The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 37.8 ~\text{mT}37.8 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 40.640.6 km/s to be undeflected. Give your answer in units of kV/m.

Answers

Answer 1

Answer:

50k/h is the answer to iy


Related Questions

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you predict about the surface elevation for 50 km thick crust with an average density of 2.8 g/cm3

Answers

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

a small bar magnet is suspended horizontally by a string. When placed in a uniform horizontal magnetic field, it will

Answers

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor? Explain your answer.

Answers

Answer:

Explanation:

Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it.  if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it

For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.

Answers

Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J

Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.

Answers

..........................................................

The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².

What is the Radiation pressure?

Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.

Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.

E.g: Black-body radiation

Given the values are,

Wattage of bulb = W = 25 W

distance = d = 6.5 cm = 0.065 m

To know the Radiation Pressure,

It can be given by

P = I/c

Where, c = 299792458 m/s is the speed of light,

I is the intensity of radiation and given by

I = W/4πd²

Where W is the Wattage of bulb and d is the distance

I = 25/4π*0.065²

I = 470.872 w/m²

so, the radiation pressure becomes

P = 470.872/299792458

P = 1.5707x10⁻⁶ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²

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A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.

Answers

Answer:

e is correct

Explanation:

When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.

What is buoyant force?

The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.

The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.

When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.

Therefore, options (d), (e) are correct.

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A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.

Answers

Answer:

4.9x10^-6T

Explanation:

See attached file

a uniform rod of 30cm is pivoted at its center.a 40N weight is hung 5cm from left.from where 50N weight be hung to maintain equilibrium?

Answers

Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is

Answers

Answer:

The magnetic field's direction is towards the north

Explanation:

The force on a positive charge in a uniform magnetic field is represented by the right hand rule. To determine the direction of the force, place your right hand with your palm up, and your thumb at 90° to the other fingers. If the fingers represent the magnetic field, and the thumb the direction of the positive charge, then the palm will push up in the direction of the force. But a negative charge like an electron pushes in exactly the opposite direction. So if you follow this rule, you will find that the magnetic field points towards the north.

The direction of the magnetic field is towards the North. This can be

determined using the right hand rule by Fleming.

The right hand rule states that to determine the direction of the magnetic

force, the right thumb should be pointed in in the direction of the velocity,

index finger in the direction of the magnetic field and middle finger in the

direction of magnetic force.

When this is applied, we will discover that the index finger will point towards

the north region.

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A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.5 ss later.
How deep is the ocean at this point? (Note: Use the bulk modulus method to determine the speed of sound in this fluid, rather than using a tabluated value.)
_____ m

Answers

Answer:

1248m

The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s

The speed of sound in seawater is 1560 m/s

Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km

In practice, a good insulator In practice, a good insulator A. slows heat flow. B. speeds negative heat flow. C. stops heat flow. D. all of the above

Answers

Answer:

The answer is  A. slows heat flow.

Explanation:

 An insulator is a material that impedes the movement of heat or electric current from flowing.

Theoretically good heat insulators stops the movement of heat, while practically this insulation can only be slowed down.

Hence from the options listed the correct answer practically is

 A. slows heat flow.

An apple falls from a tree and hits your head with a force of 9J. The apple weighs 0.22kg. How far did the apple fall?

Answers

Answer:

The apple fell at a distance of 4.17 m.

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.  

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.  

Work=Force*distance* cosine(angle)

On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:

F=m*a

where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]

In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:

Work= 9 JF=m*a=0.22 kg*9.81 m/s²= 2.1582 Ndistance= ?angle=0 → cosine(angle)= 1

Replacing:

9 J= 2.1582 N* distante* 1

Solving:

[tex]distance=\frac{9J}{2.1582 N*1}[/tex]

distance= 4.17 m

The apple fell at a distance of 4.17 m.

Which examples are simple machines?
Select all correct answers.
a hammer
an automobile
O a pulley
an inclined plane

Answers

A hammer and a pulley

A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a location a perpendicular distance of 0.053 m from the wire due to the current in the wire

Answers

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2

Answers

Answer:

650W/m²

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe

Answers

Answer:

It becomes wider

Explanation:

Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?

Answers

Answer:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

            [tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

   

The Hermes spacecraft is traveling at 0.1c(1/10 the speed of light past Mars and shines a laser in front of the ship. You would see the light traveling at c (the speed of light )away from your ship. According to Einstein's special relativity how fast with a person on Mars observe the light to be traveling?

Answers

Answer:

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

Explanation:

The special theory of relativity has two main postulates:

1- VALIDITY OF PHYSICAL LAWS

The laws of physics such as Newton's Laws and Maxwell's Equations are valid in all inertial frame of references.

2- CONSTANCY OF SPEED OF LIGHT

The speed of light in vacuum is the same for all observers in uniform translational relative motion, and it is independent of the motion of the source or the observer. Thus, speed of light is a universal constant and its value is c = 3 x 10⁸ m/s.

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors

Answers

Answer:

Current that flows through any one of the resistors has a value of 12 amperes.

Explanation:

When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:

[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]

[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]

Where:

[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.

[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.

[tex]i[/tex] - Current, measured in amperes.

If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:

[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]

Current that flows through any one of the resistors has a value of 12 amperes.

The current flows via any of the resistors should have a value of 12 amperes.

Ohm law:

At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.

Here the circuit should be

vbatt = i.(R1 + R2+ R3)

i = Vbatt/R1 + R2 + R3

here

Vbatt means the voltage of the battery

R1,R2, and R3 means the resistance of the first, second and third resistors

I means the current

So, in the case when

R1 = R2 = R3 = R

So,

i = Vbatt/3.R

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An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it

Answers

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0

So,

[tex]d=\dfrac{1}{2}(v-u)t[/tex]

[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]

Now applying the conservation of energy i.e.

