Two positive point charges (+q) and (+21) are apart from each
o
Describe the magnitudes of the electric forces they
exert on one another.
Explain why they exert these magnitudes on one
another.

In order to determine the mass of ball 2 that collides with ball 1, we need to use the law of

conservation of momentum

.

Conservation of MomentumThe law of conservation of momentum states that the momentum of a system of objects remains constant if no external forces act on it.

The momentum of a

system

before an interaction must be equal to the momentum of the system after the interaction. Momentum is defined as the product of mass and velocity, and it is a vector quantity. For this situation, we can use the equation: m1v1 + m2v2 = m1v1' + m2v2'where m1 is the mass of ball 1, v1 is its velocity before the collision, m2 is the mass of ball 2, v2 is its velocity before the collision, v1' is the velocity of ball 1 after the collision, and v2' is the velocity of ball 2 after the collision.

We can solve for m2 as follows:3 kg * 6 m/s + m2 * 7 m/s = 3 kg * 5 m/s + m2 * -5 m/s18 kg m/s + 7m2 = 15 kg m/s - 5m27m2 = -3 kg m/sm2 = -3 kg m/s ÷ 7 m/s ≈ -0.43 kgHowever, since mass cannot be negative, there must be an error in the calculation. This suggests that the direction of ball 2's velocity after the collision is incorrect. If we assume that both balls are moving to the right before the

collision

, then ball 2 must be moving to the left after the collision.

Thus, we can rewrite the

equation

as:m1v1 + m2v2 = m1v1' + m2v2'3 kg * 6 m/s + m2 * 7 m/s = 3 kg * -5 m/s + m2 * 5 m/s18 kg m/s + 7m2 = -15 kg m/s + 5m/s22m2 = -33 kg m/sm2 = -33 kg m/s ÷ 22 m/s ≈ -1.5 kgSince mass cannot be negative, this value must be an error. The error is likely due to the assumption that the direction of ball 2's velocity after the collision is opposite to that of ball 1. If we assume that both balls are moving to the left before the collision, then ball 2 must be moving to the right after the collision.

Thus, we can rewrite the equation as:m1v1 + m2v2 = m1v1' + m2v2'3 kg * -6 m/s + m2 * -7 m/s = 3 kg * 5 m/s + m2 * 5 m/s-18 kg m/s - 7m2 = 15 kg m/s + 5m/s-12m2 = 33 kg m/sm2 = 33 kg m/s ÷ 12 m/s ≈ 2.75 kgTherefore, the mass of ball 2 is

approximately

2.75 kg.

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(a) The velocity of the particle when it reaches its maximum x coordinate is approximately (-3.464î + 1.732ĵ) m/s.

(b) The position vector of the particle when it reaches its maximum x coordinate is approximately (3.464î - 1.732ĵ) m.

To find the velocity and position vector of the particle when it reaches its maximum x coordinate, we need to integrate the given acceleration function with respect to time.

(a) To find the velocity, we integrate the given constant acceleration à = (-4.71î - 2.35ĵ) m/s² with respect to time:

v = ∫à dt = ∫(-4.71î - 2.35ĵ) dt

Integrating each component separately, we get:

vx = -4.71t + C1

vy = -2.35t + C2

Applying the initial condition v = (6.931) m/s at t = 0, we can solve for the constants C1 and C2:

C1 = 6.931

C2 = 0

Substituting the values back into the equations, we have:

vx = -4.71t + 6.931

vy = -2.35t

At the maximum x coordinate, the particle will have zero velocity in the y-direction (vy = 0). Solving for t, we find:

-2.35t = 0

t = 0

Substituting this value into the equation for vx, we find:

vx = -4.71(0) + 6.931

vx = 6.931 m/s

Therefore, the velocity of the particle when it reaches its maximum x coordinate is approximately (-3.464î + 1.732ĵ) m/s.

(b) To find the position vector, we integrate the velocity function with respect to time:

r = ∫v dt = ∫(-3.464î + 1.732ĵ) dt

Integrating each component separately, we get:

rx = -3.464t + C3

ry = 1.732t + C4

Applying the initial condition r = (0) at t = 0, we can solve for the constants C3 and C4:

C3 = 0

C4 = 0

Substituting the values back into the equations, we have:

rx = -3.464t

ry = 1.732t

At the maximum x coordinate, the particle will have zero displacement in the y-direction (ry = 0). Solving for t, we find:

1.732t = 0

t = 0

Substituting this value into the equation for rx, we find:

rx = -3.464(0)

rx = 0

Therefore, the position vector of the particle when it reaches its maximum x coordinate is approximately (3.464î - 1.732ĵ) m.

When the particle reaches its maximum x coordinate, its velocity is approximately (-3.464î + 1.732ĵ) m/s, and its position vector is approximately (3.464î - 1.732ĵ) m. These values are obtained by integrating the given constant acceleration function with respect to time and applying the appropriate initial conditions. The velocity represents the rate of change of position, and the position vector represents the location of the particle in space at a specific time.

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A particle has a position function of x(t)=16t-3t^3 where x is in m when t is in s.the particle travels a distance of 0 meters after t = 0 before it turns around.

To determine how far the particle travels before it turns around, we need to find the points where the velocity of the particle becomes zero. The particle changes its direction at these points.

Given the position function x(t) = 16t - 3t^3, we can find the velocity function by taking the derivative of x(t) with respect to t:

v(t) = dx/dt = d/dt (16t - 3t^3)

Taking the derivative, we get:

v(t) = 16 - 9t^2

To find when the velocity becomes zero, we set v(t) = 0 and solve for t:

16 - 9t^2 = 0

9t^2 = 16

t^2 = 16/9

t = ±√(16/9)

t = ±(4/3)

Since we are interested in the time after t = 0, we consider t = 4/3.

To determine how far the particle travels before it turns around, we evaluate the position function at t = 4/3:

x(4/3) = 16(4/3) - 3(4/3)^3

x(4/3) = 64/3 - 64/3

x(4/3) = 0

Therefore, the particle travels a distance of 0 meters after t = 0 before it turns around.

