if a harmonic oscillator were acted upon by minor dissipative forces, how would the amplitude of the oscillations change as time goes on?

The force on the bullet of mass 21 g  is 14795.12 N.

Force is the product of mass and acceleration.

To calculate the force exerted on the bullet while it is being accelerated, we use the work energy relation below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the force on the bullet is 14795.12 N.

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The liquid is drawn up into the feed tube and forced through the nozzle by the vacuum produced by the airflow.

A vacuum is indeed a space that is devoid of all substance, sometimes known as "free-space." Only complete vacuums are actually feasible in real life. An incandescent lightbulb has a vacuum inside. The vacuum in space is almost ideal. The Moon, Mercury, or Mars have no atmospheres at all.

Scales and vacuum units commonly used. In vacuum systems, pressure is still measured in mm or inches of mercury. The Torr (from Torricelli) unit of pressure measurement is based on millimeters of mercury, and mmHg (Hg being heavy metals in the periodic table).

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None of the alternatives are correct

It's crucial to keep in mind that rolling motion is impossible on a surface with no friction. It will just continue to slide along. The sphere will slide when something is rolling down an incline if the incline's angle is very large or its coefficient of friction is very low.

The ball will continue rolling in the absence of friction. Only friction would prevent it from rolling indefinitely if the second plane's angle of inclination were reduced to zero, making it perfectly horizontal.

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(a) The formula for these two atoms' center masses in relation to the oxygen atom is. (b) The center of mass of monoxide has a numerical value of 53 pm.

The settings provided;

The carbon atom's mass is 12u.

The oxygen atom has a mass of 17 u and a spacing of 128 pm between its atoms.

These two atoms' center masses in relation to the oxygen atom are computed as follows;

where; is the atom's separation from the fixed reference point (the oxygen atom) (b) The following equation is used to calculate the center of mass of monoxide in measures of pm:

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The wave can be broken in c) the bottom interfering with its oscillatory motion.

In fluid mechanics, a breaking wave or breaker is a wave whose amplitude reaches a critical level where a large amount of wave energy is converted into turbulent kinetic energy. At this point, simple physical models describing wave dynamics, especially those that assume linear behavior, are often invalid.

The most commonly known type of breaking wave is a water wave that breaks on the shore. In general, wave breaking occurs when the amplitude reaches the point where the crest of the wave actually flips over. Another particular effect in fluid dynamics is also called "wave breaking", partly by an analogy to waves on water. In meteorology, atmospheric gravitational waves are said to break down when the wave creates a region of potential temperature drop with altitude, resulting in energy dissipation due to convective instability.

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Parabolic Up. Since the force is in the y-direction, the particle will continue to move at a constant speed in the x-direction and accelerate in the y-direction. This creates a parabolic path around y-axis.

Force is the power that changes the motion of an item. A force can purpose an object with mass to alternate its pace. A force has both value and path, making it a vector quantity.

Pressure is the push or the pull that affects us in our daily lives because, without force, people might not be capable of opening and near stuff or lifting up their arms or legs.

Variety of Forces :

* Muscular Forces, muscular tissues feature to produce an ensuing force which is referred to as 'muscular force'.

* Frictional Forces, while an item modifications its national motion, 'frictional pressure' acts upon it.

* Carried out pressure.

* Tension force.

* Spring force.

* Gravitational pressure.

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Electromagnetic (EM) waves are waves that are related to both electricity and magnetism. These waves travel over space and are made up of time-varying electric and magnetic fields.

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A ball is thrown vertically upward with an initial velocity of 64 feet per second.So, from 0.5 seconds to 3.5 seconds, the ball is at the height of 28 feet above the ground.

velocity.  . swiftness, quickness, or the speed of motion, activity, or operation A measurement of a body's rate of motion in physics is the rate at which its location changes over time in a specific direction.

Simply said, velocity is the rate of movement in a specific direction. including the speed of a car driving north on a road or the pace at which a rocket takes off. Since the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.

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Measuring devices are used as follows -1. Thermometer, 2. Heart monitor, 3. Measuring cylinder, 4. Clock, 5. Tape measure

What are measuring devices?

The instruments that are used for the measurements of various specific items based on the laws and the theories of measurement are called measuring devices.

Specific measurements have specific measurement devices. These measure the value based on the specific unit of the measurement.

For example, an inch tape measures the length of something in inches. It also has scales of centimetres and meters.

