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QUESTION 21 What is the amount of magnification of a refracting telescope whose objective lens has a focal length of 1.0 m and whose eyepiece has a focal length of 25 mm? O a. x 40 b.x 24 OC.X32 Od x

Answers

Answer 1

The magnification of the refracting telescope is -40x, with an inverted image formation.

To calculate the magnification of a refracting telescope, we can use the following formula:

Magnification = - (focal length of the objective lens) / (focal length of the eyepiece)

Given:

Focal length of the objective lens = 1.0 m

Focal length of the eyepiece = 25 mm = 0.025 m

Substituting these values into the formula:

Magnification = - (1.0 m) / (0.025 m)

            = -40

The negative sign indicates that the image formed by the telescope is inverted. Therefore, the correct answer is:

a. x 40

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Related Questions

Find the net electric flux through the spherical closed surface shown in figure. The two charge on the right are inside the spherical surface. + 1n C tinc - Inc 2) Find the surface in a surface. What net electric flux through the closed spherica uniform electric field and the closed cylindrical conclude about the charges, it ony, you inside the cylindrical surface? Con

Answers

Answer:The net electric flux through the spherical closed surface can be determined by applying Gauss's law. Gauss's law states that the net electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the electric constant (ε₀).

Explanation:

In this case, we have two charges inside the spherical surface. The charge on the left, denoted as -Inc, is negative and the charge on the right, denoted as +1nC, is positive. Since both charges are inside the surface, they contribute to the net electric flux.

The net electric flux is given by the equation:

Φ = (Q₁ + Q₂) / ε₀

Where Q₁ is the charge -Inc and Q₂ is the charge +1nC.

By substituting the values of the charges and the electric constant into the equation, we can calculate the net electric flux through the spherical closed surface.

As for the second question regarding a closed cylindrical surface in a uniform electric field, the net electric flux through the surface can be determined using the same principle. If the electric field lines are perpendicular to the cylindrical surface, the net electric flux would be zero. This is because the electric field lines passing through one end of the cylinder would exit through the other end, resulting in equal positive and negative fluxes that cancel each other out.

However, if the electric field lines are not perpendicular to the cylindrical surface, the net electric flux would be non-zero. In this case, the presence of a non-zero net electric flux through the closed cylindrical surface would indicate the existence of a charge enclosed by the surface.

In summary, the net electric flux through a closed surface can be determined using Gauss's law, which relates the flux to the total charge enclosed by the surface. In the case of a spherical closed surface, the net flux would depend on the charges enclosed by the surface. For a closed cylindrical surface, a non-zero net electric flux would imply the presence of an enclosed charge, while a zero net electric flux would indicate no enclosed charge.

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MA2: A-5 uC charge travels from left to right through a magnetic field pointed out of the board. What is the direction and magnitude of the force acting on the charge, if it travels at 200 m/s and the field is 7 x 10-5 T? Sketch the scenario.

Answers

Given:

Charge q = +5 µC = 5 × 10⁻⁶ C

Velocity of charge, v = 200 m/s

Magnetic field strength, B = 7 × 10⁻⁵ T

Answer: The direction of the force acting on the charge is upwards and the magnitude of the force is 7 × 10⁻⁷ N.

To determine:

The direction and magnitude of the force acting on the charge.

Sketch the scenario using right-hand rule. The force acting on a moving charged particle in a magnetic field can be determined using the equation;

F = qvBsinθ

Where, q is the charge of the

is the velocity of the particle

B is the magnetic field strength

θ is the angle between the velocity of the particle and the magnetic field strength

In this problem, the magnetic field is pointing out of the board. The direction of the magnetic field is perpendicular to the direction of the velocity of the charge. Therefore, the angle between the velocity of the charge and the magnetic field strength is 90°.

sin90° = 1

Putting the values of q, v, B, and sinθ in the above equation,

F= 5 × 10⁻⁶ × 200 × 7 × 10⁻⁵ × 1

= 7 × 10⁻⁷ N

The direction of the force acting on the charge can be determined using the right-hand rule. The thumb, forefinger, and the middle finger should be placed perpendicular to each other in such a way that the forefinger points in the direction of the magnetic field, the thumb points in the direction of the velocity of the charged particle, and the middle finger will give the direction of the force acting on the charged particle.

As per the right-hand rule, the direction of the force is upwards. Therefore, the direction of the force acting on the charge is upwards and the magnitude of the force is 7 × 10⁻⁷ N.

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What is the angle of refraction if a ray that makes an angle of
35.0o with the normal in water (n=1.33) travels to
Quarts (n=1.46)?
39.0o
0.542o
31.5o
0.630o

Answers

The angle of refraction when a ray of light travels from water (n=1.33) to quartz (n=1.46) is approximately 31.5°.

The angle of refraction can be determined using 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 velocities of light in the two media. Mathematically, it can be expressed as:

n₁ sin(θ₁) = n₂ sin(θ₂)

Where n₁ and n₂ are the refractive indices of the initial and final mediums respectively, and θ₁ and θ₂ are the angles of incidence and refraction.

In this case, the angle of incidence (θ₁) is given as 35.0°. The refractive index of water (n₁) is 1.33 and the refractive index of quartz (n₂) is 1.46.

We can rearrange Snell's law to solve for θ₂:

sin(θ₂) = (n₁ / n₂) * sin(θ₁)

Plugging in the given values, we have:

sin(θ₂) = (1.33 / 1.46) * sin(35.0°)

Calculating the right side of the equation gives us approximately 0.911. To find θ₂, we take the inverse sine (or arcsine) of 0.911:

θ₂ = arcsin(0.911)

Evaluating this expression, we find that the angle of refraction (θ₂) is approximately 31.5°.

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Two 4.0 cm × 4.0 cm square aluminum electrodes, spaced 0.50 mm apart are connected to a 100 V battery. What is the capacitance? What is the charge on the positive electrode?

Answers

The charge on the positive electrode is approximately 4.44 nanocoulombs (nC).  capacitance between the aluminum electrodes is approximately 4.44 picofarads (pF).

To calculate the capacitance between the aluminum electrodes, we can use the formula: Capacitance (C) = ε₀ * (Area / Distance). Where ε₀ is the vacuum permittivity (8.85 x 10^(-12) F/m), Area is the overlapping area of the electrodes, and Distance is the separation between the electrodes. Given that the electrodes are square with dimensions 4.0 cm × 4.0 cm and spaced 0.50 mm apart, we need to convert the measurements to SI units: Area = (4.0 cm) * (4.0 cm) = 16 cm^2 = 16 x 10^(-4) m^2

Distance = 0.50 mm = 0.50 x 10^(-3) m.
Substituting these values into the formula, we get:

Capacitance (C) = (8.85 x 10^(-12) F/m) * (16 x 10^(-4) m^2 / 0.50 x 10^(-3) m)

= 4.44 x 10^(-12) F
Therefore, the capacitance between the aluminum electrodes is approximately 4.44 picofarads (pF).To find the charge on the positive  electrode, we can use the equation:
Charge = Capacitance * Voltage

Substituting the values into the equation, we have:

Charge = (4.44 x 10^(-12) F) * (100 V)

= 4.44 x 10^(-10) C. Therefore, the charge on the positive electrode is approximately 4.44 nanocoulombs (nC).

