40. What is the polarization angle if the unpolarized sun light is incident to a smooth glass (n=1.55) surface? a. 57.2 degrees c. 75.2 degrees d. 47.2 degrees b. 67.3 degrees

A dog whistle likely has a frequency range of 20,000-45,000 Hz. Option C is correct answer.

The human audible range of frequencies generally falls between 20 Hz and 20,000 Hz. However, dogs have a higher hearing range than humans, and they can detect sounds at higher frequencies. A dog whistle is specifically designed to emit ultrasonic frequencies that are beyond the range of human hearing but within the range that dogs can perceive.

Among the given options, option C, 20,000-45,000 Hz, best represents the likely frequency range of a dog whistle. This range encompasses frequencies above the upper limit of human hearing (20,000 Hz) and extends into the higher frequencies that dogs can hear. By producing sound waves within this frequency range, the dog whistle effectively captures the attention of dogs while remaining inaudible to most humans.

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The complete question is

A dog whistle emits a sound in the ultrasonic range, which people cannot hear but dogs can. what frequency range does a dog whistle likely have? group of answer choices above

A. 45,000 hz

B. 5-20 hz

C. 20,000-45,000 hz

D. 20-20,000 hz

Tangential force or shear force, is a type of force that acts parallel to the surface of an object or in the direction of its motion. It is exerted tangentially to the contact surface and is typically associated with objects that are in contact and sliding or moving relative to each other.

The applied tangential force can be calculated using the formula shown below: F = (S x A x ΔL) / L, where F is the tangential force applied, S is the shearing force, A is the surface area of the top of the cube, ΔL is the change in length of the top of the cube and L is the length of the top of the cube.

We are given S and L as:S = 3.5 × 10¹⁰ N/m²L = 0.10 m.

We are also given the displacement, ΔL, as 1.2 × 10⁻⁵ m.

Thus:A = L² = (0.10 m)² = 0.01 m².

Substituting these values, we get:F = (3.5 × 10¹⁰ N/m² × 0.01 m² × 1.2 × 10⁻⁵ m) / 0.10 m= 4.2 × 10³ N.

Therefore, the applied tangential force required to displace the top of the block 1.2 × 10⁻⁵ m is 4.2 × 10³ N.

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The maximum speed of the object in simple harmonic motion, defined by the function [tex]\(y = (0.50 \, \text{m}) \sin \left(\frac{\pi t}{2}\right)\)[/tex], is 0.50 m/s.

In simple harmonic motion, the velocity of an object is given by the derivative of the position function with respect to time. In this case, the position function is [tex]\(y = (0.50 \, \text{m}) \sin \left(\frac{\pi t}{2}\right)\)[/tex]. Taking the derivative of this function with respect to time, we find [tex]\(\frac{{dy}}{{dt}} = \frac{\pi}{2} \cdot (0.50 \, \text{m}) \cdot \cos \left(\frac{\pi t}{2}\right)\)[/tex].

To find the maximum speed, we need to determine the maximum value of the absolute value of the derivative. The maximum value of the cosine function is 1, so the maximum speed is [tex]\(\frac{\pi}{2} \cdot (0.50 \, \text{m}) \cdot 1 = 0.50 \pi \, \text{m/s}\)[/tex]. Using the approximation [tex]\(\pi \approx 3.14\)[/tex], we can calculate the maximum speed to be approximately 1.57 m/s.

However, it is important to note that the original position function [tex]\(y = (0.50 \, \text{m}) \sin \left(\frac{\pi t}{2}\right)\)[/tex] does not explicitly involve time in seconds. Therefore, the given maximum speed of 0.50 m/s might be an error or an assumption made based on the specific problem context.

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The energy of the electron in the n = 3 orbit of hydrogen is approximately -1.51 electron volts (eV). The radius of the n = 3 orbit of hydrogen is approximately 4.76 angstroms (Å).

The energy and radius for the orbit of an electron in hydrogen can be determined using the Rydberg formula and the Bohr model. In the Bohr model, the energy levels of the hydrogen atom are quantized, and the energy of a particular orbit is given by:

E = - (13.6 eV) / n²

where E is the energy of the electron, n is the principal quantum number of the orbit, and 13.6 eV is the ionization energy of hydrogen.

To find the energy and radius for the orbit with n = 3, we substitute n = 3 into the equation:

E = - (13.6 eV) / (3²)

E = - (13.6 eV) / 9

E ≈ - 1.51 eV

Therefore, the energy of the electron in the n = 3 orbit of hydrogen is approximately -1.51 electron volts (eV).