For the measurement of length, scales and tape measures are used.

For the measurement of time, clocks are used.

For the measurement of volumes, graduated cylinders are used.

For the measurement of temperature, thermometers are used.

Therefore, the measuring devices vary according to the function it needs to perform. In this case, 1. Thermometer, 2. Heart monitor, 3. Measuring cylinder, 4. Clock, 5. Tape measure.

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Answer:

Answer:heart monitor

Stopwatch

Digital tape measure

Thermometer

Graduated cylinder

Explanation:

a) The starting speed of the stone was 18.0 m/s, and the cliff's height was 50.0 m.

The expression for the stone's initial x-coordinate

[tex]x_{i}[/tex] = 0

The stone's initial x-coordinate is shown here.

the definition of the top of the cliff's initial y-coordinate

[tex]y_{i}[/tex] = 50m

The stone's initial y-coordination is seen here.

Consequently, the stone's starting position's coordinates are (0, 50.0m)

b) The expression for the stone's starting speed in the x direction

[tex]v_{ix}[/tex] = 18.0m/s

The stone's initial horizontal velocity [tex]v_{ix}[/tex]  is seen here. The formula for the starting velocity in the y direction .

[tex]v_{iy}[/tex] = 0m/s

Here,  [tex]v_{iy}[/tex] is the stone's initial vertical velocity.

Consequently, the initial component of the stone's speed are [tex]v_{ix}[/tex] = 18.0m/s and  [tex]v_{iy}[/tex] = 0m/s.

c) A free fall motion with constant g-force governs the stone's vertical movement.

Particle motion with constant acceleration is what is happening in the y direction.

As a result, the acceleration in the y direction remains constant.

d) Because there is zero acceleration in the x-direction and no net force acting to modify the stone's inertia, the particle's velocity remains constant throughout the motion.

As a result, the motion in the y-direction is motion at a constant speed.

As a result, motion in the x-direction is caused by constant velocity motion.

e) Because the stone doesn't experience any acceleration in the x direction, its speed remains constant throughout the motion.

The stone's ultimate x-direction velocity is equal to its x-direction beginning velocity.

the relationship between the x-final direction's and beginning velocities,

[tex]v_{fx} = v_{ix}[/tex]

Here, [tex]v_{fx}[/tex] is the final horizontal velocity.

The speed in the x direction is independent of time.

The stone experiences a constant acceleration in the y direction, or g, which determines the stone's y-direction velocity.

The formula for the y-ultimate direction's velocity

[tex]v_{fy} = v_{iy} + at[/tex]

Here, [tex]v_{fy}[/tex] is the ultimate vertical velocity, and an is the vertical acceleration.

Substitute 0 for [tex]v_{iy}[/tex] and -g for a in the above equation to find [tex]v_{fy}[/tex]

   [tex]v_{fy}[/tex] = -gt

Consequently, the velocity's x and y components are [tex]v_{fx} = v_{ix}[/tex]  and  [tex]v_{fy}[/tex] = -gt respectively.

f) The term for the stone's x-direction position

[tex]x_{f} = x_{i} + v_{ix}t + a_{x} t^{2}[/tex]                                (i)

Here, [tex]x_{f}[/tex] is the final horizontal position and [tex]a_{x}[/tex] is the acceleration in the x-direction.

The term for the stone's y-direction position

[tex]y_{f} = y_{i} + v_{iy} t + \frac{1}{2} at^{2}[/tex]                                (ii)

Here, [tex]y_{f}[/tex] is the final vertical position.

Substitute 0 for [tex]x_{i}[/tex] and 0 for [tex]a_{x}[/tex] in the equation (i)

[tex]x_{f} = 0 + v_{ix} t + 0\\ = v_{ix} t[/tex]

Substitute 0 for [tex]v_{iy}[/tex] and -g for a in the equation (ii)

[tex]y_{f} = y_{i} + 0 - \frac{1}{2} (g)t^{2}[/tex]

    [tex]= y_{i} - \frac{1}{2} (g)t^{2}[/tex]

Consequently, the position's x and y components are [tex]v_{xi} t[/tex] and  [tex]= y_{i} - \frac{1}{2} (g)t^{2}[/tex] respectively.

g) The formula to calculate time

t = [tex]\sqrt{\frac{2h}{g} }[/tex]

Substitute 50m for h and 9.8m/[tex]s^{2}[/tex] for g in above equation to find t.

t = [tex]\sqrt{\frac{2(50m)}{9.8m/s^{2} } }[/tex]

 = 3.19s

As a result, the stone will hit the sea below the cliff in 3.19 seconds.

h) The expression to calculate velocity.