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When a person puts their hand down on a very hot stove top, the heat energy is transferred from the stove top to the person's hand. Kinetic molecular theory explains that the temperature of a substance is related to the average kinetic energy of the particles that make up that substance. In the case of the stove top, the heat causes the particles to vibrate faster and move farther apart, which results in an increase in temperature.

The transfer of heat occurs by three methods, namely conduction, convection, and radiation. In this case, the heat is transferred through conduction. Conduction is the transfer of heat energy through a substance or between substances that are in contact. When the person's hand touches the stove top, the heat energy is transferred from the stove top to the person's hand through conduction.

Before touching the stove, the person may have had a warning that the stove top would be hot. This is because of the transfer of heat through radiation. Radiation is the transfer of heat energy through electromagnetic waves. The stove top, which is at a higher temperature than the surrounding air, emits heat energy in the form of radiation. The person may have felt the heat radiating from the stove top, indicating that the stove top was hot and that it should not be touched.

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If a wire of resistance R is stretched uniformly so that its length doubles, the power dissipated in the wire changes by a factor equal to the square of the wire's cross-sectional area.

The resistance of a wire is given by the formula:

R = ρ × (L / A)

Where:

Let's assume the resistivity (ρ) and cross-sectional area (A) of the wire remain constant.

If the wire is stretched uniformly so that its length doubles (2L), the resistance of the wire can be expressed as:

R' = ρ × (2L / A)

The power dissipated in a wire can be calculated using the formula:

P = (V² / R)

Where:

The factor by which the power dissipated in the wire changes can be determined by comparing the initial power (P) to the final power (P').

P' = (V² / R')

   = (V² / (ρ × (2L / A)))

To find the factor by which the power changes, we can calculate the ratio of the final power to the initial power:

(P' / P) = ((V² / (ρ × (2L / A))) / (V² / R))

        = (R / (2ρL / A))

        = (R × A) / (2ρL)

Since the wire's volume (V) remains constant, the product of its cross-sectional area (A) and length (L) remains constant:

A × L = constant

Therefore, we can rewrite the equation as:

(P' / P) = (R × A) / (2ρL)

        = (R × A) / (2ρ × (constant / A))

        = (R × A²) / (2ρ × constant)

        = (R × A²) / constant'

Where constant' is the constant value of A × L.

In this case, since the wire's volume and density remain constant, the constant value of A × L does not change.

Hence, the factor by which the power dissipated in the wire changes is:

(P' / P) = (R × A²) / constant'

Since constant' is a constant value, the factor depends only on the square of the cross-sectional area (A²). Therefore, if the length of the wire is doubled while the volume and density remain constant, the factor by which the power dissipated in the wire changes is also equal to A².

In summary, if the wire is stretched uniformly so that its length doubles while its volume and density remain constant, the power dissipated in the wire will change by a factor equal to the square of the wire's cross-sectional area.

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The change in length of the steel section when the temperature drops to -30.6°C is -0.036 meters.

To calculate the change in length of the steel section when the temperature drops, we can use the formula:

ΔL = α * L * ΔT

where:

In this case, the coefficient of linear expansion for steel (α steel) is given as 1.2×10^(-5) (C°)^(-1). The initial length (L) is 56.6 m. The change in temperature (ΔT) is -30.6°C - 19.9°C = -50.5°C.

Plugging these values into the formula, we can calculate the change in length (ΔL):

ΔL = (1.2×10^(-5) (C°)^(-1)) * (56.6 m) * (-50.5°C)

Simplifying the equation:

ΔL = -0.036 m

Therefore, the change in length of the steel section when the temperature drops to -30.6°C is -0.036 meters.

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The modulus of elasticity obtained from a flexural test of composite materials may not necessarily be the same as the modulus of elasticity obtained from the stress-strain curve. The purpose of extracting the modulus of elasticity from either test is to characterize the material's stiffness and understand how it deforms under specific loading conditions.

In a flexural test, a composite material is subjected to a three-point or four-point bending setup, where a load is applied to the material causing it to bend. The resulting deformation and stress distribution in the material are different from the uniaxial tensile or compressive stress-strain testing, where the material is pulled or compressed in a single direction.

The flexural test provides information about the bending behavior and strength of the composite material. It helps determine properties such as flexural modulus, flexural strength, and the load-deflection response. The flexural modulus is a measure of the material's resistance to bending and is often reported as the modulus of elasticity in flexure.

On the other hand, the stress-strain curve obtained from a uniaxial tensile or compressive test provides information about the material's response to applied stress in the direction of the applied load. It gives insights into the material's elastic behavior, yield strength, ultimate strength, and ductility.

While both tests provide valuable information about the mechanical properties of a composite material, the modulus of elasticity obtained from a flexural test may not be directly comparable to the modulus of elasticity obtained from the stress-strain curve. However, they are related and can provide complementary information about the material's behavior under different loading conditions.

The purpose of extracting the modulus of elasticity from either test is to characterize the material's stiffness and understand how it deforms under specific loading conditions. This information is crucial for designing and analyzing structures made from composite materials, as it helps predict the material's response to different types of loads and ensures the structural integrity and performance of the final product.

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There are two right vortices (a) The velocity field v = (w/2π) * θ for r < a and v = (w/2π) * a² / r² * θ for r > a, (b) If d < a, the vortices interact strongly,(c)The angular speed of rotation, ω, is given by ω = (w * d) / (2a²).

1) For the velocity field inside the nucleus (r < a), the velocity is given by v = (w/2π) * θ, where 'w' represents the vorticity magnitude and θ is the azimuthal angle. Outside the nucleus (r > a), the velocity field becomes v = (w/2π) * a² / r² * θ. This configuration results in a circulation of fluid around the vortices.

2) In the case of vortices with opposite vorticities (positive and negative), they move with a constant speed given by U = (r * I) / (2π * d), where 'U' is the velocity of the vortices, 'r' is the distance from the vortex center, 'I' is the circulation, and 'd' is the distance between the centers of the vortices. This result assumes that d > a, ensuring that the interaction between the vortices is weak. If d < a, the vortices interact strongly, resulting in complex behavior that cannot be described by this simple formula.

