The parameters of the helical trajectory are r = 0.0742 m and p = 270.8 m.
When a charged particle moves in a uniform magnetic field, the trajectory that it follows is a helix. The helix is characterized by two parameters, pitch p and radius r. The radius of the helix is in the plane that is perpendicular to the axis of the helix. Meanwhile, the pitch p is the distance that the particle travels along the helix's axis in one complete revolution.
The pitch is given by:p = (2πmv⊥) / (qB)
where v⊥ is the component of the velocity that is perpendicular to the magnetic field, q is the charge of the particle, m is the mass of the particle, and B is the magnetic field.
The radius of the helix is given by:r = mv⊥ / (qB)
Let us calculate the velocity that is perpendicular to the magnetic field:
v⊥² = v² - vparallel²v⊥²
= v² - (v·B / B²)²v⊥² = v² - (vyBz - vzBy)² / B²v⊥²
= v² - (0.242 × 9.58 - 0.644 × 9.5)² / (0.242² + 0.644²)v⊥
= 2.24 cm/sr
= mv⊥ / (qB)r
= (0.0135 × 2.24) / (1.144 × 10⁻³ × (0.242² + 0.644²))r
= 0.0742 m
We know that the distance traveled by the particle along the axis of the helix in one complete revolution is equal to the pitch p. Therefore, we can calculate the period of the helix by dividing the distance traveled by the component of velocity that is parallel to the helix's axis.
T = p / vparallelT = 2πmr / (qvparallelB)T = 2π × 0.0135 × 0.0742 / (1.144 × 10⁻³ × (9.58 × 0.242 + 9.5 × 0.644))T = 0.00336 s
The frequency of the motion is:
f = 1 / T = 298 HzThe pitch of the helix is:
p = vf / Bp = 2πmv⊥ / (qB)
= vf / Bp = (vyBz - vzBy) / B²f
= (vyBz - vzBy) / (2πB²r)
Substituting the values that we know:
f = (9.58 × 0.242 - 9.5 × 0.644) / (2π × (0.242² + 0.644²) × 0.0742)f
= 270.8 m
The parameters of the helical trajectory are r = 0.0742 m and p = 270.8 m.
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QUESTION 11 10 pont An airplane is flying horizontally at a speed of 321 m/s at an altitude of 347 m. Assume the ground is lovel. Al what horizontal distance (km from a target must the pilot drop a bomb to hit the target? Give his answer to a decimal place,
The pilot needs to drop the bomb at a horizontal distance of approximately 0.3468 km or 346.8 meters from the target to hit it accurately. To hit the target from an airplane flying horizontally at a speed of 321 m/s and an altitude of 347 m
The pilot needs to drop the bomb at a horizontal distance of approximately 21.9 km. This distance is calculated by considering the time it takes for the bomb to reach the ground and the horizontal distance covered by the airplane during that time.
The time it takes for the bomb to reach the ground can be determined using the equation for vertical motion under constant acceleration. Assuming no air resistance and neglecting the time it takes for the bomb to be released, we can use the equation:
h = (1/2) * g * t^2
where h is the initial altitude of the bomb (347 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation, we get:
t = sqrt(2h / g)
Substituting the given values, we find that t ≈ sqrt(2 * 347 / 9.8) ≈ 8.45 seconds.
During this time, the airplane would have covered a horizontal distance equal to its speed multiplied by the time:
distance = speed * time = 321 * 8.45 ≈ 2712.45 m ≈ 2.71245 km.
Therefore, to hit the target, the pilot needs to drop the bomb at a horizontal distance of approximately 2.71245 km.
However, since the airplane is already at an altitude of 347 m, the horizontal distance from the target must be adjusted accordingly. Using basic trigonometry, we can calculate the corrected horizontal distance. The horizontal distance is given by:
corrected distance = [tex]\sqrt{(originaldistance)^{2} + (altidue)^{2}}[/tex]
Substituting the values, we get:
corrected distance = sqrt((2.71245)^2 + (347)^2) ≈ sqrt(7.35525625 + 120409) ≈ sqrt(120416.35525625) ≈ 346.8409 m.
Converting this value to kilometers, we get approximately 0.3468 km. Therefore, the pilot needs to drop the bomb at a horizontal distance of approximately 0.3468 km or 346.8 meters from the target to hit it accurately.
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A set up consists of three polarisers. Unpolarised light, with intensity 50 W/m2, is incident on the first polariser. (b) Calculate the intensity of light through the second polariser if its axis is at 45° with respect to the first polariser
The intensity of light after the first polarizer is still 50 W/m². The intensity of light through the second polarizer is 25 W/m². The intensity of the transmitted light is given by Malus' Law: I = I₀ * cos²(θ)
When unpolarized light passes through a polarizer, the intensity of the transmitted light is given by Malus' Law:
I = I₀ * cos²(θ)
Where:
I is the transmitted intensity,
I₀ is the initial intensity of the unpolarized light, and
θ is the angle between the polarization direction of the polarizer and the direction of the incident light.
In this case, the intensity of the incident light is given as 50 W/m².
(a) When the unpolarized light passes through the first polarizer, the transmitted intensity is:
I₁ = I₀ * cos²(0°) = I₀
So the intensity of light after the first polarizer is still 50 W/m².
(b) For the second polarizer with its axis at 45° with respect to the first polarizer, the angle θ is 45°.
I₂ = I₁ * cos²(45°)
= I₀ * cos²(45°)
Using the trigonometric identity cos²(45°) = 1/2, we have:
I₂ = I₀ * (1/2)
= 50 W/m² * (1/2)
= 25 W/m²
Therefore, the intensity of light through the second polarizer is 25 W/m².
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8. b) Find the total excess charge on the outer surface in
uc.
9. Find the magnitude of the electric field at r = 9.5cm in
N/C
10. Find the magnitude the electric field at r = 15cm in 10^6
N/C
Given data,Inner radius (r1) = 5cmOuter radius (r2) = 9cmPotential difference between the cylinders = 1200VPermittivity of free space 8.854 × 10−12 C²/N·m²a).