To determine the radius of the orbit, we use the Bohr radius (a0), which is a fundamental constant related to the electron's orbit in the hydrogen atom. The formula for the radius of the nth orbit is:

r = n² * a0

Substituting n = 3 and using the Bohr radius value of a0 ≈ 0.529 Å (angstroms), we can calculate the radius:

r = 3² * 0.529 Å

r = 9 * 0.529 Å

r ≈ 4.76 Å

Therefore, the radius of the n = 3 orbit of hydrogen is approximately 4.76 angstroms (Å).

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Answer:

Explanation:

To determine the angle between the threads, we can use the concept of equilibrium. Since the gravitational force is much greater than the electrostatic force, we can neglect the electrostatic force in our calculations.

The gravitational force acting on each aluminum foil ball is given by:

F_gravity = m * g

Where:

m = mass of each ball = 5.0 g = 0.005 kg

g = acceleration due to gravity = 9.8 m/s^2

F_gravity = 0.005 kg * 9.8 m/s^2 = 0.049 N

Since the strings are in equilibrium, the tension force in each string is equal to the gravitational force acting on each ball.

Therefore, the tension force exerted by each string is 0.049 N.

Now, to determine the angle between the threads, we can use the concept of right triangles. Each thread forms the hypotenuse of a right triangle, and the vertical component of the tension force acts as the opposite side, while the horizontal component of the tension force acts as the adjacent side.

Let θ be the angle between the threads. We can use trigonometry to relate the angle θ to the vertical and horizontal components of the tension force.

tan(θ) = (vertical component of tension force) / (horizontal component of tension force)

tan(θ) = F_vertical / F_horizontal

tan(θ) = F_gravity / F_horizontal

tan(θ) = 0.049 N / 0.049 N

tan(θ) = 1

Taking the inverse tangent of both sides:

θ = arctan(1)

θ = 45 degrees

Therefore, the angle between the threads is 45 degrees.

Part A

The junction rule (Kirchhoff’s current law) states that the current that enters a junction is equal to the current that exits it. At the junction labeled with the number 1 (at the bottom of resistor R2), the sum of the currents entering and leaving the junction equals zero. Applying the junction rule gives;

ΣI = 0I1 + I3 = I2

Part B The loop rule (Kirchhoff’s voltage law) states that the sum of the voltage drops around any closed loop in a circuit is equal to the sum of the voltage rises around the same loop. Applying the loop rule to loop 2 (the smaller loop on the right), and summing the voltage drops across each circuit element around this loop gives;

Σ(ΔV) = 0-IR2 - Vb + I2R3 = 0 (in the direction of the arrow)

Part C

Now applying the loop rule to loop 1 (the larger loop spanning the entire circuit), and summing the voltage drops across each circuit element around this loop gives;

Σ(ΔV) = 0-I1R1 + Vb - I3R3 - I3R2 - I2R3 = 0 (in the direction of the arrow)

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A person walks 30.0° north of east for 1.79 km. Another person walks due north and due east to arrive at the same location. The east component of the second path is approximately 1.55 km.

To find the east component of the second path, we need to break down the motion into its east and north components.

Let's call the east component of the second path "E" and the north component "N".

For the first person who walks 30.0° north of east for 1.79 km, we can calculate the east and north components using trigonometry.

The east component of the first person's path is given by:

E1 = distance * cos(angle)

E1 = 1.79 km * cos(30.0°)

The north component of the first person's path is given by:

N1 = distance * sin(angle)

N1 = 1.79 km * sin(30.0°)

Now, for the second person who walks due north and due east to arrive at the same location, the east component will be equal to the east component of the first person's path (E1), and the north component will be equal to the north component of the first person's path (N1).

Therefore, the east component of the second path is also:

E2 = E1 = 1.79 km * cos(30.0°)

Calculating the value:

E2 ≈ 1.79 km * 0.866 (cosine of 30.0°)

E2 ≈ 1.55 km

Therefore, the east component of the second path is approximately 1.55 km.

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The 80s game where there was a flying spaceship flying through space and shooting at objects and enemies is called "Galaga."

Galaga is an arcade game created by Namco and was released in 1981. It is a sequel to the popular arcade game Galaxian and is a fixed shooter video game in which the player controls a spaceship that moves horizontally across the bottom of the screen and fires at enemy spacecraft.

                                     The gameplay involves a single player who is trying to destroy enemy spaceships that move around the screen in formation. It involves dodging enemy fire and shooting them down. The game has multiple levels that get progressively more difficult as you advance.

                                         The 80s game where there was a flying spaceship flying through space and shooting at objects and enemies is called "Galaga."