[tex]\dfrac{1}{2}mv^2=qV[/tex]

V is voltage

[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]

So, the voltage is 451.47 V.

What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 275 km above the Earth

Answers

Answer:

   s_400 = 16.5 m , s_700 = 29.4 m

Explanation:

The limit of the human eye's solution is determined by the diffraction limit that is given by the expression

                   θ = 1.22 λ / D

where you lick the wavelength and D the mediator of the circular aperture.

In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.

by introducing these values ​​into the formula

                 

λ = 400 nm      θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad

λ = 700 nm     θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad

Now we can use the definition radians

          θ= s / R

where s is the supported arc and R is the radius. Let's find the sarcos for each case

λ = 400 nm       s_400 = θ R

                         S_400 = 6 10⁻⁵ 275 10³

                         s_400 = 16.5 m

λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³

                          s_700 = 29.4 m

Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed

Answers

Answer:

The number of turns of the wire needed is 243 turns

Explanation:

Given;

self inductance of the solenoid, L = 1.8 mH

cross sectional area of the inductor, A = 1.6 x 10⁻³ m²

length of the inductor, l = 0.066 m

The self inductance of long solenoid is given by;

L = μ₀n²Al

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

n is number of turns per length

A is the area of the solenoid

l is length of the solenoid

[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]

The number of turns is given by;

N = nL

N = (3682.74)(0.066)

N = 243 turns

Therefore, the number of turns of the wire needed is 243 turns

The cost of buying shirts is partly constant and partly varies with the number of shirts bought. When the number of shirts is 5 the cost is #240, also, 10 shirts costs #400. find the cost when 300 shirts were bought ​

Answers

Answer:

The cost of the buying the shirts is #9680

Explanation:

let the cost of buying shirt = C

let the number of shirt bought = N

The following equation can be generated based on the statement above;

C = k + Nb

When the cost, C = #240, the number of shirt = 5

240 = k + 5b ------ equation (1)

where;

k and b are constants

When the cost, C = #400, the number of shirt = 10

400 = k + 10b ------ equation (2)

From equation (1), make k the subject of the formula;

k = 240 - 5b ---- equation (3)

Substitute in the value of k into equation (2)

400 = k + 10b

400 = (240 - 5b) + 10b

400 = 240 - 5b + 10b

400 - 240 = -5b + 10b

160 = 5b

b = 160 / 5

b = 32

From equation (3), calculate k

k = 240 - 5b

k = 240 -5(32)

k = 240 - 160

k = 80

When the number of shirts bought = 300, the cost of the buying the shirts =

C =  k + Nb

C = 80 +32N

Where;

N is the number of shirts

C = 80 + 32(300)

C = 80 + 9600

C = #9680

Therefore, the cost of the buying the shirts is #9680

Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assume the orbits are circular, and estimate the mass of the object, in units of the mass of the sun (MSun = 2x1030 kg). If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.

Answers

Answer:

The mass of the object is 745000 units of the sun

Explanation:

We know that the centripetal force with which the stars orbit the object is represented as

[tex]F_{c}[/tex] = [tex]\frac{mv^{2} }{r}[/tex]

and this centripetal force is also proportional to

[tex]F_{c}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

where

m is the mass of the stars

M is the mass of the object

v is the velocity of the stars = 10^6 m/s

r is the distance between the stars and the object = 10^14 m

k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

We can equate the two centripetal force equations to give

[tex]\frac{mv^{2} }{r}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

which reduces to

[tex]v^{2}[/tex] = [tex]\frac{kM}{r}[/tex]

and then finally

M = [tex]\frac{rv^{2} }{k}[/tex]

substituting values, we have

M = [tex]\frac{10^{14}*(10^{6})^{2} }{6.67*10^{-11} }[/tex] = 1.49 x 10^36 kg

If the mass of the sun is 2 x 10^30 kg

then, the mass of the the object in units of the mass of the sun is

==> (1.49 x 10^36)/(2 x 10^30) = 745000 units of sun

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars

Answers

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.

Answers

Answer:

I think the answer probably be B

Answer :QUESTION①)

What explains why a prism separates white light into a light spectrum ?

C. The different colors that make up a white light have different refractive indexes in glass.

✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).

✔ That's why purple is more deflected so is lower than red radiation.  

A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe

Answers

Answer:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be

Answers

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.

[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.

Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:

[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]

And final speed is now calculated after clearing it:

[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]

[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which of the following statements regarding the circuit are correct?
a) The capacitor charges to its maximum value in one time constant and the current is zero at that time.
b) The potential difference across the resistor and the potential difference across the capacitor are always equal.
c) The potential difference across the resistor is always greater than the potential difference across the capacitor.
d) The potential difference across the capacitor is always greater than the potential difference across the resistor
e) Once the capacitor is essentially fully charged, There is no appreciable current in the circuit.

Answers

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J

Answers

Answer:

Explanation:

v₁² = v₁ . v₁

= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )

= 12.25 + 8.41

= 20.66 m /s

a ) kinetic energy = 1/2 m v₁²

= 1/2 x .7 x 20.66

= 7.23 J

b )

changed velocity v₂ = v₂.v₂

= (6i - 5 j ) . (6i - 5 j )

= 36 + 25

= 61 m /s

kinetic energy = 1/2 m v₂²

= 1/2 x .7 x 61

= 21.35 J

Work done = change in energy

= 21.35 - 7.23

= 14.12 J .

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