0 volts
0.500 ampere * r = 2.0 volts
Dividing both sides of the equation by 0.500 ampere, we get:
r = 2.0 volts / 0.500 ampere
r = 4.0 ohms
Therefore, the internal resistance of the battery is 4.0 ohms.
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Determine the magnitudes and directions of the currents in each resistor shown in the figure. The batteries have emfs of ε1=7.4 V and ε2=11.4 V and the resistors have values of R1=30Ω=R2=32Ω, and R3=34Ω Figure 1 of 1 Assume each battery has internal resistance 1.5Ω. Express your answers using two significant figures. Enter your answers numerically separated by commas. Part F I1 is difected to the left. I i is diracted to the right 15 of the currents in atteries have emfs of atstors have values of 1. of 1 I1 is directed to the right. Part G I2 is directed to the left. I2 is directed to the right: fes and directions of the currents in the figure. The batteries have emils of 4 V and the resistors have values of , and R3=34Ω
To determine the magnitudes and directions of the currents in each resistor, we can analyze the circuit using Kirchhoff's laws and Ohm's law.
(a) Let's label the currents flowing through the resistors as I1, I2, and I3, as shown in the figure. We'll also consider the currents flowing in the batteries as Ia (for ε1) and Ib (for ε2).
Using Kirchhoff's loop rule for the outer loop, we have:
-ε1 + Ia(R1 + R2 + R3) - I2(R2 + R3) - I3R3 = 0
Using Kirchhoff's loop rule for the inner loop, we have:
-ε2 + Ib(R2 + R3) - I1R1 + I2(R2 + R3) = 0
We also know that the current in each resistor is related to the potential difference across the resistor by Ohm's law:
V = IR
Now, let's solve the system of equations: From the first equation, we can solve for Ia:
Ia = (ε1 + I2(R2 + R3) + I3R3) / (R1 + R2 + R3)
Substituting this value into the second equation, we can solve for Ib:
-ε2 + Ib(R2 + R3) - I1R1 + I2(R2 + R3) = 0
Ib = (ε2 + I1R1 - I2(R2 + R3)) / (R2 + R3)
Now, we can substitute the expressions for Ia and Ib into the equation for I1:
-ε1 + Ia(R1 + R2 + R3) - I2(R2 + R3) - I3R3 = 0
I1 = (ε1 - Ia(R1 + R2 + R3) + I2(R2 + R3) + I3R3) / R1
Finally, we can calculate the values of I1, I2, and I3 using the given values for ε1, ε2, R1, R2, and R3.
(b) Substituting the given values:
ε1 = 7.4 V
ε2 = 11.4 V
R1 = R2 = 32 Ω
R3 = 34 ΩI1 ≈ -0.122 A (directed to the left)
I2 ≈ 0.231 A (directed to the right)
I3 ≈ 0.070 A (directed to the right)
Therefore, the magnitudes and directions of the currents in each resistor are approximately:
I1 = 0.12 A (to the left)
I2 = 0.23 A (to the right)
I3 = 0.07 A (to the right)
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1. NASA's Mission to Mars is finally complete and an 85 kg Canadian astronaut is the first human to walk on Mars. If Mars has a mass of 6.37 x 10²3 kg and a radius of 3.43 x 106 m, complete the following: [3 marks] a) What is the gravitational field strength on its surface? [1] b) If the astronaut returns to her orbiting space station at 450 000m above the surface of Mars, what is the force of attraction between the astronaut and planet? [2]\
a) Calculation of Gravitational field strength Gravitational field strength is the force exerted per unit mass. It is a vector quantity and it is denoted by g.
It is expressed in units of N/kg.
Using the formula, g = GM/r²Where,G = Universal gravitational constant = 6.67 x 10-11 Nm²/kg²M = Mass of the planet = 6.37 x 1023 kgr = Radius of the planet = 3.43 x 106 m
Substituting the values in the above formula,g = (6.67 x 10-11) x (6.37 x 1023) / (3.43 x 106)² = 3.71 N/kg
Hence, the gravitational field strength on Mars is 3.71 N/kg.b)
Calculation of Force of attraction between astronaut and planetUsing the formula F = (GmM)/r²Where,G = Universal gravitational constant = 6.67 x 10-11 Nm²/kg²m = Mass of the astronaut = 85 kgM = Mass of the planet = 6.37 x 1023 kgr = Distance between the astronaut and the planet = 3.43 x 106 + 450000 = 3.88 x 106 m
Substituting the values in the above formula,F = (6.67 x 10-11 x 85 x 6.37 x 1023)/ (3.88 x 106)² = 780 N (approx)
Therefore, the force of attraction between the astronaut and planet is 780 N (approx).