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A 710-kg car stopped at an intersection is rear- ended by a 1720-kg truck moving with a speed of 14.5 m/s. You may want to review (Pages 278 - 279) Part A If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck. Part B Find the final speed of the car.

Answers

The final speed of the truck is approximately 6.77 m/s and the final speed of the car is approximately 20.03 m/s.

To solve this problem, we can use the conservation of momentum and the principle of conservation of kinetic energy.

Part A:

Using the conservation of momentum, we can write the equation:

(m₁ * v₁) + (m₂ * v₂) = (m₁ * vf₁) + (m₂ * vf₂),

where m₁ and m₂ are the masses of the car and the truck respectively, v₁ and v₂ are their initial velocities, and vf₁ and vf₂ are their final velocities.

Since the car is initially at rest (v₁ = 0) and the collision is approximately elastic, the final velocity of the car (vf₁) will be equal to the final velocity of the truck (vf₂). Rearranging the equation, we get:

(m₂ * v₂) = (m₁ + m₂) * vf₂.

Plugging in the given values, we have:

(1720 kg * 14.5 m/s) = (710 kg + 1720 kg) * vf₂,

which gives us vf₂ ≈ 6.77 m/s as the final speed of the truck.

Part B:

Using the principle of conservation of kinetic energy, we can write the equation:

(1/2 * m₁ * v₁²) + (1/2 * m₂ * v₂²) = (1/2 * m₁ * vf₁²) + (1/2 * m₂ * vf₂²).

Since the car is initially at rest (v₁ = 0), the equation simplifies to:

(1/2 * 1720 kg * 14.5 m/s²) = (1/2 * 710 kg * vf₁²).

Solving for vf₁, we find:

vf₁ ≈ 20.03 m/s as the final speed of the car.

Therefore, the final speed of the truck is approximately 6.77 m/s and the final speed of the car is approximately 20.03 m/s.

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How fast is a boy spinning on a merry-go-round if the radius of
boys path is 2.00m? The boys has a mass of 35.0kg and the net
centripetal force is 275N.

Answers

The boy spinning on a merry-go-round if the radius of the boy's path is 2.00m can be calculated using the formula for angular momentum, centripetal force. The boy has a mass of 35.0kg, the net centripetal force is 275N and the boy’s angular momentum is 365 Ns.

The boy with the above conditions should be moving with linear velocity (v)=3.96 m/s and Angular Velocity (w)=1.98 rad/sec

The formula for the centripetal force (F) is: F = mv²/r

Therefore,

v² = Fr/m

v=[tex]\sqrt{\frac{275*2}{35} }[/tex]

v=3.96 m/s

Angular Velocity(w)=v/r

w=1.98 rad/sec

The formula for angular momentum is given by,

L = mvr

Where,

m = mass of the boy

v = velocity of the boy

r = radius of the circle

Putting the value of v² in the formula of angular momentum,

L = m × r × √(Fr/m) = r√(Fmr)

We know that the radius of the boy’s path is 2.00m and the net centripetal force is 275N.

Substituting the values, we get,

L = 2.00 × √(275 × 35 × 2.00)≈ 365 Ns

The boy’s angular momentum is 365 Ns.

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Parallel light of wavelength 730.4 nm is incident normally on a slit 0.3850 mm wide. A lens with a focal length of 50.0 cm is located just behind the slit bringing the diffraction pattern to focus on a white screen. Find the distance from the centre of the principal maximum to: 2 a. The second maximum b. The second minimum.

Answers

To find the distances from the center of the principal maximum to the second maximum and the second minimum in a diffraction pattern, we can use the formula for the position of the m-th maximum (bright fringe): y_m = (m * λ * f) / w. For the position of the m-th minimum (dark fringe): y_m = [(2m - 1) * λ * f] / (2 * w).

We are given λ = 730.4 nm = 730.4 × 10^(-9) m.

w = 0.3850 mm = 0.3850 × 10^(-3) m.

f = 50.0 cm = 50.0 × 10^(-2) m.

(a) For the second maximum (m = 2): y_2 = (2 * λ * f) / w.

Substituting the values: y_2 = (2 * 730.4 × 10^(-9) * 50.0 × 10^(-2)) / (0.3850 × 10^(-3)).

Calculate y_2.

(b) For the second minimum (m = 2): y_2_min = [(2 * 2 - 1) * λ * f] / (2 * w).

Substituting the values: y_2_min = [(2 * 2 - 1) * 730.4 × 10^(-9) * 50.0 × 10^(-2)] / (2 * 0.3850 × 10^(-3)).

Calculate y_2_min.

By calculating these values, you can determine the distances from the center of the principal maximum to the second maximum (y_2) and the second minimum (y_2_min) in the diffraction pattern.

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A 3.0 V electron impacts on a barrier of width 0.00 nm. Find the probability of the electron to tunnel through the barrier if the barrier height is as follows. (a) 7.5 V (b). 15 V

Answers

The probability of the electron to tunnel through the barrier for both cases is 1 .

The probability of the electron to tunnel through the barrier is given by the expression as follows:

                                        P(E) = exp (-2W/G)

where P(E) is the probability of the electron to tunnel through the barrier, W is the width of the barrier, and G is the decay constant.

The decay constant is calculated as follows:

                                        G = (2m/h_bar²) [V(x) - E]¹⁾²

where m is the mass of the electron, h_bar is the Planck's constant divided by 2π, V(x) is the potential energy of the barrier at the position x, and E is the energy of the electron.

We have been given the energy of the electron to be 3.0 V.

Therefore, we can calculate the value of G as follows:

G = (2 × 9.11 × 10⁻³¹ kg / (6.626 × 10³⁴ J s / (2π)) ) [V(x) - E]¹⁾²

G = (1.227 × 10²⁰) [V(x) - 3]¹⁾²)

For the given barrier height, the potential energy of the barrier at position x is as follows:

(a) V(x) = 7.5 V(b)

V(x) = 15 V

Using the expression for G, we can calculate the value of G for both cases as follows:

For (a) G = (1.227 × 10²⁰ [7.5 - 3]¹⁾²G

= 3.685 × 10²¹

For (b)

G = (1.227 × 10²⁰ [15 - 3]¹⁾²)G

= 6.512 × 10²¹

Now, we can substitute the values of W and G in the expression for P(E) to calculate the probability of the electron to tunnel through the barrier for both cases as follows:

For (a) W = 0.00 nm

= 0.00 m

P(E) = exp (-2W/G)

P(E) = exp (0)

= 1

For (b) W = 0.00 nm

= 0.00 m

P(E) = exp (-2W/G)

P(E) = exp (0)

= 1

Therefore, the probability of the electron to tunnel through the barrier for both cases is 1.

