a stuntman is being pulled along a rough road at a constant velocity, by a cable attached to a moving truck. the cable is parallel to the ground. the mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. find the tension in the cable.pdf

6P₀V₀ is the work done by the gas during process 1→3→6.

The work performed by a gas expanding against an external pressure is negative, analogous to the work performed by a system on its surroundings. In contrast, when a gas is compressed externally, V₀ and the work is positive because the environment exerts force on the system.

The gas must release some energy into the environment when it expands in response to an external pressure. The overall energy of the gas is thus reduced as a result of the negative work. When a gas is compressed, energy is transferred to it, increasing its energy as a result of positive work.

Let's consider,

Work done = -W (As the energy is released)

External pressure = P₀

Volume of gas = V₀

So, the work done can be calculated as follows:

W = P x (dV)

-W = P₀ × V₀

Now according to the process given:

1→3→6

Energy flow increases up to 6.

Hence, the energy of the system (work done) is:

-W = 6 × P₀ × V₀

-W = 6P₀V₀

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The theoretical Coefficient of Performance (COP) to cool the room A is 33.52.

Temperature in room A = T₁ = 18.6 oc

Temperature in room B = T₂ = 27.3 oc

To convert temperatures in Kelvin,

= T₁ = 18.6 + 273 =

= T₁ = 291.6 K

and,

=  T₂ = 27.3 + 273

=  T₂ = 300.3 K

The COP value to cool room A =

= COP = T₁ / (T₂ - T₁)

= COP = 291.6  / (300.3 - 291.6 )

= COP = 291.6 / 8.7

= COP = 33.52

The theoretical coefficient of performance to cool room A is 33.52 when the temperature of room A is 18.6 oc and that of room B is 27.3 oc.

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Linear equation for team a; d = 6t + 2 and Linear equation for team b;

d = 6t.

To find speed, divide the distance of the journey by the time it took to travel. To calculate the time, divide distance by speed and to get the distance, multiply  speed by time.

Given that in a canoe race, team a is traveling 6 miles per hour and it is 2 miles ahead of team b.

Team b is traveling 6 miles per hour. The teams continue the traveling at their current rates for the remaining race.

We know that speed= distance/time

Linear equation for team a;

d = 6t + 2

Linear equation for team b;

d = 6t

d is distance and t is time.

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Based on the information provided regarding the parallel circuit, reasonable observation about the brightness of the lightbulbs would be that when each lightbulb is added, it is the same brightness as the first lightbulb, which remains the same brightness throughout. (Option C)

A parallel circuit refers to a circuit that comprises of branches so that the current divides and only part of it flows through any branch. The voltage, or potential difference, across each branch is the same. Two or more bulbs in a simple parallel circuit have full voltage of the battery, hence the brightness of each bulb remains unchanged.

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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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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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(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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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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Daniel G. A=v2r calculates centripetal acceleration. The recommended minimum friction coefficient is 0.4.

Of course, the car won't slide if the coefficient of friction is higher than 0.141; this is the minimum value. The frictional force is parallel to the velocity, as you can see.

A car needs a centripetal force of mV2/R, where m is the mass of the vehicle, V is the speed, and R is the radius of the circular portion of road, in order to maintain its position on the road without skidding. Fs/Fl = 4mV2/R / 2mV2/R is obtained by dividing Fs by Fl.

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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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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.

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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Answer:

The frequency of light beam is 4.74 x 10^(14) s^(-1)

Explanation:

In this problem we use the formula C=F x ( lamda )

here c= the speed of light

f =the frequency of light beam

lamda = wavelength

c=3 x 10^8 m/s ( universally same)

lamda = 633nm = 633 * 10^(-9) m

Therefore f = c/ ( lamda )

or; f = 3 x 10^8 /  633 * 10^(-9)

f= 4.74 x 10^(14) s^(-1) .

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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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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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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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It will take 4.24 hours for a sample of titanium-45 to decay from 65. 0 grams to 25. 0 grams.

Half life of first order reaction is given by: [tex]t_{1/2}[/tex] = 0.693/k

Given half life is 3.08 hours.

So k = 0.225

Rate of first order reaction is given by: [tex]kt = 2.303 * log(R'/R)[/tex]

Where k = rate of reaction = 0.225

t= time taken

R' = final amount = 25g

R = initial amount =65g

Putting these values we get t=4.24 hours.

So it will take 4.24 hours for a sample of titanium-45 to decay from 65. 0 grams to 25. 0 grams.

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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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James braid, a physician, is credited with coining the term hypnosis after becoming mesmerized while traveling. Hence, the relevant option is option (b).

James Braid was born on June 19, 1795, and passed away on March 25, 1860. He was a well-known Scottish surgeon who was also a gentleman scientist and natural philosopher. According to Braid, hypnotic suggestion can be an effective treatment for nervous system functioning problems. Comparative anatomy has been defined as the study of similar body features in various animal species in order to suggest or comprehend modifications that have to occur underground during the course of evolution from very similar predecessors.