[tex]v_{fy}[/tex]  = -gt

Substitute 3.19s for t and 9.8 m/[tex]s^{2}[/tex] Obtain for g in the previous equation [tex]v_{fy}[/tex].

[tex]v_{fy}[/tex]  = (9.8m/[tex]s^{2}[/tex]) (3.19s)

     = -31.26m/s

     ≈ - 31.3m/s

The method for estimating the speed of stone land

v = [tex]\sqrt{v^{2} _{fx} }+ \sqrt{v^{2} _{fx} }[/tex]                             (iii)

The speed of the stone when it lands in this case is v.

The formula for calculating the stone land's angle

∅ = [tex]tan^{-1}[/tex] ( [tex]\frac{v_{fy} }{v_{fx} } )[/tex]                                   (iv)

Substitute 18m/s for [tex]v_{fx}[/tex] and -31.3m/s for [tex]v_{fy}[/tex] in equation (iii) to find v.

v = [tex]\sqrt{(18m/s)^{2} }[/tex] + [tex]\sqrt{(-31.3m/s)^{2} }[/tex]

  = 36.1m/s

As a result, the stone was moving at 36.1 m/s when it hit the ground.

Substitute 18m/s for [tex]v_{fx}[/tex] and -31.3m/s for [tex]v_{fy}[/tex] in equation (iv) to find ∅.

∅ =  [tex]tan^{-1}[/tex] ([tex]\frac{-31.3m/s}{18m/s} )[/tex]

    = -60.09°

   ≈ -60.1°

As a result, the stone terrain has an angle that is -60.1° below the horizontal.

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(a) The speed of the electron is 1.5 x 10⁹ m/s.

(b) The mass of the electron is 12.75 times greater than the rest mass energy.

The speed of the electron is calculated by applying the following equation;

K.E = ¹/₂mv²

where;

v² = 2K.E/m

v = √(2K.E/m)

where;

K.E is the kinetic energy = 6.5 MeV = 6.5 x 10⁶ x 1.6 x 10⁻¹⁹ J = 1.04 x 10⁻¹² J

v = √(2 x 1.04 x 10⁻¹²/9.11 x 10⁻³¹)

v = 1.5 x 10⁹ m/s

The rest mass energy of an electron is 0.51 MeV

0.51x = 6.5

x = 6.5/0.51

x = 12.75

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If the xy plane is grounded, Then the force on the -2q charge is derived from F=1/4πE* q²/d²  Z

So magnitude of the image charge = - 29

and location is distance d from origion at-2 axis as shown.

The charge 2q is attracted toward the plate because of the negative induced charge. Now force  felt by charge 2q All to new configuration force  is

F=1/4πE*( 2q ) ( - 2q ) /2d² Z

or

F=1/4πE* q²/d²  Z.

An object experiences a push or pull as a result of interacting with another object. any time there is a relationship between two objects. The result of the interaction between two objects, force is a push or a pull. It has both magnitude and direction because it is a vector quantity. Normal and gravitational force are the most typical types of forces. The attraction between two mass-containing objects is created by the gravitational force, which is a non-contact force.

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A front denotes the separation of two air masses with dissimilar characteristics in terms of temperature, wind, and moisture.

Distinguish between warm and cold fronts.

Both warm and cold fronts are intricate weather phenomena, necessitating a thorough explanation in order to fully comprehend their similarities and distinctions. A cold front develops on the edge of a cold air mass flowing into a warmer area, whereas a warm front develops on the edge of a warm air mass moving into a colder area. A cold front is often linked to a low-pressure system, whereas a warm front is linked to a high-pressure system.

Hence, the answer is both warm and cold front, since a front denotes the separation of two air masses with dissimilar characteristics in terms of temperature, wind, and moisture.

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Unless acted on by an imbalanced force, an object is at rest can tend to stay as rest, and a motion should tend to maintain its current speed and direction.

A pull or pull that an object experiences as a result of interacting with another item is known as a force. Every time two items touch, a force is exerted on each of objects.

Forces acting and act at a range forces are the two different types of forces. Your daily use of force is evident. Essentially, thrust and pull are forces. You exert force on an object if you push against it or pull against it.