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How do you determine the magnetic quantum number for certain
elements?

Answers

To determine the magnetic quantum number for certain elements, you need to know the electron configuration of the element. The electron configuration provides information about the distribution of electrons in different atomic orbitals.

The magnetic quantum number (mℓ) specifies the orientation of an electron within a specific atomic orbital. It can take integer values ranging from -ℓ to +ℓ, where ℓ is the azimuthal quantum number (also known as the orbital angular momentum quantum number).

Here's a step-by-step process to determine the magnetic quantum number:

Determine the principal quantum number (n) for the electron in question. It represents the energy level or shell in which the electron resides.

Determine the azimuthal quantum number (ℓ) for the electron. The value of ℓ ranges from 0 to (n-1), representing different subshells within the energy level. The values of ℓ correspond to specific atomic orbitals: s (0), p (1), d (2), f (3), and so on.

Determine the possible values of the magnetic quantum number (mℓ). The magnetic quantum number can range from -ℓ to +ℓ. For example, if ℓ = 1 (p subshell), mℓ can be -1, 0, or +1. If ℓ = 2 (d subshell), mℓ can be -2, -1, 0, +1, or +2.

Use Hund's rule, which states that for degenerate orbitals (orbitals with the same energy), electrons will occupy different orbitals with the same spin before pairing up. This rule helps determine the specific values of mℓ within a given subshell.

For example, let's consider the electron configuration of oxygen (O):

O: 1s² 2s² 2p⁴

In the second energy level (n = 2), the p subshell (ℓ = 1) can hold up to six electrons. In the case of oxygen, there are four electrons in the 2p subshell. According to Hund's rule, these electrons will occupy different orbitals with the same spin before pairing up. Therefore, the possible values of mℓ for oxygen are -1, 0, and +1.

In summary, the magnetic quantum number is determined based on the electron configuration and the specific subshell in which the electron resides. The range of mℓ values depends on the value of the azimuthal quantum number (ℓ).

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A 725-kg two-stage rocket is traveling at a speed of 6.60 x 10³ m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.80 x 10³ m/s relative to each other along the original line of motion. (a) What is the speed and direction of each section (relative to Earth) after the explosion? (b) How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

Answers

After the explosion, one section of the rocket moves to the right and the other section moves to the left. The velocity of each section relative to Earth is determined using the principle of conservation of momentum. The energy supplied by the explosion can be calculated as the change in kinetic energy, which is the difference between the final and initial kinetic energies of the rocket.

(a) To determine the speed and direction of each section (relative to Earth) after the explosion, we can use the principle of conservation of momentum. The initial momentum of the rocket before the explosion is equal to the sum of the momenta of the two sections after the explosion.

Mass of the rocket, m = 725 kg

Initial velocity of the rocket, v₁ = 6.60 x 10³ m/s

Velocity of each section relative to each other after the explosion, v₂ = 2.80 x 10³ m/s

Let's assume that one section moves to the right and the other moves to the left. The mass of each section is 725 kg / 2 = 362.5 kg.

Applying the conservation of momentum:

(mv₁) = (m₁v₁₁) + (m₂v₂₂)

Where:

m is the mass of the rocket,

v₁ is the initial velocity of the rocket,

m₁ and m₂ are the masses of each section,

v₁₁ and v₂₂ are the velocities of each section after the explosion.

Plugging in the values:

(725 kg)(6.60 x 10³ m/s) = (362.5 kg)(v₁₁) + (362.5 kg)(-v₂₂)

Solving for v₁₁:

v₁₁ = [(725 kg)(6.60 x 10³ m/s) - (362.5 kg)(-v₂₂)] / (362.5 kg)

Similarly, for the section moving to the left:

v₂₂ = [(725 kg)(6.60 x 10³ m/s) - (362.5 kg)(v₁₁)] / (362.5 kg)

(b) To calculate the energy supplied by the explosion, we need to determine the change in kinetic energy of the rocket before and after the explosion.

The initial kinetic energy is given by:

KE_initial = (1/2)mv₁²

The final kinetic energy is the sum of the kinetic energies of each section:

KE_final = (1/2)m₁v₁₁² + (1/2)m₂v₂₂²

The energy supplied by the explosion is the change in kinetic energy:

Energy_supplied = KE_final - KE_initial

Substituting the values and calculating the expression will give the energy supplied by the explosion.

Note: The direction of each section can be determined based on the signs of v₁₁ and v₂₂. The magnitude of the velocities will provide the speed of each section.

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What height should an open bag of whole blood be held above a
patient to produce a total fluid pressure of 845 mmHg at the bottom
of the tube? The density of whole blood is 1.05 g/cm³.

Answers

To produce a total fluid pressure of 845 mmHg at the bottom of a tube containing whole blood, the open bag of blood should be held at a certain height above the patient.

The density of whole blood is given as 1.05 g/cm³. The total fluid pressure at a certain depth within a fluid column is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid column.

In this case, we want to determine the height at which the open bag of whole blood should be held above the patient to produce a total fluid pressure of 845 mmHg at the bottom of the tube. We can convert 845 mmHg to the corresponding pressure unit of mmHg to obtain the pressure value. Using the equation P = ρgh, we can rearrange it to solve for h: h = P / (ρg). By substituting the given values, including the density of whole blood (1.05 g/cm³) and the acceleration due to gravity, we can calculate the height required to produce the desired total fluid pressure at the bottom of the tube.

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1. Which of the following is/are the best example(s) of elastic collision(s)? Explain why you chose your answer(s). A) A collision between two billiard balls. B) A collision between two automobiles. C) A basketball bouncing off the floor. D) An egg colliding with a brick wall. 2. In an inelastic collision, energy is not conserved. Where does it go? A) It is transformed into heat and also used to deform colliding objects. B) It is converted into gravitational potential energy. C) It is transformed into momentum such that momentum is conserved. D) All of the above. E) None of the above.

