some of the light also reflects off the surface of the water. if the incident light is initially unpolarized, the reflected light will be unpolarized somewhat horizontally polarized somewhat
vertically polarized

Using the fine focus when trying to put a specimen in sharp focus when using the 40x objective lens ensures that the specimen is brought into focus as accurately and precisely as possible.

This knob is used to adjust the specimen's focus. Additionally, it is used to highlight specific areas of the specimen. The coarse focus knob should often be used to get near before switching to the fine focus knob for fine tweaking.

The fine focus allows for a much finer adjustment of the specimen's position to be made, which is necessary when using a higher magnification objective lens like the 40x.

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Explanation:

0.75

According to the formula, P=IV

150 = I×200

Answer:

\(\boxed {\boxed {\sf 96 \ m/s}}\)

Explanation:

We are asked to find the final velocity of the car given the acceleration and time. We can use the following kinematics equation to calculate the final velocity.

\(v_f=v_i+(a \times t)\)

The car starts from rest, so the initial velocity is 0 meters per second. It accelerates a rate of 40 meters per square second over a period of time of 2.4 seconds.

Substitute the values into the formula.

\(v_f= 0 \ m/s + ( 40 \ m/s^2 \times2.4 \ s)\)

Solve inside the parentheses.

\(v_f= 0 \ m/s + (96 \ m/s)\)

Add.

\(v_f= 96 \ m/s\)

The final velocity of the car is 96 meters per second.

To determine the temperature increase of each car, we can use the principle of conservation of energy. The total kinetic energy of the cars before the collision is converted into thermal energy (heat) after the collision.

The formula to calculate the temperature increase is:
ΔT = (ΔE) / (m * c)
Where:
ΔT is the temperature increase,
ΔE is the change in thermal energy (equal to the initial kinetic energy of the cars),
m is the mass of each car, and
c is the specific heat capacity of iron.
Since the specific heat capacity of iron is approximately 450 J/(kg·°C), and the mass of each car is not given in the question, we cannot determine the specific temperature increase without that information. Therefore, the answer cannot be expressed with the appropriate units without knowing the mass of the cars.

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(a) Time to read or write an entire track is 0.0333 seconds.

(b) Time to read or write a single sector is 0.0014  seconds.

(c) it will take approximately 2.4 ms to transfer Sector 1 on Track 5 to Sector 1 on Track 6.

(d) it will take approximately 36.33 ms to transfer all the sectors of Track 5 to the corresponding sectors of Track 8.

a. To calculate how long it takes to read or write an entire track, we need to consider the disk rotation speed. The disk rotates at 1800 rpm (revolutions per minute). We can convert this to revolutions per second by dividing by 60:

1800 rpm / 60 = 30 revolutions per second

Since there are 24 sectors per track, and the head sees the sectors in ascending order, it will take one full revolution to read or write the entire track. Therefore, the time to read or write an entire track is equal to the time for one revolution:

Time to read or write an entire track,

\(t_1 = \dfrac{1} { 30 rps } \\t_1= 0.0333 seconds\)

b. To calculate how long it takes to read or write a single sector, we divide the time to read or write an entire track by the number of sectors:

Time to read or write a single sector = Time to read or write an entire track / Number of sectors

\(t_2= \dfrac{0.0333 } { 24 }\)

t₂ = 0.0014 seconds

c. To transfer sector 1 on track 5 to sector 1 on track 6, we need to consider the seek time to move the head between adjacent tracks. The seek time is given as 1 ms. Therefore, the total time to transfer the sector would be the sum of the seek time and the time to read or write a single sector:

Total time = Seek time + Time to read or write a single sector

t(total) = 1 ms + 0.0014 seconds (or 1.4 ms)

t(total) = 2.4 ms

So, it will take approximately 2.4 ms to transfer Sector 1 on Track 5 to Sector 1 on Track 6.

d. To transfer all the sectors of Track 5 to the corresponding sectors of Track 8, we need to consider both the track seek time and the time to read or write an entire track.

The track seeks time is given as 3 ms, and the time to read or write an entire track is 33.33 ms (calculated in part a).