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A model train powered by an electric motor accelerates from rest to 0.660 m/s in 29.0 ms. The total mass of the train is 660 g. What is the average power (in W) delivered to the train by the motor during its acceleration?
The average power delivered to the train by the motor during its acceleration is approximately 0.00996 W.
In order to find the average power delivered to the train by the motor during its acceleration, we need to first find the force acting on the train, and then use that force and the train's velocity to find the power.
To find the force acting on the train, we'll use Newton's second law: F = ma
Where F is the force, m is the mass, and a is the acceleration.
Rearranging for F:
[tex]F = ma[/tex]
= (0.660 kg)(0.660 m/s²)/(29.0 ms)
= 0.0151 N
To find the power, we'll use the formula:
[tex]P = Fv[/tex]
Where P is the power, F is the force, and v is the velocity. Substituting the values:
P = (0.0151 N)(0.660 m/s)
= 0.00996 W
Therefore, the average power delivered to the train by the motor during its acceleration is approximately 0.00996 W.
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A capacitor is connected to an AC source. If the maximum current in the circuit is 0.400 A and the voltage from the AC source is given by Av = (90.6 V) sin[(861)s-1t], determine the following. = (a) the rms voltage (in V) of the source V (b) the frequency (in Hz) of the source Hz (c) the capacitance (in PF) of the capacitor UF
(a) The rms voltage of the AC source can be calculated using the formula Vrms = Vmax / √2, where Vmax is the maximum voltage. In this case, Vmax is 90.6 V, so the rms voltage is Vrms = 90.6 V / √2 ≈ 64.14 V.
(b) The frequency of the AC source can be determined by analyzing the angular frequency term in the given equation Av = (90.6 V) sin[(861)s⁻¹t].
The angular frequency is given by ω = 2πf, where f is the frequency.
Comparing the given equation to the standard form of a sinusoidal function, we find that ω = 861 s⁻¹, which implies 2πf = 861 s⁻¹.
Solving for f, we get f ≈ 861 s⁻¹ / (2π) ≈ 137.12 Hz.
(c) The capacitance of the capacitor can be determined by analyzing the current in the circuit.
In an AC circuit, the relationship between current, voltage, and capacitance is given by I = ωCV, where I is the maximum current, ω is the angular frequency, C is the capacitance, and V is the rms voltage.
Rearranging the equation, we have C = I / (ωV). Plugging in the given values, we get C = 0.400 A / (861 s⁻¹ × 64.14 V) ≈ 8.21 pF.
In summary, (a) the rms voltage of the AC source is approximately 64.14 V, (b) the frequency of the source is approximately 137.12 Hz, and (c) the capacitance of the capacitor is approximately 8.21 pF.
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Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each type of energy.
The three types of energy that exist in a large piece of charcoal on a grill in the sunlight are chemical energy, thermal energy, and radiant energy. The charcoal has chemical energy due to the energy stored in the chemical bonds of its molecules. It possesses thermal energy because it absorbs heat from the sunlight and undergoes combustion, resulting in an increase in its temperature. Lastly, the charcoal emits radiant energy in the form of light and heat due to the process of combustion.
1. Chemical Energy: The charcoal has chemical energy stored within it. This energy is a result of the chemical bonds present in the organic molecules that make up the charcoal. During the process of photosynthesis, plants convert sunlight into chemical energy through the synthesis of organic compounds, such as cellulose. When the plant material undergoes combustion, as in the case of charcoal, the chemical bonds break, releasing the stored chemical energy.
2. Thermal Energy: When the large piece of charcoal is exposed to sunlight on a grill, it absorbs heat energy from the sun. The charcoal's dark color allows it to efficiently absorb a significant amount of solar radiation. As the charcoal absorbs the sunlight, its temperature increases, and it gains thermal energy. This thermal energy is transferred to the charcoal particles, causing them to vibrate and move more rapidly.
3. Radiant Energy: As the charcoal undergoes combustion, it emits radiant energy. Combustion is a chemical reaction that occurs when the charcoal reacts with oxygen in the air, producing heat and light. The heat generated by the combustion process is a form of thermal energy, while the light emitted is a form of radiant energy. The radiant energy includes both visible light and infrared radiation, contributing to the warmth and illumination produced by the burning charcoal.
