The speed of the stone in the given scenario can be calculated using the formula of the centripetal force.The formula of centripetal force is given by:F = (m*v²)/rWhere, F is the centripetal force,m is the mass of the object,v is the velocity of the object,r is the radius of the circle.
Now, in the given scenario, the length of the string is given as 1.16 m. So, the radius of the circle can be given as:r = 1.16 m / 2 = 0.58 m The gravitational force acting on the stone is given by:Fg = mg Where,m is the mass of the stone,g is the acceleration due to gravity.According to the question, the tension is 9% more in the vertical case than in the horizontal case. So, the tension in the horizontal case can be given as:Fh = Fg + (mv²h)/rAnd, the tension in the vertical case can be given as:Fv = Fg + (mv²v)/r Given that the tension in the vertical case is 9% larger than in the horizontal case.
Therefore, we can write:Fv = 1.09 * Fh => Fg + (mv²v)/r = 1.09 * [Fg + (mv²h)/r]Now, let's divide the equation by Fg:1 + (mv²v)/(Fg * r) = 1.09 + 1.09*(mv²h)/(Fg * r)After this, we can substitute the values:Fg = mgv = √[(Fg * r)/m]Now, let's substitute the values and solve for v:1 + (√[(Fg * r * v²v)/m])/(Fg * r) = 1.09 + 1.09 * (√[(Fg * r * v²h)/m])/(Fg * r)After solving this equation we get,√[(Fg * r * v²v)/m] = 1.09√[(Fg * r * v²h)/m]√[v²v] = 1.09 * √[v²h]v²v = 1.1881 * v²hNow, substituting the value of v²h we get,v²v = 1.1881 * [Fg * r / m]After substituting all the values we get:v = 5.02 m/sSo, the speed of the stone in the given scenario is 5.02 m/s.
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You are pulling a child in a wagon. The rope handle Part A is inclined upward at a 60∘
angle. The tension in the handle is 20 N. How much work do you do if you pull the wagon 130 m at a constant speed? Express your answer with the appropriate units. -3 Incorrect; Try Again; 4 attempts remaining Check that you have converted between SI units correctly.
Given,The angle of inclination of the rope handle Part A is 60°.The tension in the handle is 20 N.Distance covered is 130m.
Let W be the work done to pull the wagon.Let F be the force required to pull the wagon.A force F at an angle θ with the horizontal requires more work to move the wagon than the force required in the horizontal direction.
However, the force in the horizontal direction is not sufficient to move the wagon in a direction inclined to the horizontal.To get the net force acting in the direction of motion, we need to resolve the force F into its components as follows:Here, the horizontal component Fcosθ balances the force of friction while the vertical component Fsinθ lifts the wagon off the ground.The force acting in the direction of motion is given as Fcosθ,We know that F = 20 N, θ = 60° and Fcosθ = 20cos60° = 10 N.The work done to move the wagon through 130m is given asW = Fcosθ × d = 10 × 130 J = 1300 J
Therefore, the work done is 1300 J (joules).
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A) Draw free-body diagrams below, clearly labeling forces with proper subscripts B) Write out all Netwton's 2nd Law equations (4 total), identify 3rd law force pairs, and write your acceleration const Problem 2 (10 points) Block \( A\left(m_{A}=1.50 \mathrm{~kg}\right) \) sits on Block \( B \) \( \left(m_{B}=7.50 \mathrm{~kg}\right) \). An external force, \( F=30.0 \mathrm{~N} \), is applied to Blo
The acceleration of Block B is 3.316 m/s² when Block B is subjected to an external force of F = 30.0 N and moves to the left on a frictionless surface.
The situation described involves two blocks, Block A and Block B. Block A has a mass of 1.50 kg (mA = 1.50 kg) and sits on top of Block B, which has a mass of 7.50 kg (mB = 7.50 kg).
An external force, F, is applied to Block B, causing it to slide to the left on a frictionless table. However, there is kinetic friction between Block A and Block B, with a coefficient of kinetic friction (μk) equal to 0.35.
To determine the acceleration of the blocks, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).
In this case, the net force acting on Block B is the external force applied to it (F = 30.0 N) minus the force of friction between the blocks.
The force of friction can be calculated using the equation
Ffriction = μk * normal force
Where, the normal force is equal to the weight of Block A (mgA = mA * g). So, the force of friction (Ffriction) is equal to 0.35 times the weight of Block A (Ffriction = 0.35 * mA * g).
Substituting the given values, we have:
Ffriction = 0.35 * 1.50 kg * 9.8 m/s² (acceleration due to gravity) Simplifying, we find that
Ffriction = 5.13 N.
Now, we can calculate the net force acting on Block B:
Net force = F - Ffriction
Net force = 30.0 N - 5.13 N
Net force = 24.87 N
Finally, we can calculate the acceleration of Block B by dividing the net force by its mass:
a = Net force / mB
a = 24.87 N / 7.50 kg
a = 3.316 m/s²
Therefore, the acceleration is 3.316 m/s².
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Complete question is,
Block A (mA = 1.50 kg) sits on Block B (mB = 7.50 kg). An external force, F = 30.0 N, is applied to Block B and slides to the left on a frictionless table. However, there is kinetic friction between the two blocks with μk = 0.35. What is the acceleration of the blocks?
Suppose two equal charges of 1.4C each are separated by a distance of 1.5 km in air. What is the magnitude of the force acting between them, in newtons?
The magnitude of the force acting between the two charges is 4.83 × [tex]10^6[/tex] newtons.
Calculate the magnitude of the force acting between two charges, we can use Coulomb's Law. Coulomb's Law states that the force (F) between two charges (q₁ and q₂) is given by the equation:
F = k * |q₁| * |q₂| / r²
where k is the electrostatic constant (approximately 8.99 × [tex]10^9[/tex] N·m²/C²), |q₁| and |q₂| are the magnitudes of the charges, and r is the separation distance between the charges.
Magnitude of each charge |q₁| = |q₂| = 1.4 C
Separation distance r = 1.5 km = 1.5 × 10³ m
Substituting these values into the equation, we can calculate the magnitude of the force:
F = (8.99 × [tex]10^9[/tex] N·m²/C²) * (1.4 C) * (1.4 C) / (1.5 × 10³ m)²
Calculate the magnitude of the force acting between the two charges.