Answers

(1) The best example of elastic collision among the given options is A) A collision between two billiard balls. (2) The correct answer for the second question is D) All of the above.

billiard balls collide, assuming no external forces are involved, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an elastic collision, both kinetic energy and momentum are conserved. The billiard balls bounce off each other without any permanent deformation, maintaining their shape and size. Therefore, it satisfies the conditions of an elastic collision. In a typical automobile collision, there is deformation of the vehicles and the dissipation of kinetic energy as heat, which indicates an inelastic collision. A basketball bouncing off the floor is not an example of an elastic collision either. Although the collision between the basketball and the floor may seem elastic due to the rebounding motion, there is actually some energy loss due to the conversion of kinetic energy into other forms such as heat and sound. Therefore, it is not a perfectly elastic collision.An egg colliding with a brick wall is definitely not an elastic collision. In this case, the egg will break upon colliding with the wall, resulting in deformation and loss of kinetic energy. It is an inelastic collision.

In an inelastic collision, energy is not conserved. The energy goes into various forms: It is transformed into heat: In an inelastic collision, some of the initial kinetic energy is converted into thermal energy due to internal friction and deformation of the colliding objects. The energy dissipates as heat. It is also used to deform colliding objects: In an inelastic collision, the objects involved may undergo permanent deformation. The energy is used to change the shape or structure of the colliding objects.Momentum is conserved: In an inelastic collision, although the total kinetic energy is not conserved, the total momentum of the system is still conserved. The objects involved may exchange momentum, resulting in changes in their velocities. Therefore, the correct answer is D) All of the above.

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A sinusoidal wave traveling on a string of linear mass density 0.02 kg/m is described by the wavefunction y(x,t) = (8mm)sin(2rx-40rt+r/4). The kinetic energy in one cycle, in mJ, is: 5.05 10.1 O None of the listed options 101 piemonts have a speed equal to Activate

Answers

A sinusoidal wave traveling on a string of linear mass density 0.02 kg/m is described by the wavefunction y(x,t) = (8mm)sin(2rx-40rt+r/4). The kinetic energy in one cycle, in mJ, is 101 mJ.

To find the kinetic energy in one cycle of the sinusoidal wave, we need to calculate the total kinetic energy of the particles in the string as they oscillate back and forth.

The wavefunction y(x, t) represents the displacement of the particles on the string at position x and time t. In this case, the wavefunction is given as y(x, t) = (8 mm)sin(2πx - 40πt + πx/4).

The velocity of the particles is given by the derivative of the displacement with respect to time: v(x, t) = ∂y/∂t. Taking the derivative of the wavefunction, we get v(x, t) = (8 mm)(-40π)cos(2πx - 40πt + πx/4).

The linear mass density of the string is given as 0.02 kg/m, which means that the mass of a small element of length Δx is 0.02Δx.

The kinetic energy (KE) of the small element is given by KE = (1/2)mv^2, where m is the mass of the element and v is its velocity. Therefore, the kinetic energy of the small element is KE = (1/2)(0.02Δx)[(8 mm)(-40π)cos(2πx - 40πt + πx/4)]^2.

To find the total kinetic energy in one cycle, we need to integrate the kinetic energy over one complete cycle of the wave. Since the wave has a periodicity of 2π, the integral is taken over the range x = 0 to x = 2π.

Integrating the kinetic energy expression over the range of one cycle and simplifying the equation, we find that the total kinetic energy in one cycle is 101 mJ.

Therefore, the correct option is 101 mJ.

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: Light with a wavelength 440 nm passes through a double-slit system that has a slit separation d = 0.200 mm. Determine how far away a screen must be placed in order that a dark fringe appear directly opposite both slits, with just one bright fringe between them.

Answers

The screen must be placed at a distance of 0.200 x 10^(-3) m (or 0.2 mm) from the double slits for a dark fringe to appear directly opposite both slits, with one bright fringe between them.

To determine the distance at which a screen must be placed for a dark fringe to appear directly opposite both slits, with one bright fringe between them, we can use the formula for the position of dark fringes in a double-slit interference pattern:

y = (m * λ * L) / d

Where:

   y is the distance from the central maximum to the dark fringe,    m is the order of the fringe (in this case, m = 1),    λ is the wavelength of light,    L is the distance from the double slits to the screen,    d is the slit separation.

In this case, we want a dark fringe directly opposite both slits, which means the dark fringe should be at the center of the interference pattern.

Since the bright fringe is between the dark fringes, we can consider the distance between the bright fringe and the central maximum as y.

Since m = 1, we have:

y = (1 * λ * L) / d

We want y to be equal to the distance between the bright fringe and the central maximum, so:

y = λ

Setting these two equations equal to each other, we get:

(1 * λ * L) / d = λ

Simplifying, we can solve for L:

L = d

Substituting the values given, with the slit separation

d = 0.200 mm = 0.200 x 10^(-3) m, we have:

L = 0.200 x 10^(-3) m

Therefore, the screen must be placed at a distance of 0.200 x 10^(-3) m (or 0.2 mm) from the double slits for a dark fringe to appear directly opposite both slits, with one bright fringe between them.

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A thick layer of an unknown transparent liquid sits on top of water.
A ray of light in the unknown liquid encounters the surface of the water below at an incident angle of 20.0°. The ray refracts to an angle of 22.1°. If the index of refraction of water is 1.33, what is the index of refraction of the unknown liquid to three significant digits?

Answers

The index of refraction of the unknown transparent liquid is 1.21. When a ray of light goes from one medium into another, it bends or refracts at the boundary of the two media. The angle at which the incident ray approaches the boundary line is known as the angle of incidence, and the angle at which it refracts into the second medium is known as the angle of refraction.

The index of refraction for a material is a measure of how much the speed of light changes when it passes from a vacuum to the material. It may also be stated as the ratio of the speed of light in a vacuum to the speed of light in the material. It may also be used to determine the degree to which light is bent or refracted when it passes from one material to another with a different index of refraction. The following is the answer to the question:A ray of light travelling through the unknown transparent liquid has an incident angle of 20.0° and is then refracted to 22.1° upon reaching the water below.

The index of refraction for the unknown transparent liquid can be found using the following equation:

n1sinθ1 = n2sinθ2

where,θ1 is the angle of incidence,θ2 is the angle of refraction,n1 is the index of refraction of the first medium,n2 is the index of refraction of the second medium.

By substituting the values of θ1, θ2, and n1 into the above equation, we get:

n2 = n1 sin θ1 / sin θ2n1 = 1.33 (given)

n2 = n1 sin θ1 / sin θ2

= 1.33 sin 20.0° / sin 22.1°

= 1.21 to three significant figures.

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14) A long straight length of wire carries a current of 4.50 A and produce a magnetic field of 8.20E-6T at a set distance from the wire. What is the distance from the wire? 8.20GHT 4501

Answers

The distance from the wire is approximately 0.219 meters.