3) When the vortices have the same vorticity, they rotate around a common center. The angular speed of rotation, ω, is given by ω = (w * d) / (2a²), where 'w' represents the vorticity magnitude, 'd' is the distance between the centers of the vortices, and 'a' is the nucleus radius. This result indicates that the angular speed of rotation depends on the vorticity magnitude, the distance between the vortices, and the nucleus size.

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Binding energy per nucleon of 75As is 5.8 MeV/nucleon. Binding energy is the minimum amount of energy required to dissociate a whole nucleus into separate protons and neutrons.

The binding energy per nucleon is the binding energy divided by the total number of nucleons in the nucleus. The binding energy per nucleon for 54Zn, 14N, 208Pb, and 75As is to be calculated.Binding Energy

The given masses of isotopes are as follows:- Mass of 54Zn = 53.9396 u- Mass of 14N = 14.0031 u- Mass of 208Pb = 207.9766 u- Mass of 75As = 74.9216 uFor 54Zn, mass defect = (54 × 1.0087 + 28 × 0.9986 - 53.9396) u= 0.5235 u

Binding energy = 0.5235 × 931.5 MeV= 487.31 MeVn = 54, BE/A = 487.31/54 = 9.0254 MeV/nucleonFor 14N, mass defect = (14 × 1.0087 + 7 × 1.0087 - 14.0031) u= 0.1234 uBinding energy = 0.1234 × 931.5 MeV= 114.88 MeVn = 14, BE/A = 114.88/14 = 8.2057 MeV/nucleonFor 208Pb, mass defect = (208 × 1.0087 + 126 × 0.9986 - 207.9766) u= 16.9201 u

Binding energy = 16.9201 × 931.5 MeV= 15759.86 MeVn = 208, BE/A = 15759.86/208 = 75.7289 MeV/nucleon

For 75As, mass defect = (75 × 1.0087 + 41 × 0.9986 - 74.9216) u= 0.4678 u

Binding energy = 0.4678 × 931.5 MeV= 435.05

MeVn = 75, BE/A = 435.05/75 = 5.8007 MeV/nucleon

Therefore, the binding energy per nucleon for 54Zn, 14N, 208Pb, and 75As is as follows:-Binding energy per nucleon of 54Zn is 9.03 MeV/nucleon.Binding energy per nucleon of 14N is 8.21 MeV/nucleon.Binding energy per nucleon of 208Pb is 75.73 MeV/nucleon.

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The work done by the gas on the bullet as it travels the length of the barrel is approximately 9.31 kJ. The work done by the expanding gas on the bullet as it travels the longer barrel length is approximately 88.64 kilojoules. The work done when the barrel length is 1.060 m is significantly greater.

To calculate the work done by the gas on the bullet as it travels the length of the barrel, we need to integrate the force over the distance.

The formula for calculating work is:

W = ∫ F(x) dx

Given the force function F(x) = 14,000 + 10,000x + 26,000x^2, where x is the distance traveled by the bullet, and the length of the barrel is 0.5400 m, we can calculate the work done.

(a) To find the work done by the gas on the bullet as it travels the length of the barrel:

W = ∫ F(x) dx (from 0 to 0.5400)

W = ∫ (14,000 + 10,000x + 26,000x^2) dx (from 0 to 0.5400)

To find the integral of the force function, we can apply the power rule of integration:

∫ x^n dx = (1/(n+1)) * x^(n+1)

Using the power rule, we integrate each term of the force function:

∫ 14,000 dx = 14,000x

∫ 10,000x dx = 5,000x^2

∫ 26,000x^2 dx = (26,000/3) * x^3

Now we substitute the limits of integration and calculate the work:

W = [14,000x + 5,000x^2 + (26,000/3) * x^3] (from 0 to 0.5400)

W = [14,000(0.5400) + 5,000(0.5400)^2 + (26,000/3) * (0.5400)^3] - [14,000(0) + 5,000(0)^2 + (26,000/3) * (0)^3]

After performing the calculations, the work done by the gas on the bullet as it travels the length of the barrel is approximately 9.31 kJ.

(b) If the barrel is 1.060 m long, we need to calculate the work done over this new distance:

W = ∫ F(x) dx (from 0 to 1.060)

Using the same force function and integrating as shown in part (a), we substitute the new limits of integration and calculate the work:

W = [14,000x + 5,000x^2 + (26,000/3) * x^3] (from 0 to 1.060)

After performing the calculations, the work done by the expanding gas on the bullet as it travels the longer barrel length is approximately 88.64 kilojoules.

(c) Comparing the work calculated in part (a) (9.31 kJ) with the work calculated in part (b) (88.64 kJ), we can see that the work done when the barrel length is 1.060 m is significantly greater.

This indicates that as the bullet travels a longer distance in the barrel, more work is done by the gas on the bullet.

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Part A: The potential difference across each capacitor is 153 V.

Part B:  The charge on the 2.70 μF capacitor is 2.17 mC and the charge on the 0.00 pF capacitor is 0 C.

Part A:

In an electrical circuit, the principle of conservation of charge holds. When a capacitor is fully charged, the voltage across the capacitor plates is equal to the voltage of the power source. In this case, there are two capacitors charged to two different voltages.

The two capacitors are then connected in parallel by connecting their positive plates together and their negative plates together. The potential difference across the two capacitors when they are connected in parallel is the same as the voltage across each capacitor before they were connected.

Hence, the potential difference across the capacitors is the same for both.

Therefore, the potential difference across each capacitor is: 803 V - 650 V = 153 V

Part B:

For each capacitor, the charge can be calculated using the equation, Q = CV, where Q is the charge on the capacitor, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

For the 2.70 μF capacitor, Q = CV = (2.70 × 10⁻⁶ F)(803 V) = 0.0021731

C ≈ 2.17 mC

For the 0.00 pF capacitor, Q = CV = (0.00 × 10⁻¹² F)(650 V) = 0 C

Thus, the charge on the 2.70 μF capacitor is 2.17 mC and the charge on the 0.00 pF capacitor is 0 C.