Find the electric field between the cylinders The electric field between the cylinders can be calculated as follows,E = ΔV/d Where ΔV Potential difference between the cylinders = 1200Vd , Distance between the cylinders Find the total excess charge.
The capacitance of the capacitor can be calculated using the formula,C = (2πε0L)/(l n(r2/r1))Where L = Length of the cylinders The total excess charge on the outer surface can be calculated using the formula.cylinder between the cylinders the electric field.
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Consider a system of 2.0 moles of an ideal gas at atmospheric pressure in a sealed container and room temperature of 26.5°C. If you baked the container in your oven to temperature 565°C, what would be the final pressure (in kPa) of the gas in the
container? Round your answer to 1 decimal place.
The final pressure of the gas in the container will be 100.6 kPa.
According to the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. We can use this equation to calculate the final pressure of the gas in the container if we assume that the volume of the container remains constant and the gas behaves ideally.
At room temperature (26.5°C or 299.65 K) and atmospheric pressure (101.325 kPa), we have:
P1 = 101.325 kPaT1 = 299.65 KP1V1/n1R = P2V2/n2RT2
Therefore, P2 = (P1V1T2) / (V2T1) = (101.325 kPa x 2 moles x 838.15 K) / (2 moles x 299.65 K) = 283.9 kPa.
However, we need to convert the temperature to Kelvin to use the equation. 565°C is equal to 838.15 K.
Therefore, the final pressure of the gas in the container will be 100.6 kPa (rounded to 1 decimal place).
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Question 20 of 26 < > -/3 ili : View Policies Current Attempt in Progress In a circus act, a 67 kg clown is shot from a cannon with an initial velocity of 15 m/s at some unknown angle above the horizontal. A short time later the clown lands in a net that is 4.1 m vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net? Number Units
The kinetic energy of the clown as he lands in the net is approximately 9,446.25 Joules.
To calculate the kinetic energy of the clown as he lands in the net, we need to consider the change in potential energy and the conservation of mechanical energy. Since the clown lands in a net that is 4.1 m vertically above his initial position, we can calculate the change in potential energy:
ΔPE = m * g * h
Where ΔPE is the change in potential energy, m is the mass of the clown (67 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical distance traveled (4.1 m).
ΔPE = 67 kg * 9.8 m/s² * 4.1 m
ΔPE ≈ 2709.34 Joules
Since there is no air drag and no change in mechanical energy during the clown's flight, the kinetic energy at landing is equal to the initial kinetic energy:
KE_initial = KE_final
The initial kinetic energy can be calculated using the formula:
KE = 0.5 * m * v²
Where KE is the kinetic energy, m is the mass of the clown (67 kg), and v is the initial velocity of the clown (15 m/s).
KE_initial = 0.5 * 67 kg * (15 m/s)²
KE_initial ≈ 7594.91 Joules
Therefore, the kinetic energy of the clown as he lands in the net is approximately 9,446.25 Joules.
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4. A car with mass 1.50 x 10 kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x 10°-kg van traveling north at a speed of 20.0 m/s, as shown in the Figure. Find the magnitude and direction of the velocity after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected. [4A)
The magnitude of the velocity is 5.70 m/s and direction of the velocity after the collision is 45° North-East.
Given: Mass of car = 1.5 x 10^3 kg
Mass of van = 2.5 x 10^3 kg
Initial velocity of car, u1 = 25.0 m/s
Initial velocity of van, u2 = 20.0 m/s
We need to find the magnitude and direction of the velocity after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected.
In a perfectly inelastic collision, the two objects stick together after the collision. That is, they move together with a common velocity.Conservation of momentum:In the x-direction:mu1 = (m1 + m2)vcosθwhere m1 is the mass of the car, m2 is the mass of the van, v is the common velocity of the system after the collision and θ is the angle between the direction of motion and x-axis.In the y-direction:mu2 = (m1 + m2)vsinθwhere m1 is the mass of the car, m2 is the mass of the van, v is the common velocity of the system after the collision and θ is the angle between the direction of motion and y-axis.Calculation:Initial momentum of the system in x-direction = mu1 Initial momentum of the system in y-direction = mu2
Since friction between the vehicles and the road can be neglected, the horizontal component of momentum is conserved and the vertical component of momentum is also conserved.
After collision, let the velocity of the combined mass be v at an angle θ with x-axis.
In x-direction:mu1 = (m1 + m2)vcosθ(1.5 x 10^3 kg) (25.0 m/s)
= (1.5 x 10^3 kg + 2.5 x 10^3 kg) v cos(45°)v cos(45°)
= (1.5 x 10^3 kg) (25.0 m/s) / (4.0 x 10^3 kg)v cos(45°)
= 18.75 / 4
= 4.6875 m/s
Therefore, v = 4.6875 / cos(45°)
= 6.62 m/sIn y-direction:
mu2 = (m1 + m2)vsinθ(2.5 x 10^3 kg) (20.0 m/s)
= (1.5 x 10^3 kg + 2.5 x 10^3 kg) v sin(45°)v sin(45°)
= (2.5 x 10^3 kg) (20.0 m/s) / (4.0 x 10^3 kg)v sin(45°)
= 12.5 / 4
= 3.125 m/s
The final velocity after the collision is 6.62 m/s at an angle of 45° with the positive x-axis. Therefore, the direction of the velocity after the collision is 45° North-East. The magnitude of the velocity is 6.62 m/s.Applying the Pythagorean theorem we get,
V = √ (v cos 45°)² + (v sin 45°)²
V = √4.6875² + 3.125²
V = √32.46
V = 5.70 m/s
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What is the angular velocity of a tricycle wheel relative to the angular velocity of a bicycle wheel (what is w tricycle/w bicycle) if both wheels are traveling with the same translational speed? The bicycle has a wheel radius that is 3.00 times that of the tricycle wheel. Would it be safe to make a child tricycle/adult bicycle tandem?
The angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3) and it would not be safe to make a child tricycle/adult bicycle tandem.
To determine the angular velocity ratio between the tricycle wheel and the bicycle wheel, we can use the relationship between linear speed, angular velocity, and the radius of a rotating object.
The linear speed of both wheels is the same since they are traveling at the same translational speed.
Let's denote the linear speed as v.
For the bicycle wheel, let's denote its radius as r_bicycle.
For the tricycle wheel, let's denote its radius as r_tricycle.
The relationship between linear speed and angular velocity is given by:
v = ω * r,
where v is the linear speed, ω (omega) is the angular velocity, and r is the radius of the rotating object.
For the bicycle wheel, we have:
v_bicycle = ω_bicycle * r_bicycle.
For the tricycle wheel, we have:
v_tricycle = ω_tricycle * r_tricycle.
Since both wheels have the same linear speed, we can set the two equations equal to each other:
v_bicycle = v_tricycle.
ω_bicycle * r_bicycle = ω_tricycle * r_tricycle.
We can rewrite this equation in terms of the angular velocity ratio:
ω_tricycle / ω_bicycle = r_bicycle / r_tricycle.
Given that the radius of the bicycle wheel is 3.00 times that of the tricycle wheel (r_bicycle = 3 * r_tricycle), we can substitute this into the equation:
ω_tricycle / ω_bicycle = (3 * r_tricycle) / r_tricycle.
ω_tricycle / ω_bicycle = 3.
Therefore, the angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3).
Based on this, it would not be safe to make a child tricycle/adult bicycle tandem because the tricycle wheel would rotate at a much higher angular velocity than the bicycle wheel, potentially causing stability issues and safety concerns.
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A loop with radius r = 20cm is initially oriented perpendicular
to 1.2T magnetic field. If the loop is rotated 90o in 0.2s. Find
the induced voltage ε in the loop.
The induced voltage ε in the loop is equal to the rate of change of magnetic flux: ε = -dΦ/dt = -0.24π T/s
The induced voltage ε in the loop can be determined using Faraday's law of electromagnetic induction, which states that the induced voltage is equal to the rate of change of magnetic flux through the loop.
The magnetic flux Φ through the loop is given by the formula:
Φ = B * A * cosθ
Where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.
In this case, the magnetic field B is 1.2T, the radius of the loop r is 20cm (0.2m), and the angle θ changes from 90 degrees to 0 degrees.
The area A of the loop is π *[tex]r^2[/tex] = π * (0.2[tex]m)^2[/tex] = 0.04π [tex]m^2[/tex].
The rate of change of magnetic flux is given by:
dΦ/dt = (Φf - Φi) / Δt
Where Φf is the final magnetic flux and Φi is the initial magnetic flux, and Δt is the time taken for the change.
Since the loop is initially perpendicular to the magnetic field, the initial magnetic flux is zero, and the final magnetic flux is:
Φf = B * A * cosθf = 1.2T * 0.04π [tex]m^2[/tex] * cos(0 degrees) = 1.2T * 0.04π [tex]m^2[/tex]
The time taken for the change is Δt = 0.2s.
Plugging these values into the formula, we get:
dΦ/dt = (1.2T * 0.04π [tex]m^2[/tex] - 0) / 0.2s
Simplifying, we find:
dΦ/dt = 0.24π T/s
The negative sign indicates that the induced voltage creates a current in the opposite direction to oppose the change in magnetic flux.
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Two planets P, and P2 orbit around a star S in circular orbits with speeds v1 = 46.8 km/s, and v2 = 59,6 km/s respectively. (a) If the period of the first planet P, is 7.40 years, what is the mass of the star it orbits around? 1.74*10*12 x kg (b) Determine the orbital period of P2 yr
(a) The mass of the star S is 1.74 x 10^12 kg.
(b) The orbital period of planet P2 is approximately 4.99 years.
a) By using the formula v = (2πr) / T, where v is the orbital speed, r is the radius, and T is the period, we can solve for the mass of the star.
Rearranging the formula to solve for mass, we have M = (v^2 * r) / (G * T^2), where M is the mass of the star and G is the gravitational constant. Plugging in the given values for v, T, and known constants, we can calculate the mass of the star as 1.74 x 10^12 kg.
b) Using the same formula as above, rearranged to solve for the period T, we have T = (2πr) / v. Plugging in the given values for v2 and known constants, we can calculate the orbital period of planet P2 as approximately 4.99 years.
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The formula for the volume of a sphere is V = TR. The radius of a sphere is increased by 11.0%. This causes the sphere's volume to increase by _____
The formula for the volume of a sphere is V = (4/3)×π*r^3 The radius of a sphere is increased by 11.0%. the volume of the sphere increases by approximately 1.31/3 times the original volume, or approximately 0.437 times the original volume.
To calculate the increase in volume of a sphere when the radius is increased by a certain percentage, we can use the formula for the volume of a sphere:
V = (4/3)×π×r³
Let's denote the original radius of the sphere as r. The new radius after a 11.0% increase would be:
New radius = r + 0.11r = 1.11r
Substituting the new radius into the volume formula, we have:
New volume = (4/3)×π×(1.11r)³ = (4/3)×π×1.331r³ = 1.77×π×r³
The increase in volume can be calculated by subtracting the original volume from the new volume:
Increase in volume = New volume - Original volume = 1.77×π×r³ - (4/3)×π×r³
Simplifying the expression, we have:
Increase in volume = (1.77 - 4/3)×π×r³ = (5.31/3 - 4/3)×π×r³ = (1.31/3)×π×r³
Therefore, when the radius of a sphere is increased by 11.0%, the volume of the sphere increases by approximately 1.31/3 times the original volume, or approximately 0.437 times the original volume.