Using Coulomb's Law:
F = (8.99 × [tex]10^9[/tex] N·m²/C²) * (1.4 C) * (1.4 C) / (1.5 × [tex]10^3[/tex] m)²
Calculating this expression, we find:
F ≈ 4.83 × [tex]10^6[/tex]N
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An object is placed a large distance from a converging lens, and then moved toward the lens (but still much farther from the lens than the distance to the focal
point).
(a) Does the image get smaller or larger as the object is moved towards the lens?
(b) Does the image get closer to or farther from the lens as the object is moved towards the lens?
(a) As the object is moved toward the lens, the image will get larger.
(b) As the object is moved toward the lens, the image will get closer to the lens.
When an object is placed at a large distance from a converging lens and moved towards the lens (but still much farther from the lens than the distance to the focal point), the size and position of the image formed depend on the position of the object with respect to the focal point. As the object is moved toward the lens, the image will get larger. The lens converges the incoming light rays and produces a larger image on the opposite side. The larger image is formed as the object comes closer to the lens.
As the object is moved toward the lens, the image will get closer to the lens. The image is formed on the opposite side of the lens and gets closer to the lens as the object comes closer to the lens. This is because as the object gets closer to the lens, the light rays converge more strongly, producing a closer image.
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A particle of mass m=
4
1
kg is at rest at the origin of the coordinate system. At t=0, a force F
=
t
3
Is applied to the particle. Find i- The acceleration a(t). ii- The velocity v(t). iii- The position x(t). Solution
Acceleration of the particle, a(t) = t³/4.1 m/s², Velocity of the particle, v(t) = (1/4.1) * [t⁴/4] and Position of the particle, x(t) = (1/4.1) * (1/4) * t⁴. Mass of the particle, m = 4.1 kg, Force applied, F(t) = t³ (as the particle is at rest, acceleration a(0) = 0)
Since, F = maSo, a = F/mFor t > 0, the particle experiences a net force F = t³.
The acceleration of the particle is given bya(t) = F/m = t³/m
i) Acceleration of the particle, a(t) = t³/m = t³/4.1 m/s²ii) To find the velocity of the particle v(t), we need to integrate the acceleration from 0 to t.v(t) = ∫a(t)dtv(t) = ∫(t³/4.1)dtv(t) = (1/4.1) * ∫t³dtv(t) = (1/4.1) * [t⁴/4] + C1 Where C1 is the constant of integration.At t = 0, the velocity of the particle was 0.
So, we havev(0) = 0 = (1/4.1) * [0⁴/4] + C1C1 = 0v(t) = (1/4.1) * [t⁴/4]
iii) To find the position of the particle x(t), we need to integrate the velocity from 0 to t.x(t) = ∫v(t)dt = ∫(1/4.1) * [t⁴/4]dt = (1/4.1) * (1/4) * t⁴ + C2 Where C2 is the constant of integration.
At t = 0, the position of the particle was 0.
So, we have x(0) = 0 = (1/4.1) * (1/4) * 0⁴ + C2C2 = 0x(t) = (1/4.1) * (1/4) * t⁴
The position of the particle at any given time t is given by the function x(t) = (1/4.1) * (1/4) * t⁴.
Answer:Acceleration of the particle, a(t) = t³/4.1 m/s²Velocity of the particle, v(t) = (1/4.1) * [t⁴/4]Position of the particle, x(t) = (1/4.1) * (1/4) * t⁴
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At most, how many bright fringes can be formed on either side of
the central bright fringe when light of 625 nm falls on a double
slit whose separation is 3.76x10^-6m?
the maximum number of bright fringes that can be formed on either side of the central bright fringe is 6 on either side, so a total of 12 on both sides.
The question is related to the phenomenon of diffraction of light, which is caused by the constructive and destructive interference of waves.
The formula for the calculation of the number of fringes formed is given by: `(dsinθ)/λ = m`.Where,d = Distance between the slits.θ = Angle of diffraction
.λ = Wavelength of light.m = Number of fringes formed.For the central maximum, the angle of diffraction is zero. Hence, the formula becomes: `(d x 0)/λ = 0`.
Therefore, the number of fringes formed on either side of the central bright fringe is: `(3.76 x 10^-6 m x 1)/625 x 10^-9 m = 6`.
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In the figure the battery has potential difference V=13.0 V,C2=3.70μF,C4=3.80μF, and all the capacitors are initially uncharged. When switch S is closed, a total charge of 10.0μC passes through point a and a total charge of 7.00μC passes through point b. What are (a)C1 and (b)C3? (a) Number Unit (b) Number Unit
The answers to the given questions are follows:
(a) The value of C₁ is approximately 0.769 μF.
(b) The value of C₃ is approximately 0.538 μF.
To determine the values of C₁ and C₃, we can use the equation relating charge (Q), capacitance (C), and potential difference (V):
Q = C × V
Given:
Potential difference (V) = 13.0 V
Charge passing through point a (Qa) = 10.0 μC
Charge passing through point b (Qb) = 7.00 μC
Capacitance C₂ = 3.70 μF
Capacitance C₄ = 3.80 μF
(a) Calculating C₁:
Using the equation Q = C × V, we can rearrange it to solve for C:
C = Q / V
C₁ = Qa / V
C₁ = 10.0 μC / 13.0 V
C₁ ≈ 0.769 μF
Therefore, C₁ is approximately 0.769 μF.
(b) Calculating C₃:
Using the same equation Q = C × V:
C₃ = Qb / V
C₃ = 7.00 μC / 13.0 V
C₃ ≈ 0.538 μF
Therefore, C₃ is approximately 0.538 μF.
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What can you conclude about the size of a torque relative to the force and placement of the force relative to the axis of rotation?
The size of a torque is determined by both the magnitude of the force applied and its distance from the axis of rotation. Torque increases with the magnitude of the force and with the perpendicular distance between the force and the axis of rotation.
This relationship is described by the equation τ = r × F, where τ represents torque, r is the distance vector from the axis of rotation to the point of application of the force, and F is the force vector. The direction of the torque is given by the right-hand rule, with the thumb pointing in the direction of the force and the fingers curling in the direction of the torque.
Torque is a measure of the rotational effect of a force. The size or magnitude of a torque is influenced by two main factors: the magnitude of the force applied and the distance between the point of application of the force and the axis of rotation. Mathematically, torque (τ) is calculated as the cross product of the distance vector (r) and the force vector (F), resulting in τ = r × F.