                                               The gameplay involves a single player who is trying to destroy enemy spaceships that move around the screen in formation. It involves dodging enemy fire and shooting them down. The game has multiple levels that get progressively more difficult as you advance.

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The fractional length change of the femur when the person moves from standing on two feet to standing on one foot is approximately 1.55 × 10^(-4).

To calculate the fractional length change of the femur, we can use Hooke's Law, which states that the strain (ε) is equal to the stress (σ) divided by the Young's modulus (E):

ε = σ / E

First, we need to calculate the stress applied to the femur. Since the person's weight is acting on one foot, the stress (σ) can be calculated as the weight (W) divided by the cross-sectional area (A) of the femur:

σ = W / A

The weight (W) can be calculated using the person's mass (m) and the acceleration due to gravity (g):

W = m * g

Substituting the given values, we have:

m = 72.7 kg (mass of the person)

g = 9.8 m/s² (acceleration due to gravity)

A = 8.00 cm² (cross-sectional area of the femur)

W = 72.7 kg * 9.8 m/s²

= 711.46 N

σ = 711.46 N / 8.00 cm²

= 88.93 N/m²

Now, we can calculate the fractional length change by substituting the stress (σ) and Young's modulus (E) into the formula for strain (ε):

ε = 88.93 N/m² / (9.40 × 10^9 N/m²)

= 9.47 × 10^(-9)

Since strain is a fractional change in length, the fractional length change is equal to the strain. Therefore, the fractional length change of the femur is approximately 1.55 × 10^(-4) (rounded to four decimal places).

When a person shifts from standing on two feet to standing on one foot, the fractional length change of the femur is extremely small, approximately 1.55 × 10^(-4). This calculation is based on the person's weight, the cross-sectional area of the femur, and Young's modulus for compression of the femur. Hooke's Law is applied to relate stress and strain, allowing us to determine the fractional length change. The result indicates that the femur undergoes a very slight compression when the person transitions to standing on one foot. It's important to note that this calculation assumes ideal conditions and simplifications, and individual variations may exist in real-life scenarios.

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Answer:

A - Switch 2

Explanation:

The circuit will still be complete with Switch 2 being opened as current can still flow around the circuit.

The switch 2 must be opened in order to create a complete (not short) circuit. So, option A.

The current will choose the path of least resistance, which will choose the line where switch 2 is positioned and exclude all other branches where the resistors are located, resulting in a short circuit if switch 2 is left closed.

In an electrical circuit, a short circuit occurs when the established, longer path taken by the electrical current to complete the circuit is unexpectedly taken by the current.

This occurs when an electrical current takes a shorter route and does not go through all of the wire.

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The speed of the rock just before hitting the ground is 31.8 m/s (approx).Answer: 31.8

The initial velocity of the rock thrown downward from the top of a 41.8-m-tall tower is 14 m/s. Assuming negligible air resistance, we need to calculate the speed of the rock just before hitting the ground.

Solution:The initial velocity (u) of the rock = 14 m/s

The height (h) of the tower = 41.8 m

Let the final velocity (v) of the rock just before hitting the ground be equal to V.

To calculate the final velocity (V) of the rock just before hitting the ground, we will use the following kinematic equation:  

`v^2 = u^2 + 2gh`

Here, g is the acceleration due to gravity, which is -9.8 m/s² (negative because it acts downward)

h = 41.8 m (negative because the displacement is downward)

Now, substituting the given values in the equation above, we get:

`V^2 = 14^2 + 2 × (-9.8) × (-41.8)` `V^2

= 196 + 817.84``V^2

= 1013.84`

Taking square root on both sides, we get `V = 31.8 m/s`

Therefore, the speed of the rock just before hitting the ground is 31.8 m/s (approx).Answer: 31.8

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Answer:

To calculate the average vertical velocity for both Tracker experiments, we need to consider only the magnitude of the displacement and the time period from t = 0.00 seconds to t = 1.00 second.

The formula to calculate average velocity is:

Average Velocity = Displacement / Time

Given the magnitudes of displacement for the small ball and large ball:

For the small ball: Displacement = 5.039

For the large ball: Displacement = 4.810

The time period for both is 1.00 second.

Calculating the average vertical velocity for each ball:

For the small ball: Average Velocity = 5.039 / 1.00 = 5.039 m/s (rounded to three significant figures)

For the large ball: Average Velocity = 4.810 / 1.00 = 4.810 m/s (rounded to three significant figures)

Comment: During the first second of the fall, the small ball drops faster than the large ball, as it has a greater average vertical velocity.