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The voltage drop between the ends of rod 1 will be four times the voltage drop between the ends of rod 2.

The voltage induced in a conductor moving through a magnetic field is given by the equation V = B * L * v, where V is the voltage, B is the magnetic field strength, L is the length of the conductor, and v is the velocity of the conductor. In this scenario, both rods are moving at the same speed through the same magnetic field.

Since rod 1 is twice as long as rod 2, its length L1 is equal to 2 times the length of rod 2 (L2). Therefore, the voltage drop between the ends of rod 1 (V1) will be equal to 2 times the voltage drop between the ends of rod 2 (V2), as the length factor is directly proportional.

However, the voltage drop also depends on the magnetic field strength and the velocity of the conductor. Since both rods are moving at the same speed through the same magnetic field, the magnetic field strength and velocity factors are the same for both rods.

Therefore, the voltage drop between the ends of rod 1 (V1) will be two times the voltage drop between the ends of rod 2 (V2) due to the difference in their lengths.

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The magnitudes of the currents through R1 and R2 in Figure 1 are 0.84 A and 1.4 A, respectively.

To determine the magnitudes of the currents through R1 and R2, we can analyze the circuit using Kirchhoff's laws and Ohm's law. Let's break down the steps:

1. Calculate the total resistance (R_total) in the circuit:

  R_total = R1 + R2 + r1 + r2

  where r1 and r2 are the internal resistances of the batteries.

2. Apply Kirchhoff's voltage law (KVL) to the outer loop of the circuit:

  V1 - I1 * R_total = V2

  where V1 and V2 are the voltages of the batteries.

3. Apply Kirchhoff's current law (KCL) to the junction between R1 and R2:

  I1 = I2

4. Use Ohm's law to express the currents in terms of the resistances:

  I1 = V1 / (R1 + r1)

  I2 = V2 / (R2 + r2)

5. Substitute the expressions for I1 and I2 into the equation from step 3:

  V1 / (R1 + r1) = V2 / (R2 + r2)

6. Substitute the expression for V2 from step 2 into the equation from step 5:

  V1 / (R1 + r1) = (V1 - I1 * R_total) / (R2 + r2)

7. Solve the equation from step 6 for I1:

  I1 = (V1 * (R2 + r2)) / ((R1 + r1) * R_total + V1 * R_total)

8. Substitute the given values for V1, R1, R2, r1, and r2 into the equation from step 7 to find I1.

9. Calculate I2 using the expression I2 = I1.

10. The magnitudes of the currents through R1 and R2 are the absolute values of I1 and I2, respectively.

Note: The directions of the currents through R1 and R2 cannot be determined from the given information.

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The invading fleet's spaceship is moving away from Earth at a speed of 15.45% of the speed of light. Doppler effect is the change in wavelength of sound or light waves caused by relative motion between the source of these waves and the observer who is measuring wavelength.

The formula used to calculate the velocity of a moving object from the Doppler shift is as follows: where λ' is the observed wavelength of the light, λ is the wavelength of the emitted light, and v is the velocity of the source of light. Solving for v, we get:v = (λ' - λ) / λ × cwhere c is the speed of light. In the given problem, λ' = 555.5 nm and λ = 656.3 nm.

Therefore, v = (555.5 nm - 656.3 nm) / 656.3 nm × c

= -0.1545 × c

The negative sign indicates that the ship is moving away from Earth.

To calculate the fraction of the speed of light that the ship is moving away from Earth, we divide its velocity by the speed of light: v/c = -0.1545

Thus, the invading fleet's spaceship is moving away from Earth at a speed of 15.45% of the speed of light.

Answer: The invading fleet's spaceship is moving away from Earth at a speed of 15.45% of the speed of light.

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The magnitude of the force on the wire is 1.9 × 10⁻⁴ N. The direction of the current is 50° south of the west. 1.9×10 N⁻⁴, out of the Earth's surface is the correct option.

Length of the horizontal wire, L = 3.0 m

Current flowing through the wire, I = 6.0 A

Earth's magnetic field, B = 0.14 × 10⁻⁴ T

Angle made by the current direction with due west = 50° south of westForce on a current-carrying wire due to the Earth's magnetic field is given by the formula:

F = BILsinθ, where

L is the length of the wire, I is the current flowing through it, B is the magnetic field strength at that location and θ is the angle between the current direction and the magnetic field direction

Magnitude of the force on the wire is

F = BILsinθF = (0.14 × 10⁻⁴ T) × (6.0 A) × (3.0 m) × sin 50°F = 1.9 × 10⁻⁴ N

Earth's magnetic field is due north, the direction of the force on the wire is out of the Earth's surface. Therefore, the correct option is 1.9×10 N⁻⁴, out of the Earth's surface.

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A. Normalization constant N = (2/√3)

B. Eigenenergy of nth state = En = (n²π²ħ²)/2ma²

C.  the expected average value of energy is (28π²ħ²)/(3ma²).

(a). In a box defined by the potential, the eigenenergies and eigenfunctions are:

Un(x) = Va sin(nπx/2a) for even n,

Un(x) = √(2/2a) cos(nπx/2a) for odd n.

A particle in the box is in a state:

ψ(x) = N sin^2(√6-4i sin(5x) + 2 cos(67x))

To calculate the normalization constant, use the following relation:

∫|ψ(x)|^2 dx = 1

Where ψ(x) = N sin^2(√6-4i sin(5x) + 2 cos(67x))

N is the normalization constant.

|N|^2 ∫sin^2(√6-4i sin(5x)+2 cos(67x)) dx = 1

∫[1-cos(2(√6-4i sin(5x)+2 cos(67x)))]dx = 1

∫1dx - ∫(cos(2(√6-4i sin(5x)+2 cos(67x)))) dx = 1

x - (1/2)(sin(2(√6-4i sin(5x)+2 cos(67x))))|√6-4i sin(5x)+2 cos(67x) = a| = x - (1/2)sin(2a)0 to 2a = 1

∫2a = x - (1/2)sin(2a) = 1

x = 1 + (1/2)sin(2a)

Since the wave function is symmetric, we only need to integrate over the range 0 to a.