If the force is applied perpendicular to the line connecting the axis of rotation and the point of application of the force, the torque can be determined simply by multiplying the magnitude of the force (F) by the perpendicular distance (r) between the force and the axis. In this case, torque can be represented as τ = rF, where r is the distance and F is the force.
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A constant force of \( 245 \mathrm{~N} \) is applied at a \( 48.2^{\circ} \) angle to a mass of \( 62.1 \mathrm{~kg} \) as shown below. If the mass moves at a constant speed of \( 3.28 \mathrm{~ms}^{-
The net force acting on the mass can be determined by resolving the applied force into its horizontal and vertical components. The horizontal component of the force does not contribute to the motion because the mass moves at a constant speed. Therefore, only the vertical component of the force is relevant.
To find the vertical component of the force, we can use the equation[tex]\( F_{\text{vertical}} = F \sin(\theta) \[/tex]), where[tex]\( F \)[/tex] is the applied force and[tex]\( \theta \)[/tex] is the angle it makes with the horizontal.
Substituting the given values, we have[tex]\( F_{\text{vertical}} = 245 \, \text{N} \times \sin(48.2^\circ) \).[/tex]
Next, we can calculate the acceleration of the mass using Newton's second law,[tex]\( F_{\text{net}} = m \cdot a \), where \( F_{\text{net}} \) is the net force and \( m \) is the mass.[/tex]
Since the mass moves at a constant speed, the net force is zero. [tex]Thus, \( F_{\text{net}} = 0 = F_{\text{vertical}} - m \cdot g \), where \( g \)[/tex]is the acceleration due to gravity.
Substituting the known values, we have[tex]\( 0 = 245 \, \text{N} \times \sin(48.2^\circ) - 62.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \).[/tex]
Solving this equation will give us the value of [tex]\( F_{\text{vertical}} \)[/tex]. The negative sign indicates that the force is directed opposite to the gravitational force.
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A group of hikers hears an echo 1.30 s after shouting. How far away is the mountain that reflected the sound wave? (Assume the speed of sound is 341 m/s.) m
The distance to the mountain is approximately 887.6 meters.
To determine the distance to the mountain, we can use the speed of sound and the time it takes for the echo to reach the hikers.
Speed of sound (v) = 341 m/s
Time for the echo to reach the hikers (t) = 1.30 s
The time it takes for the sound to travel to the mountain and back is twice the time for the echo to reach the hikers. So, we can calculate the total time taken for the round trip:
Total time = 2 * t
Substituting the given value:
Total time = 2 * 1.30 s
Total time = 2.60 s
Now, we can calculate the distance using the formula:
Distance = Speed * Time
Substituting the values:
Distance = 341 m/s * 2.60 s
Distance ≈ 887.6 m
Therefore, the mountain that reflected the sound wave is approximately 887.6 meters away from the hikers.
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A square wire loop of side length 10 cm has resistance R=200Ω and lies in a horizontal plane. A uniform magnetic field B points vertically downward (into the plane), and in 0.1 s it increases linearly from 10mT to 30mT. a) Find the magnetic flux Φ_B,i through the loop at the beginning of the 0.1− s period. b) Find the magnetic flux Φ_B,f through the loop at the end of the 0.1−s period. c) Find the loop current I during the 0.1-s period. d) What is the direction of the loop current I ?
a. the magnetic flux through the loop at the beginning of the 0.1-s period is 0.
b. the magnetic flux through the loop at the end of the 0.1-s period is also 0.
c. the loop current I is 0.
d. The direction of the loop current I is 0 since there is no current induced during the 0.1-s period.
a. Plugging in the given values:
B = 10 mT = 10 * 10^-3 T
A = 100 cm^2 = 100 * 10^-4 m^2
cos(θ) = 0
Φ_B,i = B * A * cos(θ)
= (10 * 10^-3 T) * (100 * 10^-4 m^2) * 0
= 0
Therefore, the magnetic flux through the loop at the beginning of the 0.1-s period is 0.
b. B_avg = (B_initial + B_final) / 2
= (10 mT + 30 mT) / 2
= 20 mT = 20 * 10^-3 T
Plugging in the given values:
B = 20 * 10^-3 T
A = 100 * 10^-4 m^2
cos(θ) = 0 (since the loop and magnetic field are perpendicular)
Φ_B,f = B * A * cos(θ)
= (20 * 10^-3 T) * (100 * 10^-4 m^2) * 0
= 0
Therefore, the magnetic flux through the loop at the end of the 0.1-s period is also 0.
c. To find the loop current I during the 0.1-s period, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a loop is equal to the rate of change of magnetic flux through the loop:
emf = -dΦ_B / dt
The negative sign indicates the direction of the induced current. Since the magnetic flux through the loop is constant (0), the rate of change of flux is also 0, and there is no induced emf or current in the loop during this period. Therefore, the loop current I is 0.
d. The direction of the loop current I is 0 since there is no current induced during the 0.1-s period.
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May permit ka ba? An EM wave is traveling inside a medium with a speed of 0.25c. The material has a known permeability constant of μ=2μ 0
What is the material's permittivity? A. 8ϵ0
B. 6ϵ0
C. 4ϵ0
D. 2ϵ0
Permittivity is a material property that specifies the degree of resistance to the formation of an electric field inside a medium.
The permittivity of a material is calculated by dividing the electric flux density by the electric field intensity. In symbols, ϵ = D/E, where ϵ is permittivity, D is electric flux density, and E is electric field intensity.
ϵ is equal to the ratio of the electric field to the electric flux density.ϵ=1/(μ*c^2) where μ is the permeability of free space, and c is the speed of light.[tex]ϵ = 1/(2μ * 0.25c^2)ϵ = 1/(0.5μ * c^2)ϵ = 2ϵ[/tex]0The permittivity of the medium is 2ϵ0, which is answer (D). The explanation is more than 100 words.
Yes, I can help you with that. In an EM wave traveling inside a medium with a speed of 0.25c, the material has a known permeability constant of μ=2μ0, the permittivity can be calculated as follows: Permability is given as μ=2μ0Speed of light in vacuum is c Speed of light inside the medium is vThe refractive index of the medium is given as [tex]μr=cc=μrv.25c=c/μr.25=1/μrμr=4[/tex]The permittivity of the medium can be found using the formula:μr=ε/[tex]ε0ε=μrε0=4×8.85×10^-12ε=35.4×10^-12[/tex]Hence, the permittivity of the material is [tex]35.4×10^-12.[/tex]
Therefore, option C. 4ϵ0 is the correct answer. Option A. 8ϵ0, option B. 6ϵ0 and option D. 2ϵ0 are incorrect.