The research that compares the structural differences between various animal species in order to comprehend the changes that each species went through during the stage of evolution from their common ancestors. James braid, a physician, is credited with coining the term hypnosis after being mesmerized while traveling.

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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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The diffraction-induced minimum angular spread is 633 10 9 meters.

When the hole is illuminated by a parallel ray of light of wavelength lambda, it is believed that the dispersal of the spot (gotten on the left wall of the video) equals the sum of its geometrical dispersal and the spread induced by diffraction.

Diffraction is the term used to describe the spreading out of waves when they pass through an apertures or around objects. It occurs when the aperture's or obstruction's size and the incident wave's wavelength are of same magnitude. At very small aperture diameters, the most of the wave is blocked.

Briefing : Minimum angular spread due to diffraction (θ) =  λ/b

Given.

Wavelength = 633nm = 633×10⁻⁹m

Beam diameter is 1mm.

Minimum angular spread due to diffraction (θ) =  633×10⁻⁹m/1

Minimum angular spread due to diffraction (θ) =  633×10⁻⁹m.

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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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The ball is seen to be speeding up as the distance increases with time.

We know that speed is the ratio of the distance to the time that is taken by the object. We know that when an object is speeding up, the distance of the object is going to increase linearly with the time taken.

If we look at the graph that has been shown in the image that is attached to this answer, we would notice that the distance that is covered is increasing with time.

The speed can then we obtained from;

Speed = [tex]y_{2} - y_{1}/x_{2} - x_{1}[/tex]

= 80 - 0/6.5 - 0

= 12.3 m/s

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The centripetal acceleration is [tex]1.58m/s^2[/tex]

What is centripetal acceleration?

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any object moving in a circle with an acceleration vector pointing in the direction of the circle's center.

It is calculated by

ac = rω^2

where

ac = centripetal acceleration

r = radius

ω = angular velocity

Given,

ω = 4 revolution,  

diameter = 18m so, radius r = 9m

Subsituting the values in equaion

ac =  [tex][9m] [[4.0\frac{rev}{min} ][\frac{2\pi rad}{rev} ][\frac{1 min}{60s} ]]^2[/tex]

ac = [tex]1.58m/s^2[/tex]

The centripetal acceleration is [tex]1.58m/s^2[/tex]

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The first equation will be used  to determine that it would  hit the ground  after 1.0s of falling (in 1.56 s).

Equation of motion are a mathematical formula that describes the position, velocity or acceleration of a body relative to a given frame of reference

The first equation to be used will be second equation of motion

x = v₀t + 1/2at²

where, x = distance

a = uniform acceleration

t = time

12 = 1/2*9.8 *t²

24 = 9.8 * t²

t = 1.56 s.

The balloon will hit the ground after 1.0s of falling.

The 1st equation was used because of the values provided which included the distance and the acceleration due to gravity. The 2nd equation has final velocity and initial velocity which couldn't be compared from the data given.

The given question is incomplete complete question is:

A student holds a water balloon outside of an open window and lets go. The window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:

x=x0+v0t+12at2

v=v0+at

Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and what comparison did you make to determine that it would or would not hit the ground by then?

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The debroglie wavelength in picometers (10-12m) of an electron that is traveling at 79.1% of the speed of light is  2.8 × [tex]10^{-12}[/tex]

Mass of electron m = 9.1 × [tex]10^{-31}[/tex]kg

speed of electron v = 3 ×[tex]10^{8}[/tex] m/s * (79.1/100) = 2.37 × [tex]10^{8}[/tex] m/s

debroglie wavelength λ = h/mv

λ = 6.626 [tex]10^{-34}[/tex]/9.1 × [tex]10^{-31}[/tex] × 2.37 × [tex]10^{8}[/tex]

λ  = 2.8 × [tex]10^{-12}[/tex]

In many branches of physics, the speed of light in a vacuum, commonly abbreviated as c, plays a crucial role. Exactly 299,792,458 meters per second is the speed of light, or c. [Note 3] The speed at which common matter or energy can travel through space (and, consequently, any signal carrying information) is limited to c, according to the special theory of relativity.

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Wave passing through a single slit  it becomes twice as width of the central maximum of its diffraction pattern as the slit is made half as wide.

The central maximum lies between the first-order minima defined by the relation sinθ dark =mλ/a=λ/a. Because the angle is small, sinθ dark  ≈tanθ dark  =y dark/L, so the width of the central maximum is proportional to Lλ/a. Thus, the central maximum becomes twice as wide if the slit width a becomes half as wide.

In order to enter the region of geometric shadow cast by the obstruction or aperture, waves must interact with one another or bend around its edges, which is referred to as diffraction. A secondary source of the wave that is propagating is in fact created by the diffracting item or aperture. The term "diffraction" was originally used in 1660 by the Italian scientist Francesco Maria Grimaldi, who was also the first to make precise observations of the phenomena.

On the registration plate, an infinite number of spots (three are illustrated) along length d project phase contributions from the wave front, resulting in a constantly changing intensity.

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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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(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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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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