Normalization constant N = (2/√3)

(b) The probability of each eigenstate is given by |cn|^2.

Where cn is the coefficient of the nth eigenfunction in the expansion of the wave function.

We have,

ψ(x) = N sin^2(√6-4i sin(5x)+2 cos(67x) = N[(1/√3)sin(2x) - (2/√6)sin(4x) + (1/√3)sin(6x)]

Comparing with the given form, we get,

c1 = (1/√3)

c2 = - (2/√6)

c3 = (1/√3)

Probability of nth eigenstate = |cn|^2

Therefore,

Probability of first eigenstate (n = 1) = |c1|^2 = (1/3)

Probability of second eigenstate (n = 2) = |c2|^2 = (2/3)

Probability of third eigenstate (n = 3) = |c3|^2 = (1/3)

Eigenenergy of nth state = En = (n²π²ħ²)/2ma²

For even n, Un(x) = √(2/2a) cos(nπx/2a)

∴ n = 2, 4, 6, ...

For odd n, Un(x) = Va sin(nπx/2a)

∴ n = 1, 3, 5, ...

(c) The expected average value of energy is given by,

∫ψ(x)V(x)ψ(x)dx = ∫|ψ(x)|²En dx

For V(x) = E0 = 0, a < x < a

We have,

En = (n²π²ħ²)/2ma²

En for even n = 2, 4, 6...

En for odd n = 1, 3, 5...

We have already calculated |ψ(x)|² and En.

∴ ∫|ψ(x)|²En dx = ∑|cn|²En

∫(1/√3)sin²(2x)dx - (2/√6)sin²(4x)dx + (1/√3)sin²(6x)dx

= [(2/3)(π²ħ²)/(2ma²)] + [(8/3)(π²ħ²)/(2ma²)] + [(18/3)(π²ħ²)/(2ma²)]

= [(2+8+18)π²ħ²]/[3(2ma²)]

= (28π²ħ²)/(3ma²)

Hence, the expected average value of energy is (28π²ħ²)/(3ma²).

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The particle's charge is approximately 19.35 milli-Coulombs (mC).

To find the particle's charge, we can use the equation for the magnetic force on a charged particle:

F = q * v * B * sin(theta)

Where:

F is the magnetic force,

q is the charge of the particle,

v is the velocity of the particle,

B is the magnetic field,

and theta is the angle between the velocity and the magnetic field.

We are given:

F = 0.458 € N,

v = 3.68 km/s = 3.68 * 10^3 m/s,

B = 6.43 * 10^(-3) T (since 1 mT = 10^(-3) T),

and theta = 30.6°.

Let's solve the equation for q:

q = F / (v * B * sin(theta))

Substituting the given values:

q = 0.458 € N / (3.68 * 10^3 m/s * 6.43 * 10^(-3) T * sin(30.6°))

Calculating:

q = 0.458 € N / (3.68 * 6.43 * sin(30.6°)) * 10^3 C

q ≈ 0.458 € N / (23.686) * 10^3 C

q ≈ 19.35 * 10^(-3) C

Therefore, the particle's charge is approximately 19.35 milliCoulombs (mC).

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a) The depth of the tank is approximately 1.683 meters. b) On the winter solstice in Miami, the depth of the tank would need to be approximately 2.589 meters for the sun not to illuminate the bottom of the tank .

(a) In the given scenario, when the sunlight ceases to illuminate any part of the bottom of the tank, then it can be solved by following method,

The height of the tank is ='h', and the angle between the ground and the sunlight is = θ (30.5°). The radius of the tank is = 'r'.

Since the sunlight ceases to illuminate the bottom of the tank, the height 'h' will be equal to the radius 'r' of the tank. Therefore, the value of 'h should be found out.

The tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the tangent of angle θ is equal to h/r:

tan(θ) = h/r

Substituting the given values: tan(30.5°) = h/2.9

To find 'h', one can rearrange the equation:

h = tan(30.5°) × 2.9

Calculating the value of 'h':

h ≈ 2.9 × tan(30.5°) ≈ 1.683 m

So,  the depth of the tank is approximately 1.683 meters.

b)  the sun reaches a maximum altitude of 40.8° above the horizon,

The angle θ is now 40.8°, and one need to find the depth 'h' required for the sun not to illuminate the bottom of the tank.

Using the same trigonometric relationship,

tan(θ) = h/r

Substituting the given values: tan(40.8°) = h/2.9

To find 'h', rearrange the equation:

h = tan(40.8°) ×2.9

Calculating the value of 'h':

h ≈ 2.9 × tan(40.8°) ≈ 2.589 m

Therefore, on the winter solstice in Miami, the depth of the tank would need to be approximately 2.589 meters for the sun not to illuminate the bottom of the tank.

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To calculate the strength of the magnetic field at point P in the given figure, we can use Ampere's Law. Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the product of the permeability of free space (μ₀) and the current enclosed by the loop.

In this case, the loop can be chosen as a circle centered at point P with a radius equal to r2. The current enclosed by the loop is I.

Using Ampere's Law, we have:

∮ B · dl = μ₀ * I_enclosed

Since the magnetic field is assumed to be constant along the circular path, we can simplify the equation to:

B * 2πr2 = μ₀ * I

Solving for B, we get:

B = (μ₀ * I) / (2πr2)

Plugging in the given values:

B = (4π × 10^-7 T·m/A) * (5.6 A) / (2π × 0.028 m)

B ≈ 0.04 T

Therefore, the strength of the magnetic field at point P is approximately 0.04 Tesla.

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The magnitude of the force on a 1 microCoulomb charge placed at the center of the square is 21.45 N.

We know that, Force between two point charges given by:

Coulombs' law is:

F = kQq/r² where, F is the force between the charges Q and q, k is Coulomb’s constant (9 × 10⁹ Nm²/C²), r is the separation distance between the charges, measured in meters Q and q are the magnitude of charges measured in Coulombs. So, the force between the charges can be calculated as shown below:

F₁ = kQq/d² where, k = 9 × 10⁹ Nm²/C², Q = 16.63 µC, q = 1 µCd = 22.155 cm = 0.22155 m.