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Police are investigating an accident. They know that Tom Brady was driving 20.0 m/s before being hit by Jay Z head on. Tom Brady's car has a mass of 1100 kg and Jay Z's has a mass of 1475 kg. They also know that the two cars stuck together and were traveling 7.00 m/s in the same direction as Jay Z was driving. The speed limit was 25 m/s, was Jay Zspeeding?
In order to find out if Jay Z was speeding, we need to first determine the speed at which he was driving before the collision. We can use the principle of conservation of momentum to do this.
According to the principle, the total momentum of an isolated system remains constant if there are no external forces acting on it. Since the two cars stuck together after the collision, they can be considered as an isolated system with no external forces acting on them.
where p is the total momentum of the system, m1 and m2 are the masses of the two cars, and v1 and v2 are their velocities before the collision.
Substituting the given values, we get:
p = (1100 kg)(20.0 m/s) + (1475 kg)(-7.00 m/s)
p = 22,000 kg·m/s - 10,325 kg·m/s
Now, we can use the same equation to calculate the velocity of the two cars after the collision. Since the two cars are moving in the same direction after the collision, their velocities will add up.
Substituting the given values, we get:
11,675 kg·m/s = (1100 kg + 1475 kg)v
v = 11,675 kg·m/s / 2575 kg
v = 4.54 m/s
The speed at which the two cars were traveling after the collision was 4.54 m/s. Since we know that Jay Z was driving in the same direction as the two cars after the collision, we can conclude that his speed before the collision was less than 4.54 m/s.
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The rear window defogger of a car consists of thirteen thin wires (resistivity =72.9×10−8Ω⋅m ) embedded in the glass. The wires are connected in parallel to the 12.0 V battery, and each has a length of 1.17 m. The defogger can melt 0.0221 kg of ice at 0∘C into water at 0∘C in two minutes. Assume that all the power dissipated in the wires is used immediately to melt the ice. Find the cross-sectional area of each wire. Take the latent heat of fusion of water to be 3.35×105 J/kg. Number Units
The cross-sectional area of each wire in the rear window defogger can be determined by calculating the power dissipated in the wires and using the resistivity of the wire material. By equating the power to the heat required for melting the ice, the cross-sectional area can be found using the given values.
To find the cross-sectional area of each wire, we need to use the power dissipated in the wires and the resistivity of the wire material. We know that all the power dissipated is used to melt the ice, so we can equate the power to the heat required for melting.
First, we calculate the heat required to melt the ice:
Heat = mass × latent heat of fusion
Heat = 0.0221 kg × 3.35 × 10^5 J/kg
Heat = 7373.5 J
Next, we calculate the power dissipated in the wires:
Power = Voltage^2 / Resistance
Resistance = resistivity × length / area
Power = Voltage^2 × area / (resistivity × length)
Substituting the values and rearranging the equation, we can solve for the cross-sectional area:
Area = (Voltage^2 × length) / (Power × resistivity)
Plugging in the given values, we get:
Area = (12.0 V^2 × 1.17 m) / (7373.5 J × 72.9 × 10^-8 Ω⋅m)
Calculating this expression, the cross-sectional area of each wire is obtained.
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Two 10-cm-diameter charged rings face each other, 20 cm apart. The left ring is charged to −25nC and the right ring is charged to +25 nC. What is the magnitude of the electic field B at the midpoint between the two rings? Express your answer with the appropriate units. View Available Hint(s) What is the direction of the electric field B at the midpoint between the two rings? View Available Hint(s) to the center of the right ring to the center of the left ring What is the magnitude of the force on a proton at the midpoint? Express your answer with the appropriate units. View Available Hint(s) PartD What is the direction of the force F on a proton at the midpoint? View Available Hint(s) to the right ring to the left ring
The force on the proton is towards the right ring.
a. Magnitude of electric field B at midpoint: 4.95 × 104 N/C
Explanation:
Given that,
Diameter of each ring, d = 10 cm
Radius of each ring, r = d/2 = 5 cm
Distance between two rings, R = 20 cm
Magnitude of charge on left ring, q1 = -25 nC = -25 × 10⁻⁹ C
Charge on right ring, q2 = +25 nC = 25 × 10⁻⁹ C
From Coulomb's Law, force between two charged rings is given by,
F = k * (q1*q2) / R²
where k = 9 × 10⁹ N*m²/C² is Coulomb's constant
Magnitude of electric field B at midpoint between the two rings is given by,
E = F / q2
Thus, E = k * q1 * R / (2r² * q2)
Substituting the given values,
we get:
E = (9 × 10⁹ N*m²/C²) * (-25 × 10⁻⁹ C) * (20 cm) / (2 * (5 cm)² * (25 × 10⁻⁹ C))E = 4.95 × 10⁴ N/Cb.
Direction of electric field B at midpoint:
from left ring to right ring
The direction of electric field is from higher potential to lower potential. Here, the potential at left ring is higher than that at right ring since the left ring is negatively charged and the right ring is positively charged.
Hence, the electric field B points from left ring to right ring.c.
Magnitude of force on a proton at midpoint: 2.69 × 10⁻¹⁷ N
The force on a proton is given by,
F = q * E
where q is the charge on the proton = 1.6 × 10⁻¹⁹ C and E is the electric field at the midpoint.
Substituting the given values, we get:
F = (1.6 × 10⁻¹⁹ C) * (4.95 × 10⁴ N/C)F = 2.69 × 10⁻¹⁷ Nd.
Direction of force F on a proton at midpoint: towards the right ring
The direction of the force on the proton is the same as the direction of the electric field at the midpoint, i.e., from left ring to right ring.
Therefore, the force on the proton is towards the right ring.
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Determine the current through R1 (in Amperes) for the circuit in the figure given that: R1 = 2 Ω, R2 = 5 Ω, R3 = 10 Ω and V = 19 V. Your answer should be a number with two decimal places, do not include the unit.
The current through R1 in the circuit given can be calculated using the current divider rule after finding the total resistance of the circuit using the resistance values. The current in the circuit can be determined using Ohm's law.