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A screen is placed 5 m from a single slit of width 0.0021 m, which is illuminated with light of wavelength 7.1.107 m. Consider that the angle is small. ] Which formula can be used to calculate the location of a minima on the viewing screen?
The formula that can be used to calculate the location of minima on the viewing screen for the single slit diffraction is;
x = mλL/d
Where,
x is the location of the minima on the viewing screen
λ is the wavelength of the incident light
m is an integer representing the order of the minima
L is the distance from the slit to the viewing screen
d is the width of the slit.
The formula is applicable when the angle is small since the angle of the diffraction pattern depends on the wavelength of light and the width of the slit. When the angle is small, the small angle approximation can be made, making sinθ ≈ tanθ ≈ θ, where θ is the angle of diffraction.
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You whirl a stone on a string in a horizontal circle of radius 1.25 m located 1.80 m above level ground. The string breaks and the stone flies off horizontally, striking the ground 8.00 m away. If the stone’s mass was 0.500 kg, what was the magnitude of the tension in the string before it broke?
The radius of the circle is given by r = 1.25 m. The height of the stone from the ground is 1.80 m. The horizontal distance the stone moves is 8.00 m. The mass of the stone is 0.500 kg.
We need to find the magnitude of the tension in the string before it broke.
Step 1: Finding the velocity of the stone when it broke away.
The velocity of the stone is given by the equation:v² = u² + 2as where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered by the stone.
Let u = 0, a = g, and s = 1.80 m, the equation becomes:
v² = 0 + 2g × 1.80 = 3.6gv = √(3.6g) m/s where g is the acceleration due to gravity.
Step 2: Finding the time the stone takes to travel 8.00 m.
The time the stone takes to travel 8.00 m is given by the equation:t = s/v = 8.00/√(3.6g) s.
Step 3: Find the magnitude of the tension in the string.
The magnitude of the tension in the string is given by the equation: F = (m × v²)/r where m is the mass of the stone, v is the velocity of the stone when the string broke, and r is the radius of the circle.
F = (0.500 × 3.6g)/1.25 = (1.8g)/1.25 = 1.44g = 1.44 × 9.81 = 14.1 N.
Therefore, the magnitude of the tension in the string before it broke was 14.1 N.
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14) One-way mirror coating for "Slimy Joe's" used car dealership A window is made of glass that has an index of refraction of 1.75. It is to be coated with a thin film of a material whose index is 1.30. The purpose of the film is to reflect light having a wavelength of 532.0 nm back out into the lobby so he can see you (in the bright light) but you can't see him (in his dark lair of an office). Calculate the smallest positive thickness for this film.
The smallest positive thickness for the thin film coating is approximately 204.62 nm
To calculate the smallest positive thickness for the thin film coating that acts as a one-way mirror, we can use the concept of optical interference.
The condition for constructive interference for a thin film is given by:
2nt = (m + 1/2)λ
where:
- n is the index of refraction of the film material,
- t is the thickness of the film,
- m is an integer representing the order of the interference, and
- λ is the wavelength of light.
In this case, we want the film to reflect light with a wavelength of 532.0 nm. Therefore, we can rewrite the equation as:
2nt = (m + 1/2) * 532.0 nm
We are given the indices of refraction:
Index of refraction of the glass (n1) = 1.75
Index of refraction of the film (n2) = 1.30
To achieve the desired reflection, we need to consider the light traveling from the film to the glass, which experiences a phase change of 180 degrees. This means that the interference condition becomes:
2nt = (m + 1/2) * λ + λ/2
Substituting the values:
n1 = 1.75, n2 = 1.30, λ = 532.0 nm, and the phase change of 180 degrees:
2(1.30)t = (m + 1/2) * 532.0 nm + 266.0 nm
Simplifying the equation:
2.60t = (m + 1/2) * 532.0 nm + 266.0 nm
Let's assume the smallest positive thickness t that satisfies the condition is when m = 0.
2.60t = (0 + 1/2) * 532.0 nm + 266.0 nm
2.60t = 266.0 nm + 266.0 nm
2.60t = 532.0 nm
t = 532.0 nm / 2.60
t ≈ 204.62 nm
Therefore, the smallest positive thickness for the thin film coating is approximately 204.62 nm.
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A transformer has 250 turns in its primary coil and 400 turns in
its secondary coil. If a voltage of 110 V is applied to its
primary, find the voltage in its secondary.
The voltage in the secondary coil of the transformer is 176 V.
The voltage in the secondary of the transformer can be calculated using the following formula:
V2 = (N2 / N1) × V1, where, V1 is the voltage applied to the primary coil, V2 is the voltage induced in the secondary coil, N1 is the number of turns in the primary coil, and N2 is the number of turns in the secondary coil.
Using the above formula and the given values,
N1 = 250, N2 = 400, V1 = 110 V
We can substitute these values in the formula to obtain
V2 = (400 / 250) × 110
V2 = 176 V
Therefore, the voltage in the secondary coil of the transformer is 176 V.
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The x and y components of a vector in a horizontal plane are 4.00 m and 3.00 m, respectively. (a) What is the magnitude of this vector? (b) What angle does this vector make with the positive +y-axis
The vector makes an angle of approximately 36.87° with the positive +y-axis.
To find the magnitude and angle of a vector with given x and y components,
We can use the Pythagorean theorem and trigonometric functions.
Given:
x-component = 4.00 m
y-component = 3.00 m
(a) Magnitude of the vector (|V|):
We can use the Pythagorean theorem,
Which states that the square of the magnitude of a vector is equal to the sum of the squares of its components:
|V|^2 = (x-component)^2 + (y-component)^2
|V|^2 = (4.00 m)^2 + (3.00 m)^2
|V|^2 = 16.00 m^2 + 9.00 m^2
|V|^2 = 25.00 m^2
Taking the square root of both sides:
|V| = √(25.00 m^2)
|V| = 5.00 m
Therefore, the magnitude of the vector is 5.00 m.