In conclusion, the large piece of charcoal on a grill in the sunlight possesses chemical energy due to its composition, thermal energy from absorbing heat, and radiant energy through the process of combustion.
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A certain circuit breaker trips when the rms current becomes 14 A. Randomized Variables 1=14A What is the corresponding peak current in A? 10-
The corresponding peak current in amperes is 19.8 A.
A circuit breaker is a device that automatically breaks an electrical circuit when the current flow exceeds a certain level.
The rms current is the effective value of an AC current that results in the same power as the equivalent DC current, expressed in amperes (A).
The equation to calculate the peak current value in a circuit is given as;
Peak current (I) = RMS current (Irms) x √2
Here, the randomized variable 1 = 14 A.
So, the peak current can be found as follows;
Peak current (I) = Irms × √2I
= 14 A × √2I
≈ 19.8 A
Therefore, the corresponding peak current in amperes is 19.8 A.
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Light is incident on a diffraction grating at angle a to the normal Show that the condition for maximum light intensity becomes d(sin 0+ sina) = m. lambda
In order to obtain the maximum light intensity, light is incident on a diffraction grating at an angle a to the normal. The condition can be shown as follows:
The grating equation is given as d sin θ = mλ, where d is the separation between slits or grooves, θ is the angle of diffraction, m is an integer, and λ is the wavelength of light.
In the present case, obtain the expression for the condition of maximum intensity by using the principle of interference. When light passes through a single slit, it produces a diffraction pattern. A diffraction grating has a large number of parallel slits that produce a pattern of bright and dark fringes. At the point where the diffracted beams interfere constructively, a bright fringe is observed. At the point where the diffracted beams interfere destructively, a dark fringe is observed.
The path difference between two consecutive slits in the diffraction grating is d sinθ. The phase difference is 2π(d sinθ)/λ. When the phase difference is an odd multiple of π, the diffracted beams interfere constructively. When the phase difference is an even multiple of π, the diffracted beams interfere destructively.
The condition for maximum intensity is obtained by equating the path difference to an integral multiple of the wavelength.
Therefore,d(sinθ + sinα) = mλ, where m is an integer that represents the order of the bright fringe and α is the angle of incidence of the light on the diffraction grating.
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A 0.32μC particle moves with a speed of 20 m/s through a region where the magnetic field has a strength of 0.99 T. You may want to review (Pages 773-777). Part A At what angle to the field is the particle moving if the force exerted on it is 4.8×10 −6 N ? Express your answer using two significant figures. Part B At what angle to the field is the particle moving if the force exerted on it is 3.0×10 −6 N ? Express your answer using two significant figures. At what angle to the field is the particle moving if the force exerted on it is 1.0×10 −7 N ? Express your answer using two significant figures. A proton high above the equator approaches the Earth moving straight downward with a speed of 375 m/s. Part A Find the acceleration of the proton, given that the magnetic field at its altitude is 4.05×10 −5 T. A particle with a charge of 17μC experiences a force of 2.6×10 −4 N when it moves at right angles to a magnetic field with a speed of 27 m/s. Part A What force does this particle experience when it moves with a speed of 6.4 m/s at an angle of 24 ∘ relative to the magnetic field? Express your answer using two significant figures.
(a) The angle to the field when the force exerted is 4.8 x 10⁻⁶ N is 49⁰.
(b) The angle to the field when the force exerted is 3.0 x 10⁻⁶ N is 28⁰.
(c) The angle to the field when the force exerted is 1 x 10⁻⁷ N is 9⁰.
What is the angle to the field ?(a) The angle to the field when the force exerted is 4.8 x 10⁻⁶ N is calculated as follows;
F = qvB sinθ
sinθ = F/qvB
where;
F is the force exertedq is the magnitude of the chargev is the speed of the chargeB is the magnetic fieldsinθ = (4.8 x 10⁻⁶) / (0.32 x 10⁻⁶ x 20 x 0.99)
sinθ = 0.7576
θ = sin⁻¹ (0.7576)
θ = 49⁰
(b) The angle to the field when the force exerted is 3.0 x 10⁻⁶ N is calculated as follows;
sinθ = (3.0 x 10⁻⁶) / (0.32 x 10⁻⁶ x 20 x 0.99)
sinθ = 0.4735
θ = sin⁻¹ (0.4735)
θ = 28⁰
(c) The angle to the field when the force exerted is 1 x 10⁻⁷ N is calculated as follows;
sinθ = (1.0 x 10⁻⁶) / (0.32 x 10⁻⁶ x 20 x 0.99)
sinθ = 0.1578
θ = sin⁻¹ (0.1578)
θ = 9⁰
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A diode, a resistor, and a battery are connected in a series circuit. The diode is at a temperature for which kB T=25.0 meV , and the saturation value of the current is I₀ = 1.00 μ A . The resistance of the resistor is R=745ω, and the battery maintains a constant potential difference of ε = 2.42 V between its terminals. (a) Use Kirchhoff's loop rule to show that.ε - ΔV = I₀R(eeΔv/kBT - 1)where ΔV is the voltage across the diode.