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Yes, but only up to a certain height since the corner cube maintains the angle of the light's incidence. No, as any position of a corner cube sends the beam back to the window. Yes, the beam is parallel to itself when the cube travels slightly.

An optical element having the exceptional capability of returning an entering beam of light straight at its place of origin irrespective of the beam's angle of entry is a corner cube, often called as a retroreflector.

Three adjacent, mutually orthogonal plane-reflecting surfaces make up a corner cube retroreflector, also called as a CCR or trihedral prism, which is an optical structure an angle of a cube. Independent of the prism, the corner cube returns an incident ray at a certain angle.

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The brightness of bulb A will not be affected, but the brightness of bulbs B and D will become brighter.

In the given circuit, the bulbs B, C, and D are connected in series, whereas bulb A is connected in parallel.

All parts of a series circuit are wired together end to end to create a single path for current flow.

A series circuit of incandescent bulbs would have reduced brightness and increased life when a bulb is added. When a bulb is removed from the series combination, the brightness of the other bulbs will increase. Each bulb acts as a resistor reducing the current through each bulb. In a series circuit, when one bulb burns out, it acts as an open switch, and all bulbs go out.

The given question is incomplete as there is no given circuit diagram.

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A fully compressed spring may move at a maximum speed of 4.993 m/s, according to the question.

The pace at which a distance changes over time is referred to as speed. It includes a dimension of time-distance. As a result, the fundamental unit of time and also the basic unit if length are combined to form the High silica content of speed. Thus, the meter per second is the SI unit for speed.

K.E = 3RT/2 - (i)

K.E = mv²/2 - (ii)

eq(i) = eq(ii)

comparing

3RT = mv²

v² = 3RT/m

T/m = 1

v = [tex]\sqrt{3R}[/tex]

v = [tex]\sqrt{3*8.31}[/tex]

4.993m/s

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The direction of acceleration is downward, and its magnitude is 3.27 m/s².

Given:

Tension, T = 2M/3

The gravitational force is exerted downwards, and the tension force is exerted in the opposite direction of the gravitational force.

The total force is given by:

mg - T = ma

The torque is given as:

T × R = MR²/2 × α

Here, α is centripetal acceleration;

The centripetal acceleration is given by:

α = a/R

Substitute values:

MG - 2mG/3 = ma

a = g/3

The direction of acceleration is downward.

a = 9.8/3

a = 3.27 m/s²

Hence, the direction of acceleration is downward, and its magnitude is 3.27 m/s².

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The moment of inertia of  is  the object starting from rest if it has a final velocity of 5.4 m/s is 0.344 M R²

You can then use the energy conservation equation to calculate the moment of inertia of an object that starts at rest and has terminal velocity.

Then Ek1 + Ep1 = Ek2 + Ep2, where

Ek1 = kinetic energy of the object before rolling

Ep1 = potential energy of the object

Ek2 = kinetic energy when the object falls

Ep2 = potential energy of the object below.

Mgh = Iv² +Mv²

         2R²    2

19.6M = 14.58 I + 14.58 M

              R²

5.02 M R² =  14.58 I

0.344 M R² = I

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The force constant of this spring is 16.03N/m; the length of time block is in contact with the spring before it comes to rest is 1.14s; If the force constant of the spring is increased, the time required to stop the block will decrease.

What happens if a block slides on a frictionless horizontal surface?

When a block slides over a horizontal surface without friction, it has kinetic energy. Before coming to rest, it runs into an untensioned spring and compresses it.

The energy is purely kinetic when the mass first comes into contact with the spring. Energy is transformed into spring potential energy as the spring is squeezed. To find the spring force constant, equate the energies. One-fourth of a period is represented by the motion. A stiffer spring will result in a smaller stopping distance for the mass, which will result in a faster stopping time.

(a)Equations: Conservation of energy using a spring.

Ei=Ef

1/2mvo2 = 1/2kA2

 A= vo√m/k

 K is force constant

 k is 16.03N/m

(b) Time

     T = 2∏√m/k

     T = 1.14s

(c) The time to stop reduces as the force constant rises. A stronger spring and a greater stopping force result from a higher force constant, and a shorter stopping time follows. The compressed spring contains all of the energy left behind after the spring has stopped the block. The block will be accelerated back the way it came when the spring pushes back on it. The block will leave the spring moving in the opposite direction from where it first came into contact with it, but at the same speed. Equal to the stopping time is the accelerating time.