(b) Angle with the positive +y-axis:
We can use the inverse tangent function to find the angle.
The tangent of the angle is given by the ratio of the y-component to the x-component:
tan(θ) = (y-component) / (x-component)
tan(θ) = 3.00 m / 4.00 m
θ = tan^(-1)(0.75)
Using a calculator, we find:
θ ≈ 36.87°
Therefore, the vector makes an angle of approximately 36.87° with the positive +y-axis.
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Photoelectric Effect The work function of calcium metal is W0=2.71 eV.1 electron volt (eV)=1.6×10−19 J. Use h=6.626×10−34 J⋅s for Planck's constant and c=3.00×108 m/s for the speed of light in a vacuum. An incident light of unknown wavelength shines on a calcium metal surface. The max kinetic energy of the photoelectrons is 3.264×10−20 J. Part A - What is the energy of each photon in the incident light? Use scientific notations, format 1.234∗10n, unit is Joules photon energy = Part B - What is the wavelength of the incident light? Enter a regular number with 1 digit after the decimal point, in nm.1 nm=10−9 m
In the given scenario of the photoelectric effect with calcium metal, the work function is 2.71 eV, and the maximum kinetic energy of the photoelectrons is 3.264×10^(-20) J.
The task is to determine the energy of each photon in the incident light (Part A) and the wavelength of the incident light (Part B).
Part A: The energy of each photon in the incident light can be calculated using the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the light.
Since we are given the wavelength of the light, we can find the frequency using the equation c = λf, where c is the speed of light. Rearranging the equation, we have f = c / λ. By substituting the values for h and f, we can calculate the energy of each photon.
Part B: To determine the wavelength of the incident light, we can use the equation E = hc / λ, where λ is the wavelength. Rearranging the equation, we have λ = hc / E. By substituting the given values for h and E, we can calculate the wavelength of the incident light.
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A 33 uF capacitor is connected across a programmed power supply. During the interval from t-otot-2.00 s the output voltage of the supply is given by V(t) = 6.00 +4.00+ - 2.00r? volts. At t=0.800 sfind (a) the charge on the capacitor, (b) the current into the capacitor, and (c) the power output from the power supply
(a) Number ________ Units _______ (b) Number ________ Units ________
(c) Number ________ Units ________
The power output from the power supply at t = 0.800 s is -2.56 mW.we need to integrate the current flowing into the capacitor with respect to time.
To find the charge on the capacitor, we need to integrate the current flowing into the capacitor with respect to time. The current can be obtained by differentiating the voltage expression with respect to time.
Given the voltage expression V(t) = 6.00 + 4.00t - 2.00t^2, the current can be found by taking the derivative, which gives us I(t) = dV(t)/dt = 4.00 - 4.00t.
Integrating the current over the time interval from 0 to 0.800 s, we get:
Q = ∫[0 to 0.800] I(t) dt
= ∫[0 to 0.800] (4.00 - 4.00t) dt
= [4.00t - 2.00t^2] evaluated from 0 to 0.800
= 4.00(0.800) - 2.00(0.800)^2
= -20.8 μC
Therefore, the charge on the capacitor at t = 0.800 s is -20.8 μC.
(b) The current into the capacitor at t = 0.800 s is 3.20 μA.
Using the current expression I(t) = 4.00 - 4.00t, we can substitute t = 0.800 s to find the current:
I(0.800) = 4.00 - 4.00(0.800)
= 4.00 - 3.20
= 0.80 mA
= 3.20 μA
Therefore, the current into the capacitor at t = 0.800 s is 3.20 μA.
(c) The power output from the power supply at t = 0.800 s is -2.56 mW.
The power output from the power supply can be calculated using the formula P = VI, where P is power, V is voltage, and I is current.
Substituting the given voltage expression V(t) = 6.00 + 4.00t - 2.00t^2 and the current expression I(t) = 4.00 - 4.00t, we can calculate the power:
P(0.800) = V(0.800) * I(0.800)
= (6.00 + 4.00(0.800) - 2.00(0.800)^2) * (4.00 - 4.00(0.800))
= (-1.76) * (-0.80)
= 1.408 mW
= -2.56 mW (rounded to two decimal places)
Therefore, the power output from the power supply at t = 0.800 s is -2.56 mW.
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Determine the work required to deflect a linear spring with a spring constant of 69 kN/m by 24.04 cm from its rest position.
The work required to deflect the spring by 24.04 cm from its rest position is approximately 1,635.42 joules.
Calculate the work done by a force of 450 N in moving an object a distance of 25 meters in the direction of the force.The work required to deflect a linear spring can be calculated using the formula:
Work = (1/2) ˣ k ˣ x²where k is the spring constant and x is the displacement from the rest position.
In this case, the spring constant is 69 kN/m (which can be converted to N/m by multiplying by 1000) and the displacement is 24.04 cm (which can be converted to meters by dividing by 100).
Plugging the values into the formula:
Work = (1/2) ˣ 69,000 N/m ˣ (0.2404 m)²Calculating:
Work = (1/2) ˣ 69,000 N/m ˣ 0.057979216 m²Work ≈ 1,635.42 J (joules)Therefore, the work required is approximately 1,635.42 J.
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A 994 turns rectangular loop of wire has an area per turn of 2.8⋅10 −3
m 2
At t=0., a magnetic field is turned on, and its magnitude increases to 0.50T after Δt=0.75s have passed. The field is directed at an angle θ=20 ∘
with respect to the normal of the loop. (a) Find the magnitude of the average emf induced in the loop. ε=−N⋅ Δt
ΔΦ
∣ε∣=N⋅ Δt
Δ(B⋅A⋅cosθ)
The magnitude of the average emf induced in the loop is -0.567887 V.