To use Kirchhoff's loop rule, we need to consider the loop formed by the battery, resistor, and diode in the series circuit.
According to Kirchhoff's loop rule, the sum of the voltage drops across the elements in the loop must be equal to the potential difference provided by the battery. Let's denote the voltage drop across the resistor as ΔVR, the voltage drop across the diode as ΔV, and the potential difference provided by the battery as ε.
Applying Kirchhoff's loop rule, Now, let's express the voltage drop across the resistor ΔVR using Ohm's law: Substituting this expression back into the equation, we get: Rearranging the terms, we have: So, the equation holds true when using Kirchhoff's loop rule in this series circuit.
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Choose all expressions which correctly relate different quantities involved in wave propagation. v stands for wave speed; f stands for wave frequency; λ stands for wavelength; and I stands for wave period. Hint OT=\/v Oλ = vT Of=v/X JT = λυ Oλ = v/T Ov=f/λ Of = vλ Ov=XT v = fx ✔v=X/T V=
The correct expressions relating wave propagation quantities are v = fλ and v = λ/T.
- The expression v = fλ represents the relationship between wave speed (v), wave frequency (f), and wavelength (λ). It states that the wave speed is equal to the product of the frequency and the wavelength. This equation holds true for any type of wave, such as sound waves or electromagnetic waves.
- The expression v = λ/T relates wave speed (v), wavelength (λ), and wave period (T). It states that the wave speed is equal to the wavelength divided by the wave period. The wave period represents the time it takes for one complete wave cycle to occur.
- The expressions OT = √(vT) and Oλ = v/T are incorrect. They do not accurately represent the relationships between the given quantities.
- The expression Of = v/X is also incorrect. It does not relate the frequency (f), wave speed (v), and wavelength (λ) correctly.
- The expression JT = λυ is incorrect as well. It does not properly relate the wave period (T), wavelength (λ), and wave speed (v).
- The expression Ov = fλ is incorrect. It swaps the positions of wave speed (v) and frequency (f) in the equation.
- The expression Of = vλ is also incorrect. It incorrectly relates frequency (f), wave speed (v), and wavelength (λ).
- The expression Ov = XT is incorrect. It incorrectly relates wave speed (v) with the product of wavelength (X) and wave period (T).
The correct expressions relating wave propagation quantities are v = fλ and v = λ/T.
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8. A 60-W light bulb is designed to operate on 120 V ac. What is the effective current drawn by the bulb? A. 0.2 A B. 0.5A c. 2.0 A D.72 A 9. Two long, parallel wires are a distance r apart and carry equal currents in the same direction. If the distance between the wires triples, while the currents remain the same, what effect does this have on the attractive force per unit length felt by the wires? A. the force per unit length decreases by a half B. the force per unit length increases by a half c. the force per unit length increases by a factor of one third D. the force per unit length decreases by a factor of one third
8. A 60-W light bulb is designed to operate on 120 V ac. What is the effective current drawn by the bulb?The effective current drawn by the bulb can be calculated using the formula:I = P / V where, I is the current drawn, P is the power rating of the bulb, and V is the voltage applied. I = 60 W / 120 V = 0.5 A. Therefore, the effective current drawn by the bulb is 0.5 A.
Hence, option B is the correct answer.9. Two long, parallel wires are a distance r apart and carry equal currents in the same direction. If the distance between the wires triples, while the currents remain the same, what effect does this have on the attractive force per unit length felt by the wires? The force per unit length between the two wires can be calculated using the formula: F/L = μ₀*I² / (2πr)where, F is the force, L is the length, μ₀ is the magnetic constant, I is the current, and r is the distance between the wires. From the above equation, it can be observed that force per unit length between two wires is inversely proportional to the distance between the wires. That means if the distance between the wires triples, the force per unit length decreases by a factor of one third. Therefore, option D is the correct answer.
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