The force F₁ is repulsive as the charges are of the same sign. It acts along the diagonal of the square passing through the center of the square.

Now, the force on the charge at the center of the square due to the other three charges is

F = √2 F₁= √2 (kQq/d²) = √2 × (9 × 10⁹) × (16.63 × 10⁻⁶) × (1 × 10⁻⁶) / (0.22155)²= 21.45 N

The magnitude of the force on a 1 microCoulomb charge placed at the center of the square is 21.45 N.

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4kg of steam is at 100 degrees celcius and heat is removed untilthere is water at 39 degrees celcius, approximately 8,016,216 joules of heat are removed when converting 4 kg of steam at 100 degrees Celsius to water at 39 degrees Celsius.

To calculate the amount of heat removed when converting steam at 100 degrees Celsius to water at 39 degrees Celsius, we need to consider the specific heat capacities and the heat transfer equation.

The specific heat capacity of steam (C₁) is approximately 2,080 J/(kg·°C), and the specific heat capacity of water (C₂) is approximately 4,186 J/(kg·°C).

The equation for heat transfer is:

Q = m ×(C₂ × ΔT₂ + L)

Where:

Q is the heat transfer (in joules),

m is the mass of the substance (in kilograms),

C₂ is the specific heat capacity of water (in J/(kg·°C)),

ΔT₂ is the change in temperature of water (in °C), and

L is the latent heat of vaporization (in joules/kg).

In this case, since we are converting steam to water at the boiling point, we need to consider the latent heat of vaporization. The latent heat of vaporization of water (L) is approximately 2,260,000 J/kg.

Given:

Mass of steam (m) = 4 kg

Initial temperature of steam = 100°C

Final temperature of water = 39°C

ΔT₂ = Final temperature - Initial temperature

ΔT₂ = 39°C - 100°C

ΔT₂ = -61°C

Now we can calculate the heat transfer:

Q = 4 kg × (4,186 J/(kg·°C) × -61°C + 2,260,000 J/kg)

Q ≈ 4 kg × (-255,946 J + 2,260,000 J)

Q ≈ 4 kg × 2,004,054 J

Q ≈ 8,016,216 J

Therefore, approximately 8,016,216 joules of heat are removed when converting 4 kg of steam at 100 degrees Celsius to water at 39 degrees Celsius.

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The heat pump extracts more than 2.67×10⁴ W of energy from the outside air.

The coefficient of performance (COP) is a measure of the efficiency of a heat pump. It is defined as the ratio of the heat transferred into the system to the work done by the system. In this case, the COP of the heat pump is 3.80.

To determine the amount of energy extracted from the outside air, we need to use the equation:

COP = Qout / Win,

where COP is the coefficient of performance, Qout is the heat extracted from the outside air, and Win is the work done by the heat pump.

We are given that the COP is 3.80 and the power consumption is 7.03×10³W. By rearranging the equation, we can solve for Qout:

Qout = COP * Win.

Plugging in the given values, we have:

Qout = 3.80 * 7.03×10³W.

Calculating this, we find that the heat pump extracts approximately 2.67×10⁴ W of energy from the outside air. This means that for every watt of electricity consumed by the heat pump, it extracts 2.67×10⁴ watts of heat from the outside air.

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Hence as the magnetic fields due to Wires 1 and 3 are in the same direction (into the page), and the magnetic field due to Wire 2 is in the opposite direction (out of the page), we need to subtract the magnitude of the magnetic field due to Wire 2 from the sum of the magnitudes of the magnetic fields due to Wires 1 and 3 to get the net magnetic field at point C.

To calculate the magnetic field at point C, we can use the Biot-Savart Law, which relates the magnetic field generated by a current-carrying wire to the distance from the wire.

Let's assume the right triangle has sides A, B, and C, with point C being the midpoint of the hypotenuse. The wires are located along the edges of the triangle, so let's label them as follows:

Wire 1: Located along side A, with a current I1 = 2 mA

Wire 2: Located along side B, with a current I2 = 2 mA

Wire 3: Located along the hypotenuse (opposite side C), with a current I3 = 2 mA

To calculate the magnetic field at point C due to each wire, we can use the following formula:

dB = (μ₀ / 4π) * (I * dl × r) / r^3

Where:

dB is the infinitesimal magnetic field vector,

μ₀ is the permeability of free space (4π × 10^-7 T·m/A),

I is the current in the wire,

dl is an infinitesimal length element of the wire,

r is the distance from the wire element to the point where we want to calculate the magnetic field.

To calculate the net magnetic field at point C, we'll sum the magnetic fields due to each wire vectorially.

Let's first calculate the magnetic field due to Wire 1 at point C:

Using the right-hand rule, we can determine that the magnetic field at point C due to Wire 1 will be directed into the page.

Now, let's calculate the magnetic field due to Wire 2 at point C:

Using the right-hand rule, we can determine that the magnetic field at point C due to Wire 2 will be directed out of the page.

Finally, let's calculate the magnetic field due to Wire 3 at point C:

Using the right-hand rule, we can determine that the magnetic field at point C due to Wire 3 will be directed into the page.

Hence as the magnetic fields due to Wires 1 and 3 are in the same direction (into the page), and the magnetic field due to Wire 2 is in the opposite direction (out of the page), we need to subtract the magnitude of the magnetic field due to Wire 2 from the sum of the magnitudes of the magnetic fields due to Wires 1 and 3 to get the net magnetic field at point C.

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The correct altitude for a geosynchronous satellite is approximately 6.4 x 10^6 meters.

The correct option for the altitude of a geosynchronous satellite is e) 6.4 x 106 m. Geosynchronous satellites are placed in orbits at an altitude where their orbital period matches the Earth's rotation period, allowing them to remain stationary relative to a point on the Earth's surface. This altitude is approximately 35,786 kilometers or 22,236 miles above the Earth's equator. Converting this to meters, we get 35,786,000 meters or 3.6 x 107 meters. Therefore, option e) 6.4 x 106 m is not the correct altitude for a geosynchronous satellite.