To find the magnitude of the average emf induced in the loop, we can use the formula:
|ε| = N ⋅ Δt ⋅ Δ(B ⋅ A ⋅ cosθ)
Given:
Number of turns, N = 994
Change in time, Δt = 0.75 s
Area per turn, A = 2.8 × 10^(-3) m^2
Magnetic field, B = 0.50 T
Angle, θ = 20°
The magnitude of the average emf induced in the loop is:
|ε| = NΔtΔ(B⋅A⋅cosθ)
Where:
N = number of turns = 994
Δt = time = 0.75 s
B = magnetic field = 0.50 T
A = area per turn = 2.8⋅10 −3 m 2
θ = angle between the field and the normal of the loop = 20 ∘
Plugging in these values, we get:
|ε| = (994)(0.75)(0.50)(2.8⋅10 −3)(cos(20 ∘))
|ε| = -0.567887 V
Therefore, the magnitude of the average emf induced in the loop is -0.567887 V. The negative sign indicates that the induced emf opposes the change in magnetic flux.
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The 60-Hz ac source of the series circuit shown in the figure has a voltage amplitude of 120 V. The capacitive reactance is 790 Ω, the inductive reactance is 270 Ω, and the resistance is 500Ω. What is the total impedance Z?
The total impedance (Z) of the series circuit is approximately 721 Ω, given a resistance of 500 Ω, a capacitive reactance of 790 Ω, and an inductive reactance of 270 Ω.
To find the total impedance (Z) of the series circuit, we need to calculate the combined effect of the resistance (R), capacitive reactance (Xc), and inductive reactance (Xl). The impedance can be found using the formula:
Z = √(R² + (Xl - Xc)²),
where:
R is the resistance,Xl is the inductive reactance,Xc is the capacitive reactance.Substituting the given values:
R = 500 Ω,
Xc = 790 Ω,
Xl = 270 Ω,
we can calculate the total impedance:
Z = √(500² + (270 - 790)²).
Z = √(250000 + (-520)²).
Z ≈ √(250000 + 270400).
Z ≈ √520400.
Z ≈ 721 Ω.
Therefore, the total impedance (Z) of the series circuit is approximately 721 Ω.
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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 275000 kg and a velocity of 0.32 m/s in the horizontal direction, and the second having a mass of 52500 kg and a velocity of -0.15 m's in the horizontal direction What is their final velocity, in meters per second?
The final velocity of the two train cars after they are coupled together is 0.24465648854961833 m/s in the direction of the first train car's initial velocity.
We can use the following equation to calculate the final velocity of the two train cars:
v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)
Where:
v_f is the final velocity of the two train cars
m_1 is the mass of the first train car
v_1 is the initial velocity of the first train car
m_2 is the mass of the second train car
v_2 is the initial velocity of the second train car
Plugging in the values, we get:
v_f = (275000 kg * 0.32 m/s + 52500 kg * -0.15 m/s)/(275000 kg + 52500 kg) = 0.24465648854961833 m/s
Therefore, the final velocity of the two train cars together is 0.24465648854961833 m/s.
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You put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. Calculate much water is left at the end of 17 min. Please report your mass in grams to O decimal places. Hint: the latent heat of vaporisation for water is 2257 kJ/kg.
When you put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. then at the end of 17 min approximately 255 g of water are left.
To calculate the amount of water left at the end of 17 minutes, we need to consider the energy absorbed by the water from the microwave and the energy required to evaporate the water.
First, let's calculate the energy absorbed by the water from the microwave:
Energy absorbed = Power * Time = 564 W * 17 min * 60 s/min = 564 W * 1020 s = 575,280 J
Next, let's calculate the energy required to evaporate the water:
Energy required = Mass * Latent heat of vaporization
Given that the latent heat of vaporization for water is 2257 kJ/kg, we need to convert it to joules by multiplying by 1000:
Latent heat of vaporization = 2257 kJ/kg * 1000 = 2,257,000 J/kg
Now, let's calculate the mass of water using the energy absorbed and the energy required for evaporation:
Mass = Energy absorbed / Energy required
= 575,280 J / 2,257,000 J/kg
≈ 0.255 kg
Finally, let's convert the mass to grams:
Mass in grams = 0.255 kg * 1000 g/kg = 255 g
Therefore, at the end of 17 minutes, approximately 255 grams of water are left.
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a A musician with perfect pitch stands beside a roadway. She hears a pitch of 1090 Hz when a siren on an emergency vehicle approaches and a frequency of 900 Hz when it passes. a. What is the frequency of the siren if it were stationary? b. What is the speed of the vehicle?
The frequency of the siren when it is stationary is 1000 Hz and the speed of the vehicle is 34 m/s.
a) When the siren approaches, the musician hears a higher frequency of 1090 Hz. This is due to the Doppler effect, which causes the perceived frequency to increase when the source of sound is moving towards the observer. Similarly, when the siren passes, the musician hears a lower frequency of 900 Hz.
To find the frequency of the siren when it is stationary, we can calculate the average of the two observed frequencies:
[tex]\frac{(1090Hz+900Hz)}{2} =1000Hz[/tex]
b) The Doppler effect can also be used to determine the speed of the vehicle. The formula relating the observed frequency (f), source frequency ([tex]f_0[/tex]), and the speed of the source (v) is given by:
[tex]f=\frac{f_0(v+v_0)}{(v-v_s)}[/tex]
In this case, we know the observed frequencies (1090 Hz and 900 Hz), the source frequency (1000 Hz), and the speed of sound in air (343 m/s). By rearranging the formula and solving for the speed of the vehicle (v), we find:
[tex]v=\frac{(\frac{f}{f_0}-1)v_s}{\frac{f}{f_0}+1}}[/tex]
Substituting the known values, we get:
[tex]v=\frac{(\frac{1090}{1000}-1)343}{\frac{1090}{1000}+1}=34 m/s[/tex]
Therefore, the speed of the vehicle is approximately 34 m/s.
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A 6.31 µF capacitor and a 11.75 mH inductor are connected in series with an AC power source that has a frequency of 2.96 x103 Hz and a peak voltage of 71 V. Take the initial time t as zero when the instantaneous voltage equals zero. Determine the instantaneous current when t = 3.69x 10^-4 s.