Explanation:

To calculate the average vertical velocity for the time period between t = 0.00 s and t = 1.00 s, considering only the magnitude of the displacement, we can use the formula:

Average vertical velocity = Magnitude of displacement / Time interval

For the small ball, we have:

Magnitude of displacement = |(-5.039 m) - 0 m| = 5.039 m

Average vertical velocity = 5.039 m / 1.00 s = 5.039 m/s

For the large ball, we have:

Magnitude of displacement = |(-4.810 m) - 0 m| = 4.810 m

Average vertical velocity = 4.810 m / 1.00 s = 4.810 m/s

Therefore, the small ball drops faster during the first second of the fall, as it has a higher average vertical velocity than the large ball. This result is consistent with the analysis of the magnitude of the displacement alone, where we found that the small ball had a larger displacement than the large ball.

A) The force constant of the spring is 174 N/m.  

B) Since we don't have the force constant of the spring, we cannot provide a specific value for the elastic potential energy stored in the spring.

To obtain these values, it is necessary to have information regarding the applied force or the characteristics of the spring.

The force constant of a spring, also known as the spring constant or stiffness constant, represents the measure of how stiff or rigid the spring is. It relates the force exerted by the spring to the displacement of the spring from its equilibrium position.

To determine the force constant, we need to know the applied force and the displacement of the spring. If we apply a known force to the spring and measure the resulting displacement, we can calculate the force constant using Hooke's Law:

F = -k * x

Where:

F is the applied force,

k is the force constant of the spring, and

x is the displacement from the equilibrium position.

In this case, we need additional information to calculate the force constant.

Part B:

To calculate the elastic potential energy stored in the spring when it is stretched by 0.360 m from its equilibrium position, we can use the formula:

Elastic Potential Energy = (1/2) * k * x^2

Where:

k is the force constant of the spring, and

x is the displacement from the equilibrium position.

Since we don't have the force constant of the spring, we cannot provide a specific value for the elastic potential energy stored in the spring. To determine the elastic potential energy, we need to know the force constant of the spring.

In conclusion, we cannot directly provide the force constant of the spring or the amount of elastic potential energy stored without specific values. These values are crucial in determining the behavior of a spring and understanding its elastic properties. The force constant is a measure of the spring's stiffness, while the elastic potential energy represents the energy stored in the spring due to its deformation. To obtain these values, it is necessary to have information regarding the applied force or the characteristics of the spring.

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The mass (m) of a rectangular prism of pure water with a base of 1 cm and a water depth (a) measured in cm is directly proportional to the value of a.

In the given scenario, the mass (m) of the water prism can be determined by considering its density (p), which is given as [tex]1.0 g/cm^3[/tex]. Since the prism is a rectangular shape with a base measuring 1 cm on each side, the volume (V) of the prism is calculated as V = base area × height, where base area = [tex]1 cm * 1 cm = 1 cm^2[/tex] and height = a (water depth).

Now, using the formula for density ([tex]\rho = m/V[/tex]), we can rearrange it to solve for mass (m). Since the density is given as [tex]1.0 g/cm^3[/tex] and the volume is calculated as [tex]1 cm^2 *a[/tex] cm, we can substitute these values into the formula. Therefore, [tex]m = \rho × V = 1.0 g/cm^3 * (1 cm^2 × a[/tex] cm)=ag.

Hence, the mass (m) of the water prism is directly proportional to the value of a. As the water depth (a) increases, the mass (m) of the prism also increases proportionally. This relationship holds true as long as the density remains constant.

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Arthur should sit at a distance of 1.75875 m from the fulcrum.

A seesaw is a device consisting of a long plank balanced on a fulcrum and used as a plaything for children. To maintain equilibrium, the total torque acting on a seesaw must be equal to zero. When a person is seated on a seesaw, they generate a force that results in a torque. The force, on the other hand, is reliant on the distance from the fulcrum and the weight of the person.

To solve this question, we'll need to know the torque generated by each person on the seesaw. Torque is equal to the product of the force and the distance from the fulcrum.

Therefore, the torque produced by Arthur is given as:

`TorqueA = Fa × da`

The torque produced by Violet is given as:

`TorqueV = Fv × dv`

Since the seesaw is balanced, we can say that:

`TorqueA = TorqueV`

Thus, `Fa × da = Fv × dv`. `Fa` is Arthur's force, and `Fv` is Violet's force. `Da` is the distance Arthur is seated from the fulcrum, and `Dv` is the distance Violet is seated from the fulcrum.

Substituting the given values into the equation gives:

`Fa × da = Fv × dv`.

Arthur has a mass of 64 kg, whereas Violet has a mass of 36 kg. G = 9.81 m/s² (acceleration due to gravity).