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To calculate the magnitude of the magnetic field at a distance of 10 cm from the axis of the cylinder, we can use Ampere's law. The magnitude of the magnetic field at a distance of 10 cm from the axis of the cylinder is 2 × 10⁻⁵, Tesla.

Ampere's law states that the line integral of the magnetic field around a closed path is equal to the product of the current enclosed by the path and the permeability of free space (μ₀).

In this case, the current is flowing uniformly through the cylinder, so the current enclosed by the path is the product of the current density (J) and the area (A) of the cross-section of the cylinder.

First, let's calculate the current enclosed by the path:

Current enclosed = Current density × Area

The area of the cross-section of the cylinder is the difference between the areas of the outer and inner circles:

[tex]Area = \pi * (b^2 - a^2)[/tex]

Substituting the given values, we have:

[tex]Area = \pi * ((7.0 cm)^2 - (5.0 cm)^2) = 36\pi cm^2[/tex]

Now, we can calculate the current enclosed:

[tex]Current enclosed = (1.0 A/cm^2) * (36\pi cm^2) = 36\pi A[/tex]

Next, we'll apply Ampere's law:

[tex]\oint$$ B.dl = \mu_0* Current enclosed[/tex]

Since the magnetic field (B) is constant along the path, we can take it out of the line integral:

[tex]B\oint$$ dl = \mu_0 * Current enclosed[/tex]

The line integral ∮ dl is equal to the circumference of the circular path:

[tex]B * (2\pi d) = \mu_0 * Current enclosed[/tex]

Substituting the known values:

[tex]B = (\mu_0 * 36\pi A) / (2\pi * 10 cm)[/tex]

The value of the permeability of free space (μ₀) is approximately 4π × 10⁻⁷ T·m/A. Substituting this value:

[tex]B = (4\pi * 10^{-7} T.m/A * 36\pi A) / (2\pi * 10 cm)\\B = (2 * 10^{-6} T.m) / (10 cm)\\B = 2 * 10^{-5} T[/tex]

Therefore, the magnitude of the magnetic field at a distance of 10 cm from the axis of the cylinder is 2 × 10⁻⁵, Tesla.

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Part A: The electric flux through the cylinder due to the infinite line of charge is approximately 3.44 × 10^11 N·m²/C.

Part B: The flux through the cylinder remains the same when the radius is increased to 0.500 m.

Part C: The flux through the cylinder remains the same when the length is increased to 0.980 m.

Part A:

To calculate the electric flux through the cylinder due to the infinite line of charge, we can use Gauss's law. The electric flux Φ through a closed surface is given by Φ = E * A, where E is the electric field and A is the area of the surface.

In this case, the electric field due to the infinite line of charge is perpendicular to the line and has magnitude E = λ / (2πε₀r), where λ is the charge per unit length, ε₀ is the vacuum permittivity, and r is the radius of the cylinder.

The area of the cylinder's curved surface is A = 2πrl, where r is the radius and l is the length of the cylinder.

Substituting the values, we have:

Φ = (λ / (2πε₀r)) * (2πrl)

Simplifying the expression, we get:

Φ = λl / ε₀

Substituting the given values:

Φ = (7.65 μC/m) * (0.455 m) / (8.854 × 10^(-12) C²/N·m²)

Calculating the expression, we find that Φ is approximately 3.44 × 10^11 N·m²/C.

Therefore, the electric flux through the cylinder due to the infinite line of charge is approximately 3.44 × 10^11 N·m²/C.

Part B:

If the radius of the cylinder is increased to r = 0.500 m, we can use the same formula to calculate the electric flux. Substituting the new value of r into the equation, we get:

Φ = (7.65 μC/m) * (0.455 m) / (8.854 × 10^(-12) C²/N·m²)

Calculating the expression, we find that Φ is still approximately 3.44 × 10^11 N·m²/C.

Therefore, the flux through the cylinder remains the same when the radius is increased to 0.500 m.

Part C:

If the length of the cylinder is increased to l = 0.980 m, we can again use the same formula to calculate the electric flux. Substituting the new value of l into the equation, we get:

Φ = (7.65 μC/m) * (0.980 m) / (8.854 × 10^(-12) C²/N·m²)

Calculating the expression, we find that Φ is still approximately 3.44 × 10^11 N·m²/C.

Therefore, the flux through the cylinder remains the same when the length is increased to 0.980 m.

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The expression for electron density of states and its plot for Ly = Ly = 50 Å is given below.

Explanation:

To derive an expression for the electron density of states in a quantum wire, we start with the given energy dispersion relation:

E(kr, l, n) = (ħ²k²)/(2m*) + (π²ħ²n²)/(2m*Ly²)

where ħ is the reduced Planck's constant,

          k is the wave vector,

         m* is the effective mass of the electron,

        Ly is the length of the wire in the y-direction,

        l is the quantum number related to the quantized transverse modes,

        n is the quantum number related to the quantized longitudinal modes.

The electron density of states (DOS) is obtained by calculating the number of allowed states within a given energy range.

In a 1D system, the number of allowed states per unit length in the k-space is given by:

dN(k) = (LxLy)/(2π) * dk

where Lx is the length of the wire in the x-direction.

To find the density of states in energy space, we use the relation:

dN(E) = dN(k) * dk/dE

To calculate dk/dE, we differentiate the energy dispersion relation with respect to k:

dE(k)/dk = (ħ²k)/(m*)

Rearranging the above equation, we get:

dk = (m*/ħ²k) * dE(k)

Substituting this value into the expression for dN(E), we have:

dN(E) = (LxLy)/(2π) * (m*/ħ²k) * dE(k)

Now, we need to express the wave vector k in terms of energy E.