Total time = Track seek time + Time to read or write an entire track

t(total) = 3 ms + 33.33 ms

t(total) = 36.33 ms

So, it will take approximately 36.33 ms to transfer all the sectors of Track 5 to the corresponding sectors of Track 8.

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The x and y component of the vector are -18.32 m and 3.23 m respectively.

The x and y component of the vector is calculated by applying the following formula.

Bx = B cosθ

By = B sinθ

where;

The vector is located in negative x direction but positive y direction.

The angle of the vector from x axis = 90⁰ - 80⁰ = 10⁰

Bx = B cosθ  =  -18.6 m x cos ( 10 ) = -18.32 m

By = B sinθ = 18.6 m x sin ( 10 ) = 3.23 m

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The maximum velocity of a particle on the string is approximately 0.98 m/s.

To find the maximum velocity of a particle on the string, we can use the given tension, linear density, amplitude, and wavelength values.

Given:

- Tension (T) = 55.0 N

- Linear density (μ) = 4.70 g/m = 0.00470 kg/m (converted to kg/m)

- Amplitude (A) = 3.00 cm = 0.03 m (converted to meter)

- Wavelength (λ) = 2.10 m

First, we can find the wave speed (v) using the equation v = √(T/μ):

v = √(55.0 N / 0.00470 kg/m) ≈ 34.66 m/s

Next, we can find the angular frequency (ω) using the equation ω = 2πv/λ:

ω = (2π * 34.66 m/s) / 2.10 m ≈ 32.74 rad/s

Finally, we can find the maximum velocity of a particle on the string (v_max) using the equation v_max = Aω:

v_max = 0.03 m * 32.74 rad/s ≈ 0.98 m/s

So, the maximum velocity of a particle on the string is approximately 0.98 m/s.

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They will each have equal but opposite velocity.

Momentum is how we measure mass that is in motion. Any moving object will have momentum. Under the law of physics, the object’s momentum equals mass times velocity. These two skaters have just pushed off against one another. One skater has twice the mass as the other. We know that skater 1 has a smaller mass and skater 2 has a larger mass than skater 1. It implies that the force of skater 2 will be greater than that of skater 1 since force (F) varies directly proportional to the mass of an object together with its acceleration. So, when these two skaters collide, skater 2 exerts a greater force than skater 1 and the velocity of skater 1 will be increased and greater as a result of the collision with skater 2.

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Answer:

Explanation: Pules

Answer: longitudinal (compression)

Explanation:

Answer:

Explanation:

The Horizontal velocity is expressed as:

Vx = -V cos 30

Vx = -150cos30

Vx = -150(0.8660)

Vx = -129.90m/s

The vertical component of the velocity is expressed as:

Vy = V sin 30

Vy = 150sin30

Vy = 150(0.5)

Vy = 75m/s

Multiple types of friction at once are common when an object is rolling over a surface (rolling friction), which is also associated with sliding friction.

Friction is a physical force that acts as opposed to the normal movement of an object, which may depend on different types of movement and situations (i.e. rolling friction, sliding friction, fluid friction, static friction, etc).

In conclusion, multiple types of friction at once are common when an object is rolling over a surface (rolling friction), which is also associated with sliding friction.

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Answer:

Earth's larger mass results in a larger acceleration due to gravity as compared to the moon. In other words, you weigh more on Earth; you would weigh less on the moon. For instance, if you take a 10 kg bowling ball to the moon, it would still have a mass of 10 kg.

Explanation:

Hope this helps ^_^

~ KisamesAbs

The strength of the uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown is 0.32T.

Let's start by finding the torque acting on the square loop due to the magnetic field. = sinwhere  is the torque,  is the magnetic field strength,  is the current,  is the area of the square loop, and  is the angle between the plane of the loop and the magnetic field.

The square loop is in static equilibrium, which means the net force and net torque acting on it are zero. Since the loop is resting on a frictionless horizontal surface, the normal force and weight of the loop will cancel each other out.

The torque acting on the square loop due to the magnetic field is = sin= 25A × (2.5m)² × sin(60°)= 125JThe torque due to the magnetic field is balanced by an equal and opposite torque due to the tension in the wire. The tension in the wire is acting at an angle of 45° to the horizontal, so we can resolve it into horizontal and vertical components.