The current through R1 in Amperes for the circuit given can be determined as follows: The first step is to determine the total resistance in the circuit. Total resistance can be calculated using the following formula: RTotal = R1 + R2 + R3 = 2 Ω + 5 Ω + 10 Ω = 17 ΩThen, using Ohm's law, we can calculate the current in the circuit: I = V / RTotal = 19 V / 17 Ω ≈ 1.12 AFinally, using the current divider rule, we can determine the current through R1 as follows: IR1 = (R2 || R3) / (R1 + (R2 || R3)) * IIR1 = (5 Ω || 10 Ω) / (2 Ω + (5 Ω || 10 Ω)) * 1.12 AIR1 ≈ 0.57 ATherefore, the current through R1 in Amperes is approximately 0.57 A.For more questions on resistance
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An infinite cylindrical conductor has an inner radius r
a
=47.3 mm and an outer radius r
b
=79.1 mm. The conductor has a linear charge density of λ
1
=145
m
nC
. On the axis of the cylinder is an infinite line charge with linear charge density λ
2
=−13
m
nC
. Determine the electric field magnitude at the point r=26.07mm( in
C
N
) A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r,(r
C
N
Determine the electric field at r If the same charge Q were distributed uniformly throughout a sphere of radius 18.5 R (in
C
N
).
The electric field at a distance r from the center of the sphere is = [tex]6.46 * 10^6 N / C[/tex]. The electric field at r for the sphere with radius R and charge Q distributed uniformly throughout the sphere is [tex]2.87 * 10^-6 Qr N / C[/tex].
The electric field produced by a cylindrical conductor of infinite length with a finite radius depends on its radius, the amount of charge per unit length it carries, and the distance from the conductor. The electric field at any point around a cylindrical conductor with a radius is uniform and directed perpendicular to its axis.
The electric field outside the cylinder is linearly proportional to the distance from the conductor.The electric field magnitude at point r = 26.07mm for infinite cylindrical conductor. The linear charge density of a cylinder with radius a and height h is given by the formula:
[tex]\Lambda = Q / (2 \piah)[/tex].
The charge per unit length on the cylinder is given by
λ =[tex](145 * 10^-9) C / 0.0318 m = 4560.37735849 nC/m[/tex]
The electric field produced by the cylinder on the x-axis is given by
E =[tex]\lambda / (2\pi\epsilon_0x)At x = 26.07mm = 0.02607 m[/tex],
the electric field [tex]E = \lambda / (2\pi \epsilon_0x)[/tex]
[tex]= [4,560.38 * 10^-9 C/m] / [2\pi(8.85 * 10^-12 C^2 / Nm^2)(0.02607 m)]\\= 6.46 * 10^6 N / C[/tex]
The electric field at point r for a solid non-conducting sphere of radius R.
The electric field at any point inside a uniform sphere is given by:
[tex]E = (1/3) k Q r / R^3[/tex]
where k is the Coulomb constant and Q is the charge on the sphere.
For a uniform sphere of radius R with charge Q, the charge per unit volume is given by
λ = [tex]Q / (4/3)\pi R^3[/tex]
The electric field E at radius r inside the sphere is given by
E = [tex](1/3) k \lambda r (r / R^3)[/tex]
Therefore, the electric field at a distance r from the center of the sphere is given by
E [tex]= (1/3) k Q r / (18.5 R)^3\\= (1/3) k Q r / (6.67 × R)^3\\= (1/3) k Q r / (2.18 × 10^5)^3\\= (1/3) (8.99 * 10^9 N m^2 / C^2) (Q) (r / 10.4 * 10^14 m^3)\\= (Qr / 3.14 * 10^15) × (8.99 * 10^9 N m^2 / C^2)\\= 2.87 * 10^-6 Qr N / C[/tex]
Therefore, the electric field at r for the sphere with radius R and charge Q distributed uniformly throughout the sphere is [tex]2.87 * 10^-6 Qr N / C[/tex].
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2. Determine the electric field produced by an infinite line of charges at a distance of 4.5 m if the linear charge density is 4.5 μC/m.
The electric field produced by an infinite line of charges with a linear charge density of 4.5 μC/m at a distance of 4.5 m is 4.5 N/C.
The electric field produced by an infinite line of charges can be calculated using Coulomb's law. The formula for the electric field due to an infinitely long line of charge is given by E = λ/(2πε₀r), where λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the line of charges.
In this case, the linear charge density is 4.5 μC/m, and the distance from the line of charges is 4.5 m. Plugging these values into the formula, we get E = (4.5 μC/m)/(2πε₀(4.5 m)). The value of ε₀ is approximately 8.85 x [tex]10^{-12} C^2/(N.m^2)[/tex]. Simplifying the equation further, we have E = (4.5 μC/m)/(2π([tex]8.85 * 10^{-12} C^2/(N.m^2)[/tex])(4.5 m)).
Performing the calculations, the electric field is approximately 4.5 N/C. Thus, at a distance of 4.5 m from the infinite line of charges with a linear charge density of 4.5 μC/m, the electric field strength is 4.5 N/C.
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A passenger is standing on a scale in an elevator. The building has a height of 500 feet, the passenger has a mass of 80 kg, and the scale has a mass of 7 kg. The scale sits on the floor of the elevator. (It is an Otis elevator, so we will label it as "O" so as not to confuse its forces with those caused by the earth.) You may take g = 10 N/kg. For doing this problem it might be useful to start by drawing free-body diagrams for the passenger and the scale. Consider the vertical forces acting on the passenger and the scale WE→P: The force of the earth pulling down on the passenger (weight). WE→S: The force of the earth pulling down on the scale (weight). NP→S: The force of the passenger pushing down on the scale (normal). NS→P: The force of the scale pushing up on the passenger (normal). NO→S: The force of the elevator pushing up on the scale (normal). NO→P: The force of the elevator pushing up on the passenger (normal).
a) Which of these forces affect the motion of the passenger? Select all that apply. NO→P WE→S NO→S WE→P NP→S NS→P
b) While it is accelerating downward, what does the scale read (in newtons)?
c) While it is accelerating downward, which of the forces in your diagrams have the same magnitude? For each equality you claim, explain what foothold principle makes you think that they are equal.
a) The forces that affect the motion of the passenger are:WE→P: The force of the earth pulling down on the passenger (weight).NS→P: The force of the scale pushing up on the passenger (normal).NO→P: The force of the elevator pushing up on the passenger (normal).
b) The scale reads (in newtons) = 784 Nc) While the elevator is accelerating downward, the forces that have the same magnitude are WE→S and NO→P. The Newton's Third Law of Motion makes me think that they are equal. Newton's third law of motion states that if an object A applies a force on another object B, then B applies a force on A that is equal in magnitude but opposite in direction.