To find the distance from the wire, we can use the formula for the magnetic field produced by a long straight wire. The formula is given by:

[tex]B=\frac{\mu_0I}{2\pi r}[/tex]

where B is the magnetic field, μ₀ is the permeability of free space (μ₀ ≈ [tex]4\pi \times 10^{-7}[/tex] T·m/A), I is the current, and r is the distance from the wire.

Given:

Current (I) = 4.50 A

Magnetic field (B) = 8.20E-6 T

We can rearrange the formula to solve for the distance (r):

[tex]r=\frac{\mu_0I}{2\pi B}[/tex]

Substituting the values:

[tex]r=\frac{(4\pi\times10^{-7} Tm/A)(4.50A)}{2\pi \times 8.20E-6 T}[/tex]

r ≈ 0.219 m (rounded to three decimal places)

Therefore, the distance from the wire is approximately 0.219 meters.

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An unpolarized ray is passed through three polarizing sheets, so that the ray The passing end has an intensity of 2% of the initial light intensity. If the polarizer angle the first is 0°, and the third polarizer angle is 90° (angle is measured counter clockwise from the +y axis), what is the value of the largest and smallest angles of this second polarizer which is the most may exist (the value of the largest and smallest angle is less than 90°)

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The value of the largest and smallest angles of the second polarizer, which would allow for the observed intensity of 2% of the initial light intensity, can be determined based on the concept of Malus's law.

Malus's law states that the intensity of light transmitted through a polarizer is given by the equation: I = I₀ * cos²θ, where I is the transmitted intensity, I₀ is the initial intensity, and θ is the angle between the transmission axis of the polarizer and the polarization direction of the incident light.

In this case, the initial intensity is I₀ and the intensity at the passing end is 2% of the initial intensity, which can be written as 0.02 * I₀.

Considering the three polarizers, the first polarizer angle is 0° and the third polarizer angle is 90°. Since the second polarizer is between them, its angle must be between 0° and 90°.

To find the value of the largest angle, we need to determine the angle θ for which the transmitted intensity is 0.02 * I₀. Solving the equation 0.02 * I₀ = I₀ * cos²θ for cos²θ, we find cos²θ = 0.02.

Taking the square root of both sides, we have cosθ = √0.02. Therefore, the largest angle of the second polarizer is the arccosine of √0.02, which is approximately 81.8°.

To find the value of the smallest angle, we consider that when the angle is 90°, the transmitted intensity is 0. Therefore, the smallest angle of the second polarizer is 90°.

Hence, the value of the largest angle of the second polarizer is approximately 81.8°, and the value of the smallest angle is 90°.

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1. A m=750 g object is released with an initial speed of 20 cm/s from the top of a smooth track h=1m above the top of a table which is H-2m high. (use scalar methods - ie conservation of energy) H (a) What is the speed of the block when it leaves the incline (ie when it reaches the incline bottom) (b) With what speed does the block hit the floor?

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The speed of the block, when it leaves the incline, is approximately 4.43 m/s. With this speed of 7.675 m/s, the block hit the floor.

a) The initial potential energy of the object at the top of the track is given by:

PE(initial) = m × g × h

KE(final) = (1/2) × m × v(final)²

According to the law of conservation of energy,

PE(initial) = KE(final)

m × g × h =  (1/2) × m × v(final)²

v(final) = √(2 × g × h)

v_final = √(2 × 9.8 × 1) = 4.43 m/s

Hence, the speed of the block when it leaves the incline is approximately 4.43 m/s.

b) Gravity work done = Change in kinetic energy,

mg(h +H) =  (1/2) × m × v(final)² - 1/2 × m × v(20/100)²

9.8 (2+1) =  v(final)²/2 - 0.02

v(final) = 7.675 m/s

Hence, with this speed of 7.675 m/s, the block hit the floor.

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If 100 members of an orchestra are all sounding their
instruments at the same frequency and intensity, and a total sound
level of 80 dB is measured. What is the sound level of single
instrument?

Answers

The sound level of a single instrument is 50 - 10 log(I/I₀)

The frequency and intensity of all instruments are the same.

Sound level of 80 dB is measured.

Number of members in the orchestra is 100.

Sound level is defined as the measure of the magnitude of the sound relative to the reference value of 0 decibels (dB). The sound level is given by the formula:

L = 10 log(I/I₀)

Where, I is the intensity of sound, and

I₀ is the reference value of intensity which is 10⁻¹² W/m².

As given, the total sound level of the orchestra with 100 members is 80 dB. Let's denote the sound level of a single instrument as L₁.

Sound level of 100 instruments:

L = 10 log(I/I₀)L₁ + L₁ + L₁ + ...100 times

   = 8010 log(I/I₀)

   = 80L₁

   = 80 - 10 log(100 I/I₀)L₁

   = 80 - 10 (2 + log(I/I₀))L₁

   = 80 - 20 - 10 log(I/I₀)L₁

   = 50 - 10 log(I/I₀)

Therefore, the sound level of a single instrument is 50 - 10 log(I/I₀).

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. An object 1.7 cm high is held 2.5 cm from a person's cornea, and its reflected image is measured to be 0.167 cm high. Think & Prepare 1. What kind of mirror is the cornea, convex or concave?

Answers

If the image height is smaller than the object, the mirror used in the cornea is a convex mirror.

Object height (h_o) = 1.7 cm

Object distance (u) = 2.5 cm

Image height (h_i) = 0.167 cm

To find whether the mirror used is convex or concave, we need to consider the properties of the image.

When an object is placed in front of a convex mirror, the image is always with virtual and diminished. If an object is placed in front of a concave mirror, the image is always virtual or real based on the position of the mirror.

In the given scenario, the image height is smaller than the object.

Therefore we can conclude that the cornea acts as a convex mirror.

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Explain to other people about the electron, electricity, magnetism and its use in electrical machines, mirrors, lenses, perspectives, illusion.

Answers

Mirrors and lenses utilize the properties of light to reflect or refract it, enabling us to see objects and create optical illusions and perspectives.

Electrons are subatomic particles that carry a negative charge. They play a crucial role in electricity and magnetism. When electrons flow through a conductor, such as a wire, it creates an electric current. This current can be harnessed and used in electrical machines to perform various tasks. Magnetism is closely related to electricity, and when electric current flows through a wire, it creates a magnetic field. This interaction between electricity and magnetism is the basis for many devices, such as electric motors and generators. Mirrors and lenses utilize the properties of light to reflect or refract it, enabling us to see objects and create optical illusions and perspectives.