Solving the energy dispersion relation for k, we have:

k(E) = [(2m*/ħ²)(E - (π²ħ²n²)/(2m*Ly²))]^(1/2)

Substituting this value back into the expression for dN(E), we get:

dN(E) = (LxLy)/(2π) * [(m*/ħ²) / k(E)] * dE(k)

Substituting the value of k(E) in terms of E, we have:

dN(E) = (LxLy)/(2π) * [(m*/ħ²) / [(2m*/ħ²)(E - (π²ħ²n²)/(2m*Ly²))]^(1/2)] * dE

Simplifying the expression:

dN(E) = [(LxLy)/(2πħ²)] * [(2m*)^(1/2)] * [(E - (π²ħ²n²)/(2m*Ly²))^(-1/2)] * dE

Now, to obtain the total density of states (DOS), we integrate the above expression over the energy range.

Considering the limits of integration as E1 and E2, we have:

DOS(E1 to E2) = ∫[E1 to E2] dN(E)

DOS(E1 to E2) = ∫[E1 to E2] [(LxLy)/(2πħ²)] * [(2m*)^(1/2)] * [(E - (π²ħ²n²)/(2m*Ly²))^(-1/2)] * dE

Simplifying and solving the integral, we get:

DOS(E1 to E2) = (LxLy)/(πħ²) * [(2m*)^(1/2)] * [(E2 - E1 + (π²ħ²n²)/(2mLy²))^(1/2) - (E1 - (π²ħ²n²)/(2mLy²))^(1/2)]

To plot the expression for the electron density of states, we substitute the given values of Ly and calculate DOS(E) for the desired energy range (E1 to E2), and plot it against energy E.

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The L-R-C series circuit has an impedance of 250.5 Ω, current amplitude of 0.116 A, and source voltage leads the current. The voltage amplitudes across the resistor, inductor, and capacitor are approximately 26.68 V, 12.528 V, and 1.102 V, respectively.

a) The impedance of the L-R-C series circuit can be calculated using the formula:

Z = √(R^2 + (Xl - Xc)^2)

where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.

Given:

Resistance (R) = 230 Ω

Inductance (L) = 0.360 H

Capacitance (C) = 5.60 μF

Voltage amplitude (V) = 29.0 V

Angular frequency (ω) = 300 rad/s

To calculate the reactances:

Xl = ωL

Xc = 1 / (ωC)

Substituting the given values:

Xl = 300 * 0.360 = 108 Ω

Xc = 1 / (300 * 5.60 * 10^(-6)) ≈ 9.52 Ω

Now, substituting the values into the impedance formula:

Z = √(230^2 + (108 - 9.52)^2)

Z ≈ √(52900 + 9742)

Z ≈ √62642

Z ≈ 250.5 Ω

b) The current amplitude (I) can be calculated using Ohm's Law:

I = V / Z

I = 29.0 / 250.5

I ≈ 0.116 A

c) The phase angle (φ) of the source voltage with respect to the current can be determined using the formula:

φ = arctan((Xl - Xc) / R)

φ = arctan((108 - 9.52) / 230)

φ ≈ arctan(98.48 / 230)

φ ≈ arctan(0.428)

φ ≈ 23.5°

d) The source voltage leads the current because the phase angle is positive.

e) The voltage amplitude across the resistor is given by:

VR = I * R

VR ≈ 0.116 * 230

VR ≈ 26.68 V

f) The voltage amplitude across the inductor is given by:

VL = I * Xl

VL ≈ 0.116 * 108

VL ≈ 12.528 V

g) The voltage amplitude across the capacitor is given by:

VC = I * Xc

VC ≈ 0.116 * 9.52

VC ≈ 1.102 V

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The magnitude of the electric field is 18 N/C, and the true direction of the electric field is perpendicular to the ground.

In the given scenario, a small object with a mass and charge of -18.A NCs is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground.

The electric field strength, or magnitude, is given as 18 N/C. The unit "N/C" represents newtons per coulomb, indicating the force experienced by each unit of charge in the electric field. Therefore, the magnitude of the electric field is 18 N/C.

The true direction of the electric field is perpendicular to the ground. Since the object is suspended motionless, it means the electric force acting on the object is balanced by another force (such as gravity or tension) in the opposite direction.

The fact that the object remains motionless indicates that the electric force and the opposing force are equal in magnitude and opposite in direction. Therefore, the electric field points in the true direction perpendicular to the ground.

In summary, the magnitude of the electric field is 18 N/C, and the true direction of the electric field is perpendicular to the ground.

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Answer:

a.) The projectile will reach the height of 20m above the cliff after 0.4 seconds.

b.) The projectile will be in the air for 2 seconds.

c.)  The horizontal distance traveled by the projectile is 100 meters.

d.) The velocity of the projectile 3 seconds after it was shot is 20.6 m/s. The direction of the velocity is 30° below the horizontal.

Explanation:

a)  The time it takes for the projectile to reach a height of 20m above the cliff can be found using the following equation:

t = (20m - 100m) / (100m/s) * sin(60°)

t = 0.4 seconds

Therefore, the projectile will reach the height of 20m above the cliff after 0.4 seconds.

b) The time it takes for the projectile to reach the ground can be found using the following equation:

t = 2 * (100m) / (100m/s) * sin(60°)

t = 2 seconds

Therefore, the projectile will be in the air for 2 seconds.

c) The horizontal distance traveled by the projectile can be found using the following equation:

d = v * t * cos(θ)

where v is the initial velocity of the projectile, t is the time it takes for the projectile to travel the horizontal distance, and θ is the angle of projection.

v = 100 m/s

t = 2 seconds

θ = 60°

d = 100 m/s * 2 seconds * cos(60°)

d = 100 m/s * 2 seconds * 0.5

d = 100 meters

Therefore, the horizontal distance traveled by the projectile is 100 meters.

d.) The velocity of the projectile 3 seconds after it was shot can be found using the following equation:

v = v0 * cos(θ) - gt

where v is the final velocity of the projectile, v0 is the initial velocity of the projectile, θ is the angle of projection, and g is the acceleration due to gravity.

v0 = 100 m/s

θ = 60°

g = 9.8 m/s²

v = 100 m/s * cos(60°) - 9.8 m/s² * 3 seconds

v = 50 m/s - 29.4 m/s

v = 20.6 m/s

Therefore, the velocity of the projectile 3 seconds after it was shot is 20.6 m/s. The direction of the velocity is 30° below the horizontal.

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