The horizontal component is equal to the magnetic torque, and the vertical component is equal to the weight of the loop.Using trigonometry, we can find the tension in the wire.Tcos(45°) = T = /cos(45°)= 125J/cos(45°)= 177JThe weight of the square loop is = = 2.0kg × 9.8m/s²= 19.6NTherefore, the vertical component of the tension in the wire is equal to the weight of the square loop.

Tsin(45°) = Tsin(45°) = 19.6NT = 27.7NThe horizontal component of the tension in the wire is equal to the magnetic torque.Tcos(45°) = Tcos(45°) = 125JT = 177JThe magnetic field strength is = /(sin)= 125J/(25A × (2.5m)² × sin(60°))= 0.32TTherefore, the strength of the uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown is 0.32T.

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We are given the following information about an inelastic collision.

Mass of 1st object = 4 kg

Mass of 2nd object = 10 kg

Initial velocity of 1st object = 25 m/s

Final velocity of both objects = 14 m/s

Initial velocity of 2nd object = ?

In an inelastic collision, the momentum is conserved but the kinetic energy is not conserved.

Recall that the total momentum is conserved and given by

Let us substitute the given values and solve for initial velocity of the body (u2)

Therefore, the initial velocity of the body of mass 10 kg is 9.6 m/s

Answer:

Hi

Explanation:

A H20

Both springs have a total compression of 2x = 1.5 m.

The force acting on the block is equal to the sum of the forces from both springs. Let x be the compression of each spring.

The force from each spring is given by Hooke's law: F = kx, where k is the spring constant. Since there are two springs, the total force is:

F = 2kx

The force on the block is 600 N, so set up the equation:

2kx = 600

Substituting k = 400 N/m:

2(400 N/m)x = 600 N

Solving for x:

x = 0.75 m

Therefore, the total compression of both springs is 2x = 1.5 m.

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Since it is practically in the plane of its orbit, its ring system occasionally appears to be made up of near-circles. The majority of scientists think that Uranus' rotation is the result of a violent collision.

In contrast to other planets, Uranus has an axis that nearly parallels its orbital plane. This causes Uranus to rotate virtually on its side as it travels through its orbit, with its poles alternately pointing toward the Sun.

Venus and Uranus' rotation on their axes is remarkable in some ways. All other planets revolve from the West to the East, but they do the opposite.

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Analyzing the motion of a rubber ball thrown upward, with an initial speed of 15.8 m/s, we find that it takes approximately 1.61 seconds to reach its highest point and reaches a maximum height of approximately 12.74 meters.

When analyzing the motion of the rubber ball, we can use the equations of motion to find various parameters. Let's break it down:

First, the initial speed of the ball is given as 15.8 m/s directly upward. Since we have chosen upward as the positive direction, the initial velocity of the ball will be +15.8 m/s.

Next, let's find the time it takes for the ball to reach its highest point. When the ball reaches its highest point, its vertical velocity becomes zero. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity is 0 m/s, the initial velocity is +15.8 m/s, and the acceleration is the acceleration due to gravity, which is -9.8 m/s² (negative because it acts in the opposite direction to the initial velocity). Plugging these values into the equation, we get:
0 = 15.8 - 9.8t
9.8t = 15.8
t ≈ 1.61 s

Therefore, it takes approximately 1.61 seconds for the ball to reach its highest point.

Now, let's find the maximum height reached by the ball. We can use the equation v² = u² + 2as, where s is the displacement. At the highest point, the final velocity is 0 m/s, the initial velocity is +15.8 m/s, the acceleration is -9.8 m/s², and we want to find the displacement. Plugging these values into the equation, we get:
0 = (15.8)² + 2(-9.8)s
0 = 249.64 - 19.6s
19.6s = 249.64
s ≈ 12.74 m

Therefore, the maximum height reached by the ball is approximately 12.74 meters.

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The back of the seat exerts a greater force on Jonathan when he accelerates away from the stop sign.

This is because force is directly related to mass, and Jonathan's mass is likely greater than that of his eight-year-old daughter.