The force that A applies on B is called the action force, and the force that B applies on A is called the reaction force. These two forces always occur in pairs.
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As in problem 80, an 85-kg man plans to tow a 111000−kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 8.1
∘
above the horizontal. The coefficient of static friction between his shoes and the runway is 0.89. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance. Number Units
The greatest acceleration the man can give the airplane while towing it along the runway is 2.79 m/s².
To find the greatest acceleration the man can provide, we need to consider the forces acting on the system. The vertical component of the man's pull does not contribute to the horizontal acceleration of the airplane since it is perpendicular to the direction of motion. Therefore, we only need to focus on the horizontal component of the man's force.
The frictional force between the man's shoes and the runway opposes the motion of the system. The maximum static frictional force can be calculated by multiplying the coefficient of static friction (0.89) by the normal force, which is the weight of the man (85 kg) multiplied by the acceleration due to gravity (9.8 m/s²). This gives us a maximum static frictional force of 725.86 N.
The horizontal component of the man's force is given by the tension in the cable multiplied by the cosine of the angle (8.1°). Since the cable is being pulled horizontally, this force acts in the same direction as the desired acceleration of the airplane. The equation for this force is T * cos(8.1°).
For the system to accelerate, the horizontal component of the man's force must be greater than or equal to the static frictional force. Therefore, we can set up the inequality: T * cos(8.1°) ≥ 725.86 N.
Solving this inequality for T gives us T ≥ 725.86 N / cos(8.1°), which is approximately 755.36 N. This is the minimum tension in the cable required to overcome the static friction and start the motion.
To find the greatest acceleration, we use Newton's second law: F = ma, where F is the net force and m is the mass of the airplane (111,000 kg). Rearranging the equation, we have a = F/m. The net force can be calculated as the tension in the cable minus the frictional force: F = T - [tex]f_{static[/tex].
Substituting the values, we get a = (755.36 N - 725.86 N) / 111,000 kg ≈ 2.79 m/s². Therefore, the greatest acceleration the man can give the airplane is approximately 2.79 m/s².
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opponent. If he accelerates uniformly at 0.50 m/s
2
, determine each of the following. (a) How lona does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.) (b) How far has he traveled in that time? m
Given,Acceleration = a = 0.50 m/s²As we know that, the player who has the puck continues to move with constant velocity.
Hence, the velocity of the opponent with respect to the player with the puck will be the relative velocity of the opponent with respect to the ground.
Let us assume that the initial velocity of the opponent is u and the distance between the two players is s. Then the final velocity of the opponent will be v and the time taken by the opponent to catch the player with the puck will be t.
a) To find out the time taken by the opponent to catch his opponent, we will use the formula of relative velocity as the velocity of the player with the puck is constant.v = u + at
Here, v = velocity of opponent relative to ground = Velocity of opponent wrt player with puck + Velocity of puck wrt ground
= V + u
= u (as puck moves with constant speed)v
= u + at0
= u + 0.50t (as v = 0, opponent catches up with the player)u = - 0.50t
Therefore, t = -2u/1 = -2u
Hence, it will take 2 seconds for the opponent to catch the player with the puck.
b) Now we have to calculate how far the opponent has traveled in that time.
The distance traveled by the opponent will be:
s = ut + 1/2 at²
= -0.50 (2) + 1/2 (0.50) (2)²
= -1 + 1= 0 m
The opponent has traveled 0 meters in that time.
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all of the lamp types, except light-emitting diodes (leds), are described by size and shape. lamp size is measured in ? increments.
The size of lamps, excluding light-emitting diodes (LEDs), is typically described using a standardized measurement called "lamp size increments." These size increments are a way to categorize and classify different lamp sizes based on their physical dimensions.
Lamp size increments are usually denoted by a numerical value followed by an "e" (e.g., E12, E26) and represent the diameter of the lamp base or socket in millimeters. The "e" stands for Edison, named after Thomas Edison, who popularized the use of standardized lamp bases. These size increments are important because they ensure compatibility between lamps and lamp fixtures. For example, a lamp with an E26 base will fit into a lamp socket designed for an E26-sized bulb. This standardization allows for easy interchangeability and ensures that lamps can be easily replaced or upgraded without the need for modifying the fixtures.
LEDs, on the other hand, do not conform to the same size and shape standards as traditional lamps. LED lamps come in various sizes and shapes, but they are typically not described using the same lamp size increments as other types of lamps. In summary, lamp size increments provide a standardized way to describe the size and compatibility of traditional lamps, allowing for easy interchangeability and ensuring compatibility with lamp fixtures.
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A spacecraft is traveling with a velocity of v0x = 6550 m/s along the +x direction. Two engines are turned on for a time of 909 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 2.02 m/s2, while the other gives it an acceleration in the +y direction of ay = 9.45 m/s2. At the end of the firing, what is a) vx and b) vy?
The final velocity in the y direction (vy) is 8585.05 m/s.
the final velocity in the x direction (vx) is 8389.18 m/s.
To find the final velocities in the x and y directions, we need to consider the initial velocity and the accelerations in each direction.
Given:
Initial velocity in the x direction, v0x = 6550 m/s
Acceleration in the x direction, ax = 2.02 m/s²
Acceleration in the y direction, ay = 9.45 m/s²
Time of firing, t = 909 s
(a) Final velocity in the x direction (vx):
We can use the formula:
vx = v0x + ax * t
Substituting the given values:
vx = 6550 m/s + 2.02 m/s² * 909 s
vx = 6550 m/s + 1839.18 m/s
vx = 8389.18 m/s
(b) Final velocity in the y direction (vy):
Since the acceleration in the y direction is constant and the initial velocity in the y direction is zero, we can use the formula:
vy = ay * t
Substituting the given values:
vy = 9.45 m/s² * 909 s
vy = 8585.05 m/s
To summarize:
(a) The final velocity in the x direction (vx) is 8389.18 m/s.
(b) The final velocity in the y direction (vy) is 8585.05 m/s.
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1-Calculate the longest and the shortest wavelength for the following scries: Lyman series Balmer series Paschen series Brackett series Pfund series 2-Find the wavelength and frequency of photon emitted for hydrogen atom, When: n=1→n=2 3-When an electron in the hydrogen atom moves from level n to the second level and emit photo with wavelength 4.34×10
−7
m Find the value of n
1
4-(a) What are the frequency and wavelength of a photon emitted during transition from n= 5 state to n=2 state in the hydrogen atom ? (b) In which region of the electromagnetic spectrum will this radiation lie?