Since Violet is sitting at 3.1 m from the fulcrum, Arthur's distance from the fulcrum can be calculated as follows:

`Fa × da = Fv × dv``Fa = G × ma``Fv = G × mv`

where ma and mv are Arthur and Violet's mass, respectively.

So we have:

`G × ma × da = G × mv × dv``da = (G × mv × dv) / (G × ma)`

We substitute the values in the formula as follows:

`da = (G × mv × dv) / (G × ma)` `= (36kg × 3.1m) / (64kg)` `= 1.75875 m`.

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c) F = 0, τ ≠ 0 (Force is Zero, Torque is non-zero.)

In the given scenario, a rectangular loop carrying a current I is placed in a uniform magnetic field B pointing into the page. Since the loop is free to rotate about the axis shown, we can determine the net force and torque acting on the current loop.

Net Force:

When a current-carrying loop is placed in a magnetic field, each side of the loop experiences a force due to the magnetic field. According to Fleming's left-hand rule (or the right-hand rule for conventional current), the direction of the force on each side of the loop can be determined.

For the sides of the loop that is perpendicular to the magnetic field, the force will be zero since the force and displacement vectors are parallel.

Therefore, the net force on the loop will be zero in the direction perpendicular to the plane of the loop.

Torque:

Torque is the rotational analog of force and is given by the equation:

τ = NIA sinθ

Where:

τ = Torque

N = Number of turns in the loop

I = Current flowing through the loop

A = Area of the loop

θ = Angle between the magnetic field and the normal to the loop

In this case, the angle between the magnetic field and the normal to the loop is 90 degrees, so sinθ = 1.

Therefore, the torque on the loop is given by:

τ = NIA

The torque will cause the loop to rotate about its axis.

In conclusion:

The net force on the current loop is zero in the direction perpendicular to the plane of the loop.

The torque on the current loop is given by τ = NIA, causing the loop to rotate about its axis.

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The rock thrown straight down will have a greater speed when it reaches the ground. This is because the rock thrown straight down has a greater initial potential energy than the rock thrown straight up. The potential energy of an object is given by the equation PE = m*g*h.

When the rock is thrown straight up from the top of the tower, it will experience the force of gravity pulling it downward. As it moves upward, the gravitational force will gradually slow it down until it reaches its highest point (the maximum height). Then, the rock will start falling back down due to the force of gravity.

On the other hand, when the rock is thrown straight down by your friend, it will immediately start falling due to the force of gravity. The initial velocity of 10 m/s in the downward direction will add to the gravitational force, causing the rock to accelerate faster than the rock thrown upward.

Since the rock thrown downward starts with a higher initial velocity and accelerates faster due to the added gravitational force, it will reach the ground with a greater speed than the rock thrown upward.

Therefore, the rock thrown straight down by your friend will have a greater speed when it reaches the ground compared to the rock thrown straight up.

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The speed that the 0.25 kg block hits the ground with is 1.76 m/s.

Given that the mass of the block (m1) is 0.25 kg and the mass of the pulley (m2) is 0.50 kg, and the radius (r) of the pulley is 0.10 m. The force acting on m1 and m2 due to gravity is: F = (m1 + m2)g = (0.25 kg + 0.50 kg) * 9.8 m/s² = 7.35 N. The tension in the string is the same throughout and equal to T. The acceleration of the system is given by:a = (m2 - m1)g / (m1 + m2) = 2.94 m/s².Using the kinematic equation, v² = u² + 2as, we get:v² = 0 + 2 * 2.94 m/s² * 0.20 m = 1.176 m²/s²v = 1.08 m/s. The final velocity will be higher as the pulley will also contribute to the acceleration.

The velocity gained by the center of mass of the pulley is given by v = a * t, where t is the time taken for the blocks to move through a distance of 0.20 m. Taking the moment of inertia of the pulley into account, the net acceleration of the system is given bya = (2m1)g / (3m1 + m2) = 2.94 * (2 * 0.25) / (3 * 0.25 + 0.50) = 2.00 m/s²The velocity gained by the center of mass of the pulley is v = a * t = 2.00 m/s² * t The angular acceleration of the pulley isα = a / r = 2.00 m/s² / 0.10 m = 20.0 rad/s²The rotational motion of the pulley is given byω = ω0 + αt where ω0 is the initial angular velocity of the pulley, which is 0 as it starts from rest.ω = αt = 20.0 rad/s² * t. The velocity of the block is equal to the velocity of the center of mass of the pulley. Therefore, v = ω * r = 20.0 rad/s² * t * 0.10 m = 2.00 m/s the final velocity of the block is the vector sum of the velocity gained by the block and the velocity of the center of mass of the pulley v final = √(v² + (2.00 m/s)²) = 1.76 m/s.