When t = 3.69x10^-4 s, the instantaneous current in the series circuit is approximately 0.34 A. We need to use the concepts of impedance and phase difference. With the impedance known, we can then calculate the magnitude and phase of the current at the given time t = 3.69 x 10^-4 s.
In a series circuit containing a capacitor and an inductor, the total impedance Z of the circuit is given by Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The reactances can be calculated using the formulas XL = 2πfL and XC = 1 / (2πfC), where f is the frequency, L is the inductance, and C is the capacitance.
The magnitude of the current I can be determined using Ohm's law, where I = Vpeak / Z, and the phase angle φ between the voltage and current can be calculated as φ = arctan((XL - XC) / R).
By plugging in the given values of frequency (2.96 x 10^3 Hz), capacitance (6.31 µF), inductance (11.75 mH), and peak voltage (71 V), we can calculate the impedance Z. When t = 3.69x10^-4 s, the instantaneous current in the series circuit is approximately 0.34 A.
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A skydiver has a mass of 73 kg. Suppose that the air resistive force acting on the diver increases in direct proportion to his velocity such that for every 10 m/s that the diver’s velocity increases, the force of air resistance increases by 82 N. Use g = 9.8 m/s^2. Let F1 be the net force acting on the skydiver when his velocity is 39. Let a1 be the acceleration of the skydiver at that moment. Let vT be the terminal velocity of the skydiver. Compute F1+2*a1+3*vT.
A skydiver's net force, acceleration, and terminal velocity are calculated using air resistance proportional to velocity. F1 + 2a1 + 3vT = 392.12 N is obtained using given values.
Let's begin by finding the net force, F1, acting on the skydiver when his velocity is 39 m/s. We can use the formula for net force, F = ma, where m is the mass of the skydiver and a is his acceleration. The force of air resistance, Fr, is given by Fr = kv, where v is the velocity of the skydiver and k is the constant of proportionality.
From the problem statement, we know that for every 10 m/s increase in velocity, the air resistive force increases by 82 N. This means that k = 8.2 Ns/m. Therefore, the force of air resistance on the skydiver when his velocity is 39 m/s is given by Fr = 8.2(39) = 319.8 N.
The net force acting on the skydiver is the difference between the force of gravity and the force of air resistance:
F1 = mg - Fr = (73 kg)(9.8 m/s^2) - 319.8 N = 422.6 N
Next, we can find the acceleration of the skydiver at that moment, a1, by dividing the net force by the mass:
a1 = F1/m = 422.6 N / 73 kg = 5.7959 m/s^2
To find the terminal velocity, we can set the force of air resistance equal to the force of gravity, since the net force is zero when the skydiver reaches terminal velocity:
Fr = mg
8.2vT = (73 kg)(9.8 m/s^2)
vT = 28.6804 m/s
Finally, we can substitute the values we have found into the expression F1 + 2a1 + 3vT and simplify:
F1 + 2a1 + 3vT = 422.6 N + 2(5.7959 m/s^2)(2) + 3(28.6804 m/s)(3) = 392.12 N
Therefore, F1 + 2a1 + 3vT = 392.12 N.
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"A ray of light strikes a surface at ninety degrees, that is, it
is parallel to the normal. The angle of refraction is
A. one hundred and eighty degrees, 180°
B. ninety degrees, 90°.
C. forty-five degrees
When a ray of light strikes a surface at a 90-degree angle, which means it is parallel to the normal, the angle of refraction is 90 degrees (Option B).
When light passes from one medium to another, it usually undergoes refraction, which is the bending of light due to the change in its speed. The angle of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media.
However, when a ray of light strikes a surface at a ninety-degree angle, it is parallel to the normal of the surface. In this case, the light does not change its direction upon entering the new medium, and no refraction occurs. The angle of refraction is undefined because there is no bending or change in the direction of the light ray.
Option A (180 degrees) is incorrect because an angle of 180 degrees would mean that the refracted ray is opposite in direction to the incident ray, which is not possible in this case. Option C (45 degrees) is also incorrect because it does not apply to the scenario described, where the incident ray is parallel to the normal.
When a ray of light strikes a surface at a 90-degree angle, the angle of refraction is also 90 degrees. This occurs because the incident ray, being parallel to the normal, does not undergo any change in direction as it passes from one medium to another.
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A curling stone slides on ice with a speed of 2.0 m/s and collides inelastically with an identical, stationary curling stone. After the collision, the first stone is deflected by a counterclockwise angle of 28° from its original direction of travel, and the second stone moves in a direction that makes a 42° clockwise angle with the original direction of travel of the first stone. What fraction of the initial energy is lost in this collision? A) 0.12 B) 0.24 C) 0.48 D) 0.64 E) 0.36
The fraction of initial energy lost in this collision is 0. This implies that no energy is lost, indicating an elastic collision.
To determine the fraction of initial energy lost in the collision, we need to compare the initial kinetic energy with the final kinetic energy after the collision.
Given:
Initial speed of the first stone (v_1) = 2.0 m/s
Angle of deflection for the first stone (θ_1) = 28°
Angle of deflection for the second stone (θ_2) = 42°
Let's calculate the final speeds of the first and second stones using the given information:
Using trigonometry, we can find the components of the final velocities in the x and y directions for both stones.
For the first stone:
vx_1 = v_1 * cos(θ_1)
vy_1 = v_1 * sin(θ_1)
For the second stone:
vx_2 = v_2 * cos(θ_2)
vy_2 = v_2 * sin(θ_2)
Since the second stone is initially stationary, its initial velocity is zero (v_2 = 0).