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Option (4) is correct: There are ten times more craters on the older highlands than the younger maria.

Who are astronomers?

Astronomers are scientists in the field of astronomy, who focus their research on a particular problem or area outside the realm of the Earth. Observe celestial bodies such as stars, planets, moons, comets, and galaxies either by observation (by analyzing data) or by theoretical astronomy. Examples of topics and fields studied by astronomers include planetary science, solar astronomy, the origin and evolution of stars, and the formation of galaxies. A related but different field is physical cosmology, which studies the universe as a whole.

Therefore, Option (4) is correct: There are ten times more craters on the older highlands than the younger maria.

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Answer:

The particle model, because the metal is only absorbing radiation. of specific frequencies.

Explanation:

When a piece of metal is heated to high temperatures, it begins to glow red, and then white, as its temperature increases.

2725 Watt, is the power needed to change the speed of a sport utility vehicle.

What is the Power?

The quantity of energy moved or converted per unit of time is known as power. The watt, or one joule per second, is the unit of power in the International System of Units also it is a scalar quantity.

What are the Calculations?

Mass of the sport utility = 1600kg

Initial velocity ([tex]v1[/tex]) - 15m/s, ([tex]15^{2}[/tex] = 225)

Final velocity ([tex]v2[/tex]) - 40m/s, ( [tex]40^{2}[/tex] = 1600)

To find the energy required, Kinetic energy (K.E) = [tex]\frac{1}{2}[/tex] [tex]mv^{2}[/tex]

K.E = final K.E - initial K.E

      = [tex]\frac{1}{2} (v2 - v1 )[/tex]

      = [tex]\frac{1}{2} 1600 * (1600 - 225)[/tex]

      = 80 * 1375

      = 11000 J

Power = [tex]\frac{Energy (E)}{Time (t)}[/tex]  = [tex]\frac{1100 joule}{ 4 sec}[/tex]

           = 2725Watt.

Hence, 2725Watt, is the power needed to change the speed.

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Finally, Joe turns his attention to the mountain in the distance but finds that he cannot bring the mountain into focus. This is because he is nearsighted. But when Joe puts on his glasses, he can see the mountain clearly. To adjust for his nearsightedness, his glasses must contain diverging lenses.

An image of an object is formed when its light rays are recorded on the retina. The lenses in the eye focus these rays onto the retina, making the phenomenon of vision possible.

Joe is nearsighted- or in other words, he has myopia. Myopia is a defect in normal vision where one cannot see nearby objects clearly and distinctly.

The image of the mountain (distant object) is formed at the retina properly for Joe as he can see distant objects distinctly. But for nearby objects, he cannot see clearly, because he suffers from nearsightedness.

The inability of the ciliary muscles to arbitrarily increase or decrease the focal length of the lens results in nearsightedness and farsightedness. Therefore, his eyes must be able to focus on the image of the distant mountains naturally due to the corrective diverging lenses.

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54.85N is the tension in a string of mass [tex]$0.83 \mathrm{~kg}$[/tex] and length [tex]$12.29 \mathrm{~m}$[/tex] if when plucked, waves travel at [tex]$28.5 \mathrm{~m} / \mathrm{s}$[/tex].

What is  tension?

In physics, tension is defined as the pulling force that is transmitted axially by a string, rope, chain, or similar object, or by each end of a rod, truss member, or other comparable three-dimensional object. Tension can also be defined as the action-reaction pair of forces acting at each end of the aforementioned elements.

[tex]\begin{aligned}& m=0.83 \mathrm{~kg} \\& l=12.29 \mathrm{~m} \\& v=28.5 \mathrm{~m} / \mathrm{s} \\& \mu=\frac{m}{l} \\& v=\sqrt{\frac{T}{\mu}}\end{aligned}[/tex]

[tex]\Rightarrow T=v^2 \mu=[/tex] [tex](28.5)^2 \times \frac{0.83}{12.29}[/tex] = [tex]54.85 \mathrm{~N}[/tex]

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The term "Group factor" refers to an intermediate class of factors that are shared by a group of activities but not by all in the context of Spearman's two-factor theory of intelligence.

What is the two-factor hypothesis of intelligence proposed by Spearman?