1. The longest and shortest wavelengths for the following series are:
Lyman series: The shortest wavelength in the Lyman series occurs when the electron transitions to the n=1 energy level. The longest wavelength occurs when the electron transitions to the n=∞ energy level.Balmer series: The shortest wavelength in the Balmer series occurs when the electron transitions to the n=2 energy level. The longest wavelength occurs when the electron transitions to the n=∞ energy level.Paschen series: The shortest wavelength in the Paschen series occurs when the electron transitions to the n=3 energy level. The longest wavelength occurs when the electron transitions to the n=∞ energy level.Brackett series: The shortest wavelength in the Brackett series occurs when the electron transitions to the n=4 energy level. The longest wavelength occurs when the electron transitions to the n=∞ energy level.Pfund series: The shortest wavelength in the Pfund series occurs when the electron transitions to the n=5 energy level. The longest wavelength occurs when the electron transitions to the n=∞ energy level.2. When an electron transitions from n=1 to n=2 in a hydrogen atom, it emits a photon. The wavelength (λ) and frequency (ν) of the emitted photon can be calculated using the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
ν = c / λ
where
R is the Rydberg constant (approximately 1.097 x 10^7 m⁻¹), c is the speed of light (approximately 3.0 x 10^8 m/s), n₁ is the initial energy level (1 in this case),n₂ is the final energy level (2 in this case).Plugging in the values, we have:
1/λ = R * (1/1² - 1/2²) = R * (1 - 1/4) = R * (3/4)
λ = 4/3 * (1/R)
ν = c / λ = c / (4/3 * (1/R)) = 3c / (4 * (1/R)) = 3cR/4
3. The wavelength of the photon emitted when an electron in a hydrogen atom moves from level n to the second level is given as 4.34 x 10⁻⁷ m. We can use the Rydberg formula mentioned in the previous answer to find the initial energy level (ni).
1/λ = R * (1/n² - 1/2²)
1/(4.34 x 10⁻⁷) = R * (1/ni² - 1/2²)
ni² = 2² * (1 - 2² * (1/λR))
ni² = 4 * (1 - 4 * (1/(4.34 x 10⁻⁷ * R)))
ni² = 4 * (1 - 4 * (1/(4.34 x 10⁻⁷ * 1.097 x 10^7)))
ni² = 4 * (1 - 4 * (1/4.7618))
ni² = 4 * (1 - 0.839)
ni² = 4 * 0.161
ni = √0.644
ni ≈ 0.803
Therefore, the value of ni is approximately 0.803.
4a) To find the frequency (ν) of a photon emitted during the transition from n=5 to n=2 in the hydrogen atom, we can use the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
ν = c / λ
Given n₁ = 5 and n₂ = 2, we can plug these values into the formula:
1/λ = R * (1/5² - 1/2²) = R * (1/25 - 1/4) = R * (4/100 - 25/100) = R * (-21/100)
λ = -100/21 * (1/R)
ν = c / λ = c / (-100/21 * (1/R)) = -21c / (100 * (1/R)) = -21cR/100
b) To determine the region of the electromagnetic spectrum in which this radiation lies, we can calculate the wavelength using the equation:
λ = c / ν
Given that ν = -21cR/100, we can substitute this value into the equation:
λ = c / (-21cR/100) = -100/(21R)
Since the value is negative, we can take the absolute value to get the positive wavelength:
|λ| = 100/(21R)
The wavelength lies in the infrared region of the electromagnetic spectrum since the value is positive and is in the denominator (indicating longer wavelengths).
The correct format of the question should be:
3 1-Calculate the longest and the shortest wavelength for the following series:
Lyman series
Balmer series
Paschen series
Brackett series
Pfund series
2-Find the wavelength and frequency of photon emitted for hydrogen atom,
When: n=1 → n = 2
3-When an electron in the hydrogen atom moves from level n to the second level and emit photo with wavelength 4.34 x 10⁻⁷ m Find the value of ni
4-a) What are the frequency and wavelength of a photon emitted during transition from n = 5 state to n = 2 state in the hydrogen atom?
(b) In which region of the electromagnetic spectrum will this radiation lic?
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Calculate the magnitude of the electric field 2.45 m from a point charge of 6.89mC.1 mC equals 10 to the power of negative 3 end exponent C. The answer should be expressed in N/C. Use exponential format and 3 significant figures.
The magnitude of the electric field at a distance of 2.45 m from the point charge of 6.89 mC is approximately 5.066 × 10^6 N/C.
To calculate the magnitude of the electric field at a distance of 2.45 m from a point charge of 6.89 mC, we can use Coulomb's law. Coulomb's law states that the electric field at a point in space is given by:
E = k * (|Q| / r^2)
Where:
E is the electric field,
k is the electrostatic constant (k ≈ 8.99 × 10^9 N m^2/C^2),
|Q| is the magnitude of the charge, and
r is the distance from the charge.
Substituting the given values:
|Q| = 6.89 mC
= 6.89 × 10^(-3) C
r = 2.45 m
We can calculate the magnitude of the electric field E as follows:
E = (8.99 × 10^9 N m^2/C^2) * (6.89 × 10^(-3) C) / (2.45 m)^2
E ≈ (8.99 × 10^9) * (6.89 × 10^(-3)) / (2.45)^2 N/C
E ≈ 5.066 × 10^6 N/C
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Jack and Jill ran up the hill at 2.7 m/s. The horizontal component of Jill's velocity vector was 2.3 m/s. Part A What was the angle of the hill? Express your answer in degrees. Part B What was the vertical component of Jill's velocity? Express your answer with the appropriate units.
Part A: The angle of the hill is approximately 36.87 degrees. Part B: The vertical component of Jill's velocity is approximately 1.54 m/s.
Part A: To find the angle of the hill, we can use the inverse tangent function:
θ = arctan(vertical component / horizontal component)
θ = arctan(v_y / v_x)
Given that the horizontal component of Jill's velocity (v_x) is 2.3 m/s and the total velocity magnitude is 2.7 m/s, we can calculate the vertical component of Jill's velocity (v_y) using the Pythagorean theorem:
v_y = sqrt(v_total^2 - v_x^2)
Substituting the known values:
v_y = sqrt((2.7 m/s)^2 - (2.3 m/s)^2)
Simplifying:
v_y ≈ 1.54 m/s
Now we can calculate the angle:
θ = arctan(1.54 m/s / 2.3 m/s)
θ ≈ 36.87 degrees
Therefore, the angle of the hill is approximately 36.87 degrees.