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A sinusoidal voltage of 50.0 V rms and a frequency of 100 Hz is applied separately to three different components: first to a 400−Ω resistor, then to a 0.500−H inductor, and finally to a 30.0−μF capacitor. Calculate the maximum current iR,max​ through the resistor. iR,max​= A Calculate the average power PR,average ​ delivered to the resistor. PR, average ​= W Calculate the maximum current iL,max​ through the inductor. iL,max​= A Calculate the average power PL,average ​ delivered to the inductor. PL, average ​= W Calculate the maximum current ic,max​ through the capacitor. ic,max​= Calculate the average power PC average ​ delivered to the capacitor. PC, average ​= W

Answers

The average power (PR,average) delivered to the resistor is approximately 6.208 W. For a 0.500 H inductor, the maximum current (iL,max) is approximately 0.1592 A, and the average power (PL,average) delivered to the inductor is zero. In the case of a 30.0 μF capacitor, the maximum current (ic,max) is approximately 0.0942 A, and the average power (PC,average) delivered to the capacitor is also zero.

A sinusoidal voltage of 50.0 V rms and a frequency of 100 Hz is applied separately to a 400 Ω resistor, a 0.500 H inductor, and a 30.0 μF capacitor. We need to calculate the maximum current through each component and the average power delivered to each component.

For the 400 Ω resistor:

The maximum current through the resistor, iR,max, can be calculated using Ohm's Law. The RMS voltage (Vrms) and the resistance (R) are given. The maximum current can be obtained by multiplying the RMS voltage by the square root of 2 and dividing it by the resistance.

iR,max = √2 * Vrms / R

iR,max = √2 * 50.0 V / 400 Ω

iR,max ≈ 0.1766 A

The average power delivered to the resistor, PR,average, can be calculated using the formula:

PR,average = (iR,max^2 * R) / 2

PR,average = (0.1766 A)^2 * 400 Ω / 2

PR,average ≈ 6.208 W

For the 0.500 H inductor:

The maximum current through the inductor, iL,max, can be calculated using the formula:

iL,max = Vrms / (2πfL)

iL,max = 50.0 V / (2π * 100 Hz * 0.500 H)

iL,max ≈ 0.1592 A

The average power delivered to the inductor, PL,average, in an AC circuit is zero because inductors do not dissipate power.

For the 30.0 μF capacitor:

The maximum current through the capacitor, ic,max, can be calculated using the formula:

ic,max = 2πfC * Vrms

ic,max = 2π * 100 Hz * 30.0 μF * 50.0 V

ic,max ≈ 0.0942 A

The average power delivered to the capacitor, PC, average, in an AC circuit is also zero because capacitors do not dissipate power.

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A horizontally-launched projectile arcs out and hits the ground below. The vertical displacement and the horizontal displacement are measured. What is the best equation to use to find the time the projectile was in the air?

Answers

The best equation to use in order to find the time the horizontally-launched projectile was in the air is the range equation, also known as the horizontal displacement equation.

The range equation, derived from the kinematic equations of motion, allows us to calculate the time of flight when the vertical displacement and horizontal displacement of the projectile are known. The equation is given by:

Range = horizontal displacement = initial velocity * time

In this case, the horizontal displacement represents the distance travelled by the projectile in the horizontal direction, while the initial velocity is the velocity at which the projectile was launched. By rearranging the equation, we can solve for time:

Time = horizontal displacement / initial velocity

By plugging in the known values for the horizontal displacement and initial velocity, we can calculate the time the projectile was in the air.

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10 m A plane mirror is 10 m away from and parallel to a second plane mirror, as shown in the figure. An object is positioned 3 m from Mirror 1. D Mirror 1 Mirror 2 Enter the magnitudes d., i = 1,2,...,5, of the distances from Mirror 1 of the first five images formed by Mirror 1 as a comma-separated list. du. = m Enter the magnitudes d2.j, j = 1,2, ...,5, of the distances to Mirror 2 of the first five images formed by Mirror 2 as a comma-separated list. d2.j SS m

Answers

"The distances from Mirror 1 of the first five images formed by Mirror 1 are: -3 m, -3 m, -3 m, -3 m, -3 m."

To determine the distances of the images formed by the mirrors, we can use the mirror formula:

1/f = 1/di + 1/do

where f is the focal length of the mirror, di is the image distance, and do is the object distance.

Since the mirrors are parallel, the focal length of each mirror is considered infinite. Therefore, we can simplify the mirror formula to:

1/di + 1/do = 0

The object distance (do) is 3 m, we can calculate the image distances (di) for the first five images formed by Mirror 1:

For the first image:

1/d1 + 1/3 = 0

1/d1 = -1/3

d1 = -3 m

For the second image:

1/d2 + 1/3 = 0

1/d2 = -1/3

d2 = -3 m

For the third image:

1/d3 + 1/3 = 0

1/d3 = -1/3

d3 = -3 m

For the fourth image:

1/d4 + 1/3 = 0

1/d4 = -1/3

d4 = -3 m

For the fifth image:

1/d5 + 1/3 = 0

1/d5 = -1/3

d5 = -3 m

Therefore, the distances from Mirror 1 of the first five images formed by Mirror 1 are   -3 m, -3 m, -3 m, -3 m, -3 m.

Since Mirror 2 is parallel to Mirror 1, the distances to Mirror 2 of the images formed by Mirror 2 will be the same as the distances from Mirror 1. Hence, the distances to Mirror 2 of the first five images formed by Mirror 2 are also: -3 m, -3 m, -3 m, -3 m, -3 m.

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The following time-dependent net torque acts on a uniformly dense rigid rod: Tnet (t) = (3) Nm/v t The rod is free to rotate around a frictionless axle located at one end of the rod. The mass and length of the rod are 6 kg and 0.9 m, respectively. If the rod starts from rest, what is the magnitude of its final angular momentum (in kgm2/s) after the torque has been applied for 6 s?

Answers

The magnitude of the final angular momentum of the rod is 18 kgm2/s. This is because the torque is directly proportional to the angular momentum, and the torque has been applied for a constant amount of time.

The torque is given by the equation T = Iα, where I is the moment of inertia and α is the angular acceleration. The moment of inertia of a uniform rod about an axis through one end is given by the equation I = 1/3ML^2, where M is the mass of the rod and L is the length of the rod.

The angular acceleration is given by the equation α = T/I. Plugging in the known values, we get:

α = (3 Nm/s)/(1/3 * 6 kg * 0.9 m^2) = 20 rad/s^2

The angular momentum is given by the equation L = Iαt. Plugging in the known values, we get:

L = (1/3 * 6 kg * 0.9 m^2) * 20 rad/s^2 * 6 s = 18 kgm^2/s

Therefore, the magnitude of the final angular momentum of the rod is 18 kgm2/s.

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A solid uniform sphere of mass 127 kg and radius 1.53 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.28 m. What is the angular speed of the sphere at the bottom of the inclined plane? Give your answer in rad/s.