According to Newton's second law of motion, force (F) equals mass (m) times acceleration (a), or F = ma.

Since Jonathan's mass is greater, the force exerted on him by the back of the seat will also be greater.

Thus, the back of the seat exerts a greater force on Jonathan when he accelerates away from the stop sign.

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Answer: 10 seconds for the car to go from a velocity of 10 m/s to 30 m/s with an acceleration of 2 m/s^2.

Explanation:

To calculate the time it takes for a car with an acceleration of 2 m/s^2 to go from a velocity of 10 m/s to a velocity of 30 m/s, we can use the equation:

time = (final velocity - initial velocity) / acceleration

In this case:

time = (30 m/s - 10 m/s) / 2 m/s^2

time = 20 m/s / 2 m/s^2

time = 10 s

Therefore, it takes 10 seconds for the car to go from a velocity of 10 m/s to 30 m/s with an acceleration of 2 m/s^2.

To calculate the work required to set up the four-charge configuration, we need to consider the electric potential energy. The electric potential energy of a system of point charges is given by:

U = k * (q1 * q2 / r12 + q1 * q3 / r13 + q1 * q4 / r14 + q2 * q3 / r23 + q2 * q4 / r24 + q3 * q4 / r34)

where:

U is the electric potential energy,

k is Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2),

q1, q2, q3, q4 are the charges (in Coulombs),

r12, r13, r14, r23, r24, r34 are the distances between the charges (in meters).

Now, we can calculate the distances between the charges using the given information about the separation a:

r12 = r34 = a (since they are horizontally aligned)

r13 = r14 = r23 = r24 = √(a^2 + a^2) = √(2a^2) = a√2 (since they form a square)

Substituting the values into the electric potential energy formula:

U = k * (q1 * q2 / r12 + q1 * q3 / r13 + q1 * q4 / r14 + q2 * q3 / r23 + q2 * q4 / r24 + q3 * q4 / r34)

U = (8.99 × 10^9 N m^2/C^2) * (q1 * q2 / a + q1 * q3 / (a√2) + q1 * q4 / (a√2) + q2 * q3 / a√2 + q2 * q4 / a√2 + q3 * q4 / a)

Now, substitute q = 7.31 × 10^(-12) C into the equation:

U = (8.99 × 10^9 N m^2/C^2) * (7.31 × 10^(-12) C * 7.31 × 10^(-12) C / a + 7.31 × 10^(-12) C * 7.31 × 10^(-12) C / (a√2) + 7.31 × 10^(-12) C * 7.31 × 10^(-12) C / (a√2) + 7.31 × 10^(-12) C * 7.31 × 10^(-12) C / a√2 + 7.31 × 10^(-12) C * 7.31 × 10^(-12) C / a√2 + 7.31 × 10^(-12) C * 7.31 × 10^(-12) C /

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The work required to set up the four-charge configuration can be calculated by summing the work done in bringing each charge from infinity. This utilizes concepts from Physics, especially Electrostatics, and utilizes Coulomb's Law equation.

This problem can be solved using electric potential energy concepts from Physics. In order to calculate the work done, we need to use Coulomb's Law (a central equation in Electrostatics). The work needed to assemble a charge configuration from infinity is the potential energy of that configuration.

In this case, the system involves four charges and their linkages can be thought of as those in a square. Each pair's work to bring them together is given by W=k*q*q/r where k is the electrostatic constant (9x10^9 N*m^2/C^2), q is the charge (7.31e-12 C) and r is the separation distance (78.8*10^-2 m). There are 6 unique pairings, so you would calculate the work for each pair and multiply by 6.

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Answer:

100 newton

Explanation:

newton third law of motion says to every action there is an always an equal and opposite reaction so the magnitude will stay equal but opposite direction

Answer:

- 100N is the answer of this question

Answer:

F= 1800N

Explanation:

the equation for force is F= ma

so plug in the numbers: F= (120)(15)

solve this to get F= 1800N

tip: don't forget to add the units when writing your answer :)

The new period (T2) is equal to the square root of 2 times the original period (T1) when the mass is doubled in simple harmonic motion.