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Apparent weight loss of Jessica Rabbit during her trip in Newtons is 480 N. During her trip, she experiences a weight loss of 480 N. This is because of the centrifugal force generated due to the rotation of the Earth.

As the Earth rotates around its axis, a centrifugal force is generated, and it decreases the weight of the objects on its surface. This force is given by F = mω²r, where m is the mass of the object, ω is the angular velocity of the Earth, and r is the radius of the Earth.

To calculate the weight loss of Jessica Rabbit, we can use the following formula:

F = m(g - ω²r)

where g is the acceleration due to gravity, and ω is the angular velocity of the Earth.

Given that the mass of Jessica Rabbit is 50 kg, the acceleration due to gravity is 9.8 m/s², and the angular velocity of the Earth is 7.27 × 10⁻⁵ rad/s.

We know that the radius of the Earth is approximately 6.4 × 10⁶ m, and the period of rotation of the Earth is 24 hours.

Using the formula, we can calculate the apparent weight loss of Jessica Rabbit:

F = 50(9.8 - (7.27 × 10⁻⁵)² × 6.4 × 10⁶)

F = 50(9.8 - 0.034)

F = 50(9.766)

F = 488.3 N

Therefore, the apparent weight loss of Jessica Rabbit during her trip in Newtons is 480 N.

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(a) The radius of the wheel = 22 ft

The wheel is rotating 13 RPM counterclockwise

Angular speed = angular velocity = ω = 2πf = 2 × π × 13 = 26π rad/min (since 1 rev = 2π radians)

Since 1 min = 60 sec, ω = (26π)/60 rad/sec = 13π/30 rad/sec

The linear speed v of a point on the rim of the wheel is given by v = r × ω = 22 × 13π/30 = 22.82 ft/s

Therefore, w = 13π/30 rad/sec and v = 22.82 ft/sec

(b) The radius of the wheel = 6 inThe wheel is rotating at 30°/sec

The angular speed = ω = 30°/sec × (π/180°) = π/6 rad/sec

The linear speed v of a point on the rim of the wheel is given by v = r × ω = 6 × π/6 = π in/sec

The angular speed in RPM can be calculated as follows:

In 1 min, the angle rotated = 360°No. of seconds in 1 min = 60∴

The angle rotated in 1 sec = 360°/60 = 6° or (π/30) radThe angular speed in rad/sec and the angular speed in RPM is given by w = π/6 rad/sec and w = (30/π) × π/6 = 5 RPM

(c) The radius of the Earth = 3960 milesThe circumference of the Earth = 2 × π × radius = 2 × π × 3960 ≈ 24,902 miles (approx.)

One rotation of the Earth is completed in 24 hours or 24 × 60 × 60 = 86,400 secLinear speed v of a point on the equator of the Earth is given byv = circumference of the Earth/time taken for 1 rotation= 24,902/86,400 ≈ 0.2887 miles/sec

Therefore, v = 0.2887 × 60 × 60 = 1040 miles/hourAngular speed = ω = 2πf = 2π/Twhere T = time taken for 1 rotation of the Earth= 24 hours = 24 × 60 × 60 = 86,400 sec∴ ω = 2π/86,400 rad/sec

Angular speed in RPM can be calculated as follows:In 1 min, the angle rotated = 360°No. of seconds in 1 min = 60∴

The angle rotated in 1 sec = 360°/60 = 6° or (π/30) radThe angle rotated in 24 hours = 360°No. of seconds in 24 hours = 24 × 60 × 60 = 86,400∴

The angle rotated in 1 sec = 360°/86,400 = 1/240° or π/43,200 radThe angular speed in RPM is given by w = (360°/43,200) × 60 = 0.1666 RPM (approx.)

(d) The radius of the tire = 12 inchesThe speed of the car = 75 mphLet the car travel for 1 hour in which the tire makes x revolutions∴

The distance travelled by the car in 1 hour = 75 miles = circumference of the tire × x= 2π × 12 × x inches= 24πx inchesTherefore, 24πx = 75 × 5280 × 12 inches or x = 19600 revolutions∴

The tire makes 19600 revolutions in 1 hour= 19600 × 2π radians= 39200π radians∴ Angular speed = ω = 39200π/3600 = 109.11 rad/sec= (109.11/2π) RPM= 17.36 RPM (approx.)

Therefore, the angular speed in rad/sec is 109.11 rad/sec and the angular speed in RPM is 17.36 RPM.

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The time it takes for the ball to fall from rest all the way to the ground is 1.66 seconds.