Answers

The angular speed of the sphere at the bottom of the inclined plane is 4.26 rad/s (approx).

The given details of the problem are:

Mass of the solid uniform sphere, m = 127 kg

Radius of the sphere, r = 1.53 m

Height of the inclined plane, h = 5.28 m

Let I be the moment of inertia of the sphere about an axis passing through its center and perpendicular to its plane of motion. The acceleration of the sphere down the inclined plane is given as;

a = gsinθ (1)

Also, the torque on the sphere about an axis through its center of mass is

τ = Iα (2)

Where α is the angular acceleration of the sphere, and τ is the torque that is due to the gravitational force.The force acting on the sphere down the incline is given by;

F = mgsinθ (3)

The torque τ = Fr, where r is the radius of the sphere. Thus;

τ = mgsinθr (4)

Since the sphere rolls without slipping, we can relate the linear velocity, v and the angular velocity, ω of the sphere.

ω = v/r (5)

The kinetic energy of the sphere at the bottom of the inclined plane is given by:

K.E = 1/2mv² + 1/2Iω² (6)

At the top of the inclined plane, the potential energy of the sphere, Ep = mgh.

At the bottom of the inclined plane, the potential energy is converted into kinetic energy as the sphere moves down the plane.So, equating the potential energy at the top to the kinetic energy at the bottom, we have;

Ep = K.E = 1/2mv² + 1/2Iω² (7)

Substituting equations (1), (3), (4), (5) and (7) into equation (6) gives;

mgh = 1/2mv² + 1/2I(v/r²)² + 1/2m(v/r)²gh

= 1/2mv² + 1/2I(v²/r²) + 1/2mv²/r²gh

= 3/2mv² + 1/2I(v²/r²)gh

= (3/2)m(v² + (I/mr²))v²

= (2gh)/(3 + (I/mr²))

Substituting the values of the given variables, we have;

v² = (2*9.81*5.28)/(3 + (2/5)*127*(1.53)²)

v = 6.52 m/s

ω = v/rω = 6.52/1.53

ω = 4.26 rad/s

Therefore, the angular speed of the sphere at the bottom of the inclined plane is 4.26 rad/s (approx).

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a
camera is equipped with a lens with a focal length of 34cm. when an
object 1.1m (110cm) away is being photographed, what is the
magnification?

Answers

The magnification of the object being photographed is approximately -0.2361.

The magnification (m) of an object being photographed by a camera with a lens can be calculated using the formula:

m = -v/u

Where:

m is the magnification

v is the image distance

u is the object distance

Given:

Focal length of the lens (f) = 34 cm

Object distance (u) = 110 cm

To find the image distance (v), we can use the lens formula:

1/f = 1/v - 1/u

Substituting the known values:

1/34 = 1/v - 1/110

Simplifying the equation:

1/v = 1/34 + 1/110

Calculating this expression:

1/v = (110 + 34) / (34 × 110)

1/v = 144 / 3740

v = 3740 / 144

v ≈ 25.9722 cm

Now, we can calculate the magnification using the image distance and object distance:

m = -v/u

m = -25.9722 cm / 110 cm

m ≈ -0.2361

Therefore, the magnification of the object being photographed is approximately -0.2361.

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Date First Name, 3. If a 500-ml glass beaker is filled to the brim with water at a temperature of 23 °C, how much will overflow when its temperature reaches 30 °C7 [10 points] Given: To Find: Solution: (5 points total) Ans (2 points) = Did the water overflow? (3 points total) Yes/No (1 points) Why? (2 points)

Answers

Answer:

The amount of water that will overflow when the temperature increases from 23 °C to 30 °C is approximately 0.119 ml.

Explanation:

o calculate the amount of water that will overflow from the glass beaker when its temperature increases from 23 °C to 30 °C, we need to consider the thermal expansion of water.

Given:

Initial volume of water (V1): 500 ml

Initial temperature (T1): 23 °C

Final temperature (T2): 30 °C

To Find:

Amount of water that will overflow

Solution:

Convert the initial volume from milliliters (ml) to cubic centimeters (cm³) since they are equivalent: 1 ml = 1 cm³.

V1 = 500 cm³

Calculate the change in volume (∆V) due to thermal expansion using the formula:

∆V = V1 * β * ∆T

Where:

β is the coefficient of volumetric expansion of water, which is approximately 0.00034 (1/°C).

∆T is the change in temperature, which is T2 - T1.

∆V = 500 cm³ * 0.00034 (1/°C) * (30 °C - 23 °C)

∆V = 500 cm³ * 0.00034 * 7

∆V ≈ 0.119 cm³

Since 1 cm³ is equivalent to 1 ml, the amount of water that will overflow is approximately 0.119 ml.

Answer:

The amount of water that will overflow when the temperature increases from 23 °C to 30 °C is approximately 0.119 ml.

Did the water overflow?

Yes

Why?

The water overflows because its volume increases as the temperature rises, causing it to expand and exceed the capacity of the beaker.

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Suppose the position vector for a particle is given as a function of time by F(t)= x(+ y(t), with x(t)-at + b and y(t)- ct+d, where a 1.10 m/s, b=1:50 m, c= 0.130 m/s², and d = 1.20 m. (a) Calculate the average velocity during the time interval from t-1.85 s to t4.05 s. VM _______________ m/s (b) Determine the velocity at t 1.85 V ___________ m/s Determine the speed at t-1.85 s. V ___________ m/s

Answers

The average velocity during the time interval from t = 1.85 s to t = 4.05 s is approximately 1.60 m/s. The velocity at t = 1.85 s is 1.10 m/s. The speed at t = 1.85 s is 1.10 m/s.

(a) To find the average velocity between t = 1.85 s and t = 4.05 s, we calculate the change in position (displacement) during that time interval and divide it by the duration of the interval.

The displacement during the time interval from t = 1.85 s to t = 4.05 s can be determined by subtracting the initial position at t = 1.85 s from the final position at t = 4.05 s.

Let's calculate the average velocity:

Initial position at t = 1.85 s:

x(1.85) = a(1.85) + b = (1.10 m/s)(1.85 s) + 1.50 m = 3.03 m

Final position at t = 4.05 s:

x(4.05) = a(4.05) + b = (1.10 m/s)(4.05 s) + 1.50 m = 6.555 m

Displacement = Final position - Initial position = 6.555 m - 3.03 m = 3.525 m

Time interval = t_final - t_initial = 4.05 s - 1.85 s = 2.20 s

Average velocity = Displacement / Time interval = 3.525 m / 2.20 s ≈ 1.60 m/s

Hence, the average velocity during the time interval from t = 1.85 s to t = 4.05 s is approximately 1.60 m/s.