Now, we can calculate the final velocities:
vx_1 = v1 * cos(θ_1)
vy_1 = v1 * sin(θ_1)
vx_2 = 0 (as v_2 = 0)
vy_2 = 0 (as v_2 = 0)
The final kinetic energy (Kf) can be calculated using the formula:
Kf = (1/2) * m * (vx1^2 + vy1^2) + (1/2) * m * (vx2^2 + vy2^2)
Since the second stone is initially stationary, its final kinetic energy is zero:
Kf = (1/2) * m * (vx_1^2 + vy_1^2)
The initial kinetic energy (Ki) can be calculated using the formula:
Ki = (1/2) * m * v_1^2
Now, we can determine the fraction of initial energy lost in the collision:
Fraction of initial energy lost = (K_i - K_f) / K_i
Substituting the expressions for K_i and K_f:
[tex]Fraction of initial energy lost = [(1/2) * m * v1^2 - (1/2) * m * (vx_1^2 + vy_1^2)] / [(1/2) * m * v_1^2]Simplifying and canceling out the mass (m):Fraction of initial energy lost = (v_1^2 - vx_1^2 - vy_1^2) / v_1^2Using the trigonometric identities sin^2(θ) + cos^2(θ) = 1, we can simplify further:[/tex]
Therefore, the fraction of initial energy lost in this collision is 0. This implies that no energy is lost, indicating an elastic collision.
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#SPJ11[tex]Fraction of initial energy lost = (v_1^2 - vx_1^2 - vy_1^2) / v_1^2Fraction of initial energy lost = (v_1^2 - v_1^2 * cos^2(θ_1) - v_1^2 * sin^2(θ_1)) / v_1^2Fraction of initial energy lost = (v_1^2 * (1 - cos^2(θ_1) - sin^2(θ_1))) / v_1^2Fraction of initial energy lost = (v_1^2 * (1 - 1)) / v1^2Fraction of initial energy lost = 0[/tex]
For what frequencies does a 17.0−μF capacitor have a reactance below 150Ω ?
The frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
To determine the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω, we can use the formula for capacitive reactance:
Xc = 1 / (2πfC)
Where:
Xc is the capacitive reactance in ohms,
f is the frequency in hertz (Hz),
C is the capacitance in farads (F).
In this case, we want to find the frequencies at which Xc is below 150Ω. We can rearrange the formula to solve for f:
f = 1 / (2πXcC)
Substituting Xc = 150Ω and C = 17.0-μF (which is equal to 17.0 × 10^(-6) F), we can calculate the frequencies.
f = 1 / (2π × 150Ω × 17.0 × 10^(-6) F)
f ≈ 590.64 Hz
Therefore, the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
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The sonar unit on a boat is designed to measure the depth of fresh water ( = 1.00 x 103 kg/m3, Bad = 2.20 x 109 Pa). When the boat moves into salt water ( = 1025 kg/m3, Bad = 2.37 x 109 Pa), the sonar unit is no longer calibrated properly. In salt water, the sonar unit indicates the water depth to be 7.96 m. What is the actual depth (in m) of the water?
The actual depth of the water in saltwater is 240.3 m.
The sonar unit on a boat is designed to measure the depth of fresh water, but when the boat moves into salt water the sonar unit is no longer calibrated properly.
Given, Depth indicated by sonar in saltwater=7.96 m
Density of freshwater =1.00 x 10³ kg/m³
Density of saltwater =1025 kg/m³
Pressure of freshwater=2.20 x 10⁹ Pa
Pressure of saltwater=2.37 x 10⁹ Pa.
To find out the actual depth of water in m we need to use the relationship between pressure and depth which is given as follows : ρgh = P
where ρ is the density of the fluid
g is the acceleration due to gravity
h is the depth of the fluid
P is the pressure of the fluid in N/m²
For freshwater, ρ = 1.00 x 10³ kg/m³ and P = 2.20 x 10⁹ Pa and
For saltwater, ρ = 1025 kg/m³ and P = 2.37 x 10⁹ Pa.
So, ρgh = P
⇒h = P/(ρg)
For freshwater, h = 2.20 x 10⁹/(1.00 x 10³ x 9.8) = 224.5 m
For saltwater , h = 2.37 x 10⁹/(1025 x 9.8) = 240.3 m
So, the actual depth is 240.3 m.
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Required information A 35.0-nC charge is placed at the origin and a 57.0 nC charge is placed on the +x-axis, 2.20 cm from the origin. What is the electric potential at a point halfway between these two charges?
V =
The electric potential at a point halfway between the 35.0 nC charge at the origin and the 57.0 nC charge on the +x-axis is 1.83 kV.
To calculate the electric potential at a point halfway between the two charges, we need to consider the contributions from each charge and sum them together.
Given:
Charge q1 = 35.0 nC at the origin (0, 0).
Charge q2 = 57.0 nC on the +x-axis, 2.20 cm from the origin.
The electric potential due to a point charge at a distance r is given by the formula:
V = k * (q / r),
where V is the electric potential, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance.
Let's calculate the electric potential due to each charge:
For q1 at the origin (0, 0):
V1 = k * (q1 / r1),
where r1 is the distance from the point halfway between the charges to the origin (0, 0).
For q2 on the +x-axis, 2.20 cm from the origin:
V2 = k * (q2 / r2),
where r2 is the distance from the point halfway between the charges to the charge q2.
Since the point halfway between the charges is equidistant from each charge, r1 = r2.
Now, let's calculate the distances:
r1 = r2 = 2.20 cm / 2 = 1.10 cm = 0.0110 m.
Substituting the values into the formula:
V1 = k * (35.0 x 10^(-9) C) / (0.0110 m),
V2 = k * (57.0 x 10^(-9) C) / (0.0110 m).
Calculating the electric potentials:
V1 ≈ 2863.64 V,
V2 ≈ 4660.18 V.
To find the electric potential at the point halfway between the charges, we need to sum the contributions from each charge:
V = V1 + V2.
Substituting the calculated values:
V ≈ 2863.64 V + 4660.18 V.
Calculating the sum:
V ≈ 7523.82 V.
Therefore, the electric potential at a point halfway between the two charges is approximately 7523.82 volts.
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