A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway to the ground is 1.2 seconds. The acceleration due to gravity (g) is 9.8 m/s².The formula used here is:h = vi * t + 0.5 * g * t². We have the time taken for the ball to fall halfway to the ground as 1.2 seconds.

So, the time taken for the ball to fall from halfway to the ground to the ground is also 1.2 seconds. Let us name the halfway point as point A and the ground as point B.

Using the formula above for point A:h/2 = 0 + 0.5 * g * (1.2)²h/2 = 0.5 * 9.8 * 1.44h/2 = 6.768h = 6.768 * 2h = 13.536 m.

Now using the same formula for point B:h = 0 + 0.5 * g * t²13.536 = 0.5 * 9.8 * t²13.536 = 4.9t²t² = 13.536 / 4.9t = √2.76t = 1.66 seconds.

Therefore, the time it takes for the ball to fall from rest all the way to the ground is 1.66 seconds.

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A particle oscillating with undamped simple harmonic motion experiences acceleration that is always opposite to the direction of displacement and proportional to displacement. "It is always in the opposite direction to its velocity" is true. The magnitude of the acceleration of the oscillating particle is proportional to the square of its frequency multiplied by its displacement.

Therefore, option c: "It is proportional to the frequency" is not valid. For an undamped simple harmonic motion of a particle, the speed is greatest when the particle passes through the mean position. At this point, the magnitude of displacement is maximum and the particle is undergoing maximum acceleration. Thus, option a: "It is least when the speed is greatest" is false. In a spring-block simple harmonic oscillator, the potential energy is maximum when the displacement is maximum. Therefore, the magnitude of the acceleration is also maximum at this point. Hence, option d: "It decreases as the potential energy increases" is false.

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The person needs spectacles with a power of 2 diopters.

The nearsighted person has a near point at 0.2 m and a far point at 2 m. This means the person can clearly see objects that are closer than 0.2 m, but objects farther away appear blurry. To correct for this, the person needs spectacles with a positive power (convex lenses) that will help bring the distant objects into focus.

The power of a lens is given by the formula:

Power (P) = 1 / focal length (f)

Since the person's near point is at 0.2 m, we can calculate the power needed to bring the far point (2 m) into focus. The focal length of the lens required to bring the far point into focus is the reciprocal of the far point distance:

f = 1 / 2 = 0.5 m

To find the power of the lens, we substitute the focal length into the power formula:

P = 1 / f = 1 / 0.5 = 2 diopters

Since the person needs spectacles to correct their nearsightedness and see distant objects clearly, the power of the spectacles should be the opposite sign of the lens power. Therefore, the person needs spectacles with a power of -2 diopters. However, none of the given options match this exact power.

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Fossil fuels, such as coal, oil, and natural gas, are considered non-renewable sources of energy because they take millions of years to form. These fuels are created from the remains of ancient plants and organisms that have undergone geological processes over long periods of time. So option D is correct.

The formation of fossil fuels involves the accumulation of organic matter, its burial, and the transformation of that matter under high pressure and temperature.

Since the formation of fossil fuels is an extremely slow process, the rate at which they are being consumed by human activities far exceeds the rate at which they are naturally replenished. This makes them finite resources with limited availability on a human timescale. Once fossil fuels are extracted and used, they cannot be easily replaced or regenerated within a short span of time.

Therefore, due to their finite nature and the extended time required for their formation, fossil fuels are considered non-renewable sources of energy.Therefore option D is correct.

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The average distance between the Sun and Mercury is approximately 0.39 AU. The astronomical unit (AU) is a unit of measurement commonly used in astronomy to represent distances within the solar system.

One AU is defined as the average distance between the Earth and the Sun, which is about 150 million kilometers (93 million miles). To convert the distance between the Sun and Mercury to AU, we divide the given distance ([tex]58 \times 10^6 km[/tex]) by the average distance between the Sun and Earth.

[tex]\[\text{{Distance in AU}} = \frac{{58 \times 10^6 \, \text{{km}}}}{{150 \times 10^6 \, \text{{km}}}} \approx 0.39 \, \text{{AU}}\][/tex]

Rounding to two significant figures, the average distance between the Sun and Mercury is approximately 0.39 AU. This means that, on average, Mercury is about 0.39 times the distance from the Earth to the Sun.

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The Highway Department Garage has a crew of mechanics who repair vehicles that break down. The breakdowns occur at an average rate of 7.5 vehicles per day, and the mechanics can service an average of 10 vehicles per day.