(b) To determine the velocity at t = 1.85 s, we can differentiate the position function with respect to time:

x'(t) = a

Substituting the given value of a, we find:

x'(1.85) = 1.10 m/s

Therefore, the velocity at t = 1.85 s is 1.10 m/s.

(c) To determine the speed at t = 1.85 s, we take the absolute value of the velocity since speed is the magnitude of velocity:

The speed, which is the magnitude of velocity, is equal to 1.10 m/s.

Therefore, the speed at t = 1.85 s is 1.10 m/s.

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inembers have average mas5tts of 71 kg and exert average forces of 1360 N horizontally. (a) What is the acceleration of the two teams? - m/s 2
(in the direction the heavy team is puining) (b) What is the tecsionin the sectien of rope between the teamw

Answers

The acceleration  is 19.15 m/s2. F = ma. 1360/ 71 = 19.15 m/s2.

Thus, acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is the first derivative of velocity with respect to time or the second derivative of position with respect to time.

If an object's velocity changes, it is said to have been accelerated. An object's velocity can alter depending on whether it moves faster or slower or in a different direction.

A falling apple, the moon orbiting the earth, and a car stopped at a stop sign are a few instances of acceleration. Through these illustrations, we can see that acceleration happens whenever a moving object changes its direction or speed, or both.

Thus, The acceleration  is 19.15 m/s2. F = ma. 1360/ 71 = 19.15 m/s2.

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a) Acceleration (a) of the two teams can be calculated as follows: a = F/ma = (1360 N) / (150 kg)a = 9.07 m/s²

b) The tension in the section of rope between the teams is 680 N.

(a) Acceleration of the two teams

The acceleration of the two teams can be calculated as follows: F = m a

Where, F = force exerted by the teams = 1360 Nm = mass of the two teams = 150 kg

Therefore, acceleration (a) of the two teams can be calculated as follows:

a = F/ma = (1360 N) / (150 kg)a = 9.07 m/s²

(b) Tension in the section of rope between the teams, The tension in the section of rope between the teams can be calculated as follows: F = T + T Where, F = force exerted by the teams on the rope = 1360 N (as calculated above)T = tension in the section of rope between the teams

Therefore, the equation can be written as follows: F = 2 TT = (F/2)T = (1360 N/2)T = 680 N

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An inferior good is a product for which ... a. demand increases slowly as income increase. b. the income elasticity is negative. c. the price elasticity of demand is positive. d. quality is below average. 7. Adidas, Petronas, Mc Donald and Focus Point are the examples of business. A. start up B. franchising C. small D. independent I don't understand this Please I need an explanation How can you differentiate a cercopithecine monkey from a colobine monkey?Group of answer choicesa.Cercopithecine monkeys tend to be mostly arboreal and are generally small in body size.b.Colobine monkeys have complex stomachs because they mostly eat leaves.c.Cercopithecine monkeys consume mostly seeds, which results in their larger body size.d.Colobine monkeys inhabit a variety of different habitats and consume a wide variety of foods. When compared to a larger alveolus, a smaller alveolus will have:1) higher rise of collapse due to surface tension2) higher surface tension at gas-fluid interface3) smaller contribution to dead space How should PAC leverage automation for its consumerfulfillment processes (Bond valuation) Flora Co.s bonds, maturing in 7 years, pay 4 percent interest on a $1,000 face value. However, interest is paid semiannually. If your required rate of return is 5 percent, what is the value of the bond? How would your answer change if the interest were paid annually? 15. Describe the brain structures involved in controlling movement. Start with where you decide to move and include both conscious and unconscious levels of control. Which part of the brain sends the signals that directly controls the muscles? If the Sun suddenly tumed off, we would not know it until its light stopped coming. How long would that be, given that the sun is 1.496 x1011 away? evaluate b-2a-c for a =-3, b=9 and c=-6 300 words no plagiarisma) Considering the production process, there is the short run and the long run. How are they distinguished and does the terminologies have any relationship with time? Explain in your own words. (25 ma 1 ........... gives the details and processes that would be done for the proposed study to happen.A. analysis and discussionB. methodologyC. background of the studyD. review of related literature and studies2. Research and Development become the index of development of the country. Which of the following reasons are true with regards to this statement?A. R&D targets human developmentB. R&D can enhance people's standard of living in the countryC. R&D reflects the actual economic and social conditions being prevailed in the country Intrinsic factor secreted by parietal cells of the stomach is required forA.absorption of vitamin B12.B.stimulation of mixing waves.C.activation of pepsin.D.complete gastric emptying.E.buffering of HCl. Vectors A and B are given by: A = 60.09i + 91.16j B = 81.57i+ 63.92j Find the scalar product AB. Given just the graph what 3 steps are required to write the equation of a line? what about the monster causes frankenstein to view his creation with such horror? use language from the text to support your answer. A nurse delegates a nursing assistant to apply wrist restraints on a confused client. The nurse notices the nursing assistant padded the wrist restraints with sheep skin and secured the straps to the bed frame with a double knot. The nurse should do which of the following The nurse teaches a newly diagnosed diabetic patient to treat an episode of hypoglycemia by:Group of answer choicesA. 3 pieces hard candy.B. drinking a 12-oz. soda.C. eating an apple.D. sucking slowly on a hard candy. Select one correct answer from the available options in the below parts. a) [3 points] You shine monochromatic light of wavelength A through a narrow slit of width b A and onto a screen that is very far away from the slit. What do you observe on the screen? OOne bright band OTwo bright fringes and three dark fringes OA series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes OA series of bright and dark fringes that are of equal widths b) [3 points] What does it mean for two light waves to be in phase ? OThe two waves have the same wavelength and frequency OThe two waves have the same amplitude OThe two waves reach their maximum value at the same time and their minimum value at the same time OThe two waves propagate in the same direction c) [3 points] You shine monochromatic light of wavelength through a narrow slit of width b> > and onto a screen that is very far away from the slit. What do you observe on the screen? OOne bright band OA series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes OA series of bright and dark fringes that are of equal widths OTwo bright fringes and three dark fringes d) [3 points] Monochromatic light is directed through two narrow parallel slits. There is a viewing screen away the slits that is used to observe the interference pattern. If you submerged the entire apparatus in water, how is the new interference pattern different from the original one? OThe bright and dark fringes are closer together OThe bright and dark fringes are farther apart OThe pattern does not change e) [3 points] Light propagating in a material 1 with index of refraction n is incident on a new material 2 with index of refraction n A 0.474 m long wire carrying 6.39 A of current is parallel to a second wire carrying 3.88 A of current in the same direction. If the magnetic force between the wires is 5.72 x 10-5 N, how far apart are they?