In simple harmonic motion, the period (T) is defined as the time taken for one complete cycle of oscillation. The period is inversely proportional to the square root of the mass attached to the spring.

Let's denote the original mass as m1 and the new mass as m2. The period of oscillation with the original mass is T1, and we need to find the new period T2 with the doubled mass.

Mathematically, the period is given by:

\(T = 2\pi * \sqrt {(m/k)}\)

where k is the spring constant.

Since the spring constant remains the same, we can write the equation as:

\(T = 2\pi * \sqrt {(m/k)\)

\(T1 = 2\pi * \sqrt {(m1/k)\)

To find the new period T2 with the doubled mass (2m1), we substitute m2 = 2m1 into the equation:

\(T2 = 2\pi * \sqrt {(2m1/k)}\)

Simplifying the equation, we get:

\(T2 = 2\pi * (\sqrt {2 * \sqrt {(m1/k))\\T2 = \sqrt2 * T1\)

Therefore, the new period (T2) is equal to the square root of 2 times the original period (T1) when the mass is doubled in simple harmonic motion.

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B. Good cleave.  A good cleave is required to obtain minimal loss from a prepolished /splice connection. A prepolished/splice connection is a type of fibre optic connector .

With a prepolished ferrule that provides for rapid and uncomplicated fibre optic cable field terminations. Typically, the ferrule is pre-polished at the manufacturer to achieve a good surface smoothness and reduced insertion loss. Nonetheless, a good cleave is required to obtain low loss in the field. Cleave is the technique of cutting a fibre optic cable using a precision tool to create a smooth, flat endface perpendicular to the fiber's axis. A proper cleave guarantees that the fibre endface is devoid of imperfections or abnormalities that might scatter light and increase insertion loss.

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The equations mentioned by Emily in the valuation analysis for Aspeon are as follows:

1) Equation (1): This equation represents the value of equity (S) and calculates it based on the EBIT (earnings before interest and taxes), the tax shield provided by debt (D), and the required return on debt (kd) and equity (ks). It implies that the value of equity is equal to the adjusted EBIT after deducting the tax shield from debt.

2) Equation (2): This equation calculates the total enterprise value (V) by adding the value of equity (S) and debt (D). It represents the total worth of the company, considering both equity and debt.

3) Equation (3): This equation calculates the price per share (P) by dividing the total enterprise value (V) minus the initial debt (D0) by the number of shares (n0). It represents the price per share based on the valuation of the company.

4) Equation (4): This equation calculates the new number of shares (n1) by subtracting the dividend (D) from the initial number of shares (n0) divided by the price per share (P). It represents the adjusted number of shares after the payment of dividends.

5) Equation (5): This equation calculates the levered value (VL) by adding the unlevered value (VU) with the tax shield value (TD). It represents the value of the company after considering the tax advantages of debt.

These equations provide a framework for valuation analysis, considering factors such as earnings, taxes, debt, and equity. They help assess the value and financial implications of Aspeon's growth prospects.

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Answer:

The wrong items are;

1) The normal for FN equals the weight Fmg

2) The force of friction, Ff, equals the applied force Fpull

The corrected statements are;

1) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ)

Explanation:

The given statement was;

The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull

By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;

The normal force is equal to the weight less the vertical component of the applied force;

That is we have, FN = Fmg - Fpull × sin(θ)

The friction force similarly, is equal to the horizontal component of the applied force;

Ff = Fpull × cos(θ)

The wrong items are therefore as follows;

1) The normal for FN equals the weight Fmg

1 i) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction, Ff, equals the applied force Fpull

2 i) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ).

Each item is sorted into whether it is more common in spiral arms or about equally common within and between the spiral arms in a spiral galaxy's disk

More common in spiral arms

Equally common within and between spiral arms

According to spiral density wave theory, spiral arms are a long-lived structure and rotate like a rigid body with a given pattern speed, while the stars rotate around the Galactic center with a velocity that depends on the radius of the orbit.

Therefore, each item is sorted above into whether it is more common in spiral arms or about equally common within and between the spiral arms in a spiral galaxy's disk

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I think that the awsand is 147.2 .