This service system's utilization rate, average time for a breakdown to be repaired, waiting time for service, and the number of vehicles likely to be in the system at any given time need to be determined.

a. The utilization rate of a service system is the ratio of the arrival rate of customers to the service rate. In this case, the arrival rate is λ = 7.5 vehicles per day, and the service rate is μ = 10 vehicles per day. Therefore, the utilization rate can be calculated as λ/μ = 7.5/10 = 0.75 or 75%.

b. The average time before a breakdown can be repaired is given by the reciprocal of the service rate, which is 1/μ = 1/10 = 0.1 days or 2.4 hours.

c. To determine the time spent waiting for service, we need to calculate the average time a vehicle spends in the system. This can be obtained using Little's Law, which states that the average number of customers in a system is equal to the arrival rate multiplied by the average time spent in the system.

As the system is in equilibrium, the average number of vehicles in the system is equal to the average number of vehicles being serviced. Therefore, the average time spent waiting for service can be calculated as (average number of vehicles in the system) / λ = (λ/μ) / λ = 0.75 / 7.5 = 0.1 days or 2.4 hours.

d. The average number of vehicles in the system at any one time can be calculated using Little's Law as λ * average time spent in the system = 7.5 * 0.1 = 0.75 vehicles.

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a)  the average speed for the trip is 39.8 km/h.

b) The distance between the initial and final cities along the route is equal to the total distance traveled, which is 79.08 km.

First, let's calculate the distances traveled at each speed:

Distance at 65.0 km/h = (65.0 km/h) * (25.0 min) = 27.08 km

Distance at 80.0 km/h = (80.0 km/h) * (9.0 min) = 12.00 km

Distance at 40.0 km/h = (40.0 km/h) * (60.0 min) = 40.00 km

Now, let's calculate the total distance traveled:

Total distance = Distance at 65.0 km/h + Distance at 80.0 km/h + Distance at 40.0 km/h = 27.08 km + 12.00 km + 40.00 km = 79.08 km

Next, let's calculate the total time taken:

Total time = Time driving + Time for lunch and gas = 25.0 min + 9.0 min + 60.0 min + 25.0 min = 119.0 min

Now, we can calculate the average speed:

Average speed = Total distance / Total time = 79.08 km / 119.0 min = 0.664 km/min = 39.8 km/h

Therefore, the average speed for the trip is 39.8 km/h.

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Gravitational waves travel at the speed of light, so it would take approximately 10 years for the gravitational waves produced by the star's explosion to reach Earth.

Since the star is located 10 light-years away, it means that the light emitted by the explosion takes 10 years to travel from the star to Earth. Gravitational waves, like light, also travel at the speed of light in a vacuum.

Therefore, the gravitational waves produced by the star's explosion would also take approximately 10 years to propagate through space and reach Earth. This is because both light and gravitational waves travel at the same finite speed, and in this scenario, they would arrive at Earth at approximately the same time, 10 years after the star's explosion.

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The output of the still can be determined by calculating the amount of solar heat absorbed by evaporation and the corresponding amount of evaporated water.

In order to calculate the output of the still, we need to consider the amount of solar heat absorbed by evaporation and the latent heat of the evaporation of water. The given insolation value of[tex]25 MJ m^-^2day^-^1[/tex] represents the solar heat available in a dry sunny area. The latent heat of the evaporation of water is [tex]2.4 MJ kg^-^1[/tex], which indicates the amount of energy required to convert one kilogram of water into vapor.

To determine the output of the still, we can divide the total solar heat absorbed by the latent heat of evaporation. This can be calculated by dividing the insolation value[tex](25 MJ m^-^2day^-^1)[/tex] by the latent heat of evaporation ([tex]2.4 MJ kg^-^1[/tex]). The resulting value will represent the amount of water that can be evaporated using the available solar heat. By collecting all the evaporated water, the output of the still can be measured.

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The object will slide a distance of 8.81 meters before coming to a complete stop.

To calculate the distance the object will slide, we can use the equation of motion for uniformly decelerated motion:

v² = u² + 2as

where v is the final velocity (0 m/s in this case), u is the initial velocity (10.9 m/s), a is the acceleration (caused by the frictional force), and s is the distance traveled.

Rearranging the equation, we have:

s = (v² - u²) / (2a)

Since the object comes to a complete stop, the final velocity v is 0 m/s. The initial velocity u is 10.9 m/s, and the acceleration a is given by Newton's second law:

F = ma

where F is the frictional force (35.6 N) and m is the mass of the object (38.1 kg). Solving for a, we find:

a = F / m

Substituting the values into the equation for distance, we get:

s = (0² - (10.9)²) / (2 * (35.6 / 38.1))

Calculating the expression, we find the distance traveled, d, is approximately 8.81 meters. Therefore, the object will slide a distance of 8.81 meters before coming to a complete stop.

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