According to Spearman's two-factor theory, intelligence is divided into two categories: general intelligence ("g") and specific ability ("s"). Spearman proposed that the "s" component was particular to a certain feature of intelligence to account for variations in performance on various tasks.

What exactly are general and specific factors?

The "g" factor is associated with general aptitude, whereas the "s" factor is associated with aptitude for a particular task. The "s" factor is in charge of identifying a person's specific abilities, while the "g" factor measures a person's capacity for performing tasks requiring general mental ability.

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The automobile accelerates at 0.9g.

acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight forward is accelerated when it accelerates or decelerates.

1 g equals 9.8 m/s2, as we are aware. Therefore, we must first convert 100 km/h to m/s. So that we may compare units like meters and seconds (we'll deal with s2 eventually).

100 km/h * (1 km) * (1 hour) * (1 minute) * (60 sec) = 27.778 m/s.

We are aware that this automobile needed three seconds to reach this speed. Since acceleration is a change in velocity, and we know that the car's velocity increased from 0 to about 27.8 m/s in 3 seconds, we may calculate the accelerationknow that our acceleration is:

(27.8 m/s - 0 m/s) / (3 s) = 9.259 m/s^2.

Now to find how many g’s, we know that 1 g = 9.8 m/s^2.

So we know that 9.259 m/s^2 / 9.8 m/s^2 = about 0.9. This will be 0.9 of 1 g, which is just 0.9 g.

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Answer: 0.0016 kg

Explanation: To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of moles of gas present. The ideal gas law is given by the following equation: PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas present, R is the ideal gas constant, and T is the temperature of the gas.

In this case, we are given the pressure of the gas inside the tire (760 kPa and 550 kPa), the temperature of the gas (20 °C), and the diameter and circumference of the tire (2.2 cm and 2.0 m). We need to use this information to calculate the mass of air that was lost from the tire.

First, we need to find the volume of the gas inside the tire. Since the circumference of the tire is 2.0 m and the diameter is 2.2 cm, we can use the formula for the circumference of a circle to calculate the radius of the tire:

C = 2 * pi * r

2.0 m = 2 * pi * r

r = 0.3 m

Next, we can use the formula for the volume of a circle to calculate the volume of the gas inside the tire:

V = pi * r^2 * h

V = pi * (0.3 m)^2 * 2.2 cm

V = 0.0183 m^3

Now that we have the volume of the gas inside the tire, we can use the ideal gas law to calculate the number of moles of air inside the tire:

PV = nRT

n = PV / RT

n = (760 kPa * 0.0183 m^3) / (8.314 J/molK * 293 K)

n = 0.0059 mol

Next, we can use the ideal gas law to calculate the number of moles of air lost from the tire when the pressure decreased from 760 kPa to 550 kPa:

n = PV / RT

n = (550 kPa * 0.0183 m^3) / (8.314 J/molK * 293 K)

n = 0.0043 mol

Finally, we can use the molar mass of air to calculate the mass of air that was lost from the tire:

m = n * M

m = (0.0059 mol - 0.0043 mol) * 28.97 g/mol

m = 0.0016 kg

Therefore, the mass of air lost from the tire when the pressure decreased from 760 kPa to 550 kPa is 0.0016 kg.

Through p3, a light of equal power (60 W) is transmitted.

The angle between the direction of polarization of the light and the polarizer is, where I and Io are the transmitted and incident intensities, respectively.

Half of the unpolarized light that enters the first polarizer (I0 I1 = 12 300 I1 = 150 W) is transmitted.

I2 = I1 cos2 1 I2 = 150 cos2 45 I2 = 75 W is the transmission through the second polarizer (P2).

I3 = I2 cos2 2 I3 = 75 cos2 30 I3 = 56.25 W is the third polarizer's (P3) transmission.

The closest response, A, corresponds to 60 W and represents the light that the 56 W system transmits.

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Easy student solutions manual a 5.00-kg ball, moving to the right at a velocity of 2.00 m/s on a frictionless table, collides head-on with a stationary 7.50-kg ball. find the final velocities of the balls if the collision is

(a)elastic and (b)completely inelastic

An act or instance of colliding : conflict. : an encounter among particles (which includes atoms or molecules) resulting in alternate or transformation of strength. collisional.

Velocity (vɪˈlɒsɪtɪ) / noun plural -ties. pace of movement, action, or operation; rapidity; swiftness. physics a measure of the fee of movement of a frame expressed because the rate of exchange of its function in a specific route with time.

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