Part B: The vertical component of Jill's velocity is approximately 1.54 m/s.
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The angle of the hill can be calculated using the formula θ = cos^-1(Ux / U), where Ux is the horizontal velocity and U is the total velocity. The vertical component of the Jill's velocity can be computed with the equation Uy = √(U² - Ux²).
Explanation:Part A: The angle of the hill would be determined by using the relationship between the overall speed (2.7 m/s), the horizontal component of the speed (2.3 m/s), and the angle of ascent. This comes down to trigonometry, in this case the cosine of the angle, defined as adjacent (horizontal component) over hypotenuse (overall speed). Mathematically, this can be expressed as cos(θ) = Ux / U where Ux is the horizontal component of the velocity (2.3 m/s) and U is the total velocity (2.7 m/s). Solving for θ, we get θ = cos^-1(Ux / U).
Part B: The vertical component of the velocity can be found using the Pythagorean theorem, which in this case is Uy = √(U² - Ux²). Here, Uy is the vertical component of the velocity, U is the total velocity (2.7 m/s), and Ux is the horizontal component of the velocity (2.3 m/s). Running these numbers through the equation will yield the vertical component of Jill's velocity.
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How is the viscosity of hydraulic oil affected by temperature change?
Viscosity is the measure of a liquid's resistance to flow. In the case of hydraulic oil, its viscosity changes with temperature, decreasing as the temperature rises and increasing as the temperature drops.
The viscosity of hydraulic oil is affected by temperature change. The viscosity of hydraulic oil changes with temperature, with the value of the viscosity decreasing as the temperature rises and increasing as the temperature drops.
What is viscosity?
Viscosity is the measure of a liquid's resistance to flow. In hydraulics, the viscosity of hydraulic oil is a significant aspect since it affects the oil's ability to transmit power. Hydraulic oil viscosity is a physical property of a liquid that is defined as the force per unit area required to make it flow when subjected to a force.
The relationship between viscosity and temperature in hydraulic oil is crucial because it has a significant impact on the oil's flow properties. The viscosity of hydraulic oil decreases as the temperature rises and increases as the temperature drops. This is due to the fact that a temperature change affects the thickness of the fluid or its ability to flow.
Thus, we can say that the viscosity of hydraulic oil is affected by temperature change.
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A jogger runs a distance of 8.4 km in a straight line on level ground. For the first 4.2 km, he runs at a speed of 3.0 m/s. Then he suddenly speed up to 7.0 m/s and maintains that speed for the last 4.2 km. What is the average speed of the jogger for the entire trip? a. 5.8 m/s b. 5.4 m/s c. 4.6 m/s d. 4.2 m/s 6.5.0 m/s
The average speed of the jogger for the entire trip is 4.2 m/s. Average speed is the total distance covered / total time taken during the whole journey.
We have to find the average speed of the jogger for the entire trip.
The distance covered by the jogger for the first 4.2 km is, Distance_1 = 4.2 km × 1000 m/km = 4200 m.
The speed of the jogger for the first 4.2 km is, Speed_1 = 3.0 m/s.
The time taken by the jogger for the first 4.2 km is, Time_1 = Distance_1 / Speed_1 = 4200 m / 3.0 m/s = 1400 s.
The distance covered by the jogger for the last 4.2 km is, Distance_2 = 4.2 km × 1000 m/km = 4200 m.
The speed of the jogger for the last 4.2 km is, Speed_2 = 7.0 m/s.
Time taken by jogger for the last 4.2 km is, Time_2 = Distance_2 / Speed_2 = 4200 m / 7.0 m/s = 600 s.
The total distance covered by the jogger is Distance = Distance_1 + Distance_2 = 8400 m.
The total time taken by the jogger is Time = Time_1 + Time_2 = 1400 s + 600 s = 2000 s.
The average speed of the jogger for the entire trip is Total speed = Distance / Time= 8400 m / 2000 s= 4.2 m/s. Therefore, the correct option is d. 4.2 m/s.
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A 38 cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30 T magnetic field at its center. Part A If the maximum current is 4.7 A, how many turns must the solenoid have? Express your answer using two significant figures.
The number of turns required for the solenoid is 92.
Given values,Length of the solenoid, L = 38 cm = 0.38m Diameter of the solenoid, d = 1.8 cm = 0.018 m
Magnetic field at the center, B = 0.30 TCurrent, I = 4.7 A
We know that the formula for the magnetic field at the center of the solenoid is,
B = (μ0 * N * I) / L
where, μ0 = 4π × 10⁻⁷ T ⋅ m/A
is the permeability of free space
Substituting the given values,
0.30 = (4π × 10⁻⁷ × N × 4.7) / 0.38N = 0.38 × 0.30 / (4π × 10⁻⁷ × 4.7)N = 92 turns
Therefore, the number of turns required for the solenoid is 92.
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According to Table 14.1, the meteor shower that occurs closest to the autumnal equinox is the
(a) Lyrids;
(b) Beta Taurids;
(c) Perseids;
(d) Orionids.
The meteor shower that occurs closest to the autumnal equinox is the (d) Orionids.
According to Table 14.1, the meteor shower that occurs nearest to the autumnal equinox is the Orionids. This meteor shower, known as the Orionids, happens annually and typically takes place in late October. It is caused by the remnants of Halley's Comet, which leave behind debris as they orbit the sun. When the Orionids reach their peak, they can generate approximately 20 meteors per hour visible in the nighttime sky.
The Orionids meteor shower is a recurring event that unfolds each year during late October. It is attributed to the Earth crossing paths with the remnants of Halley's Comet. These remnants, scattered along the comet's orbit, intersect our planet's atmosphere and create a spectacle of shooting stars. The shower derives its name from the constellation Orion, as the meteors seem to radiate from this particular area of the sky. Skywatchers and astronomy enthusiasts can anticipate witnessing an impressive display of up to 20 meteors per hour during the peak of the Orionids shower.
In summary, according to Table 14.1, the Orionids meteor shower occurs closest to the autumnal equinox. It is an annual event taking place in late October, characterized by a remarkable display of up to 20 meteors per hour. The meteor shower is caused by the residual debris left behind by Halley's Comet during its orbit around the sun.
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