A single slit 1.5 mm wide is illuminated by 420- nm light. Part A What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.5 m away

Answers

Answer 1

Answer:

The width is [tex]w_c = 0.00252 \ m[/tex]

Explanation:

From the question we are told that

  The  width of the single slit is  [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 420 *10^{-9} \ m[/tex]

   The distance of the screen is  [tex]D = 4.5 \ m[/tex]

Generally the width of the central maximum is  

        [tex]w_c = 2 * y[/tex]

where y is the width of the first maxima which is mathematically represented as

       [tex]y = \frac{\lambda * D}{a}[/tex]

=>   [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]

=>  [tex]y = 0.00126 \ m[/tex]

So

    [tex]w_c = 2 *0.00126[/tex]

    [tex]w_c = 0.00252 \ m[/tex]


Related Questions

White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (λ = 640 nm) appear in first order?

Answers

Answer:

3.67°

Explanation: Given that λ=640nm , m = 1

Considering the slit separation

d = 1cm/1000

= 1.000×10^-3cm

= 1.000×10-5m

We then have

Sinθ = mλ/d

Sinθ= (1×640×10^-9)/1.000×10-5m

Sinθ = 0.064

θ= sin-1 0.064

θ= 3.669°

= 3.67°

When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass

Answers

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits. Find the distance along the screen between adjacent bright fringes.

Answers

Answer:

The distance is [tex]y = 0.00162 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]

   The  slit spacing is  [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 2.0 \ m[/tex]

 Generally the distance along the screen between adjacent bright fringes is mathematically represented as  

        [tex]y = \frac{\lambda * D}{d}[/tex]

substituting values

         [tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]

         [tex]y = 0.00162 \ m[/tex]

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

Answers

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]

where [tex]M_{0}[/tex] is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]

where

[tex]\beta = V/c[/tex]

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

the equation reduces to

1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]

square both sides of the equation, we have

3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]

3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1

2.0625 = 3.0625[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 0.67

β = 0.819

substitute for  [tex]\beta = V/c[/tex]

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

When a magnet moves above a conducting ladder, the currents induced in the ladder produces a magnetic field. This field interacts with the magnetic field of the magnet to produce a force that A) pulls the ladder in the direction opposite to the direction of the moving magnet. B) pulls the ladder in the same direction as the moving magnet. C) brakes the ladder. D) None of the above.

Answers

Answer:

) pulls the ladder in the direction opposite

Explanation:

This is in line with lenz law that states that the magnetic field induced in a conductor act to oppose the magnetic field that produced it

A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.

Answers

Answer:

5.29×10^-7

Explanation:

shear stress τ = F/ A

shear deformation δ = (VL)/ (AG)

= (τL)/ G

V=shear force

L=height of disk=6.50×10^-2

A=cross sectional area

G= shear modulus= (1.60x10^9N/m^2)

A=πd^2/4

Then substitute the values we have

4×(375N)(0.00750m)

________________ = δ

(π*0.00650^2)(1.60x10^9N/m^2)

= 5.29×10^-7

Using a conventional two-slit apparatus with light of wavelength 605 nm, 34 bright fringes per centimeter are observed on a screen 3.1 m away. What is the slit separation

Answers

Answer:

d = 6.38 x 10⁻³ m = 6.38 mm

Explanation:

Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:

Fringe Spacing = Δx = 1 cm/34

Δx = 0.0294 cm = 2.94 x 10⁻⁴ m

But, the formula for fringe spacing in a double slit experiment is:

Δx = λL/d

where,

λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m

L = Distance between screen and slits = 3.1 m

d = slit separation = ?

Therefore,

2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d

d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)

d = 6.38 x 10⁻³ m = 6.38 mm

In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window

Answers

Answer:

Post indicator valve

Explanation:

Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.

3 QUESTIONS PLEASE ANSWER!

Answers

1. Meters (m)
2. Liters (L)
3. Gm/cm3

Answer:

1. A

2. C

3. D

Explanation:

Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequency of 544 Hz. The other cyclist hears the frequency as 563 Hz. If the speed of sound in air is 344 m/s, what is the speed of the motorcycles

Answers

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

Difference between matter and energy

Answers

Answer:

Energy is the strength and vitality required for sustained physical or mental activity.

Matter occupies space and possesses rest mass, especially as distinct from energy.

Hope this helps! (づ ̄3 ̄)づ╭❤~

In _____ research, a group of people of one age is compared to a group of people who are another age.

Answers

Answer:

cross-sectional

Explanation:

The full definition of this is ''a research design in which several different age-groups of participants are studied at one particular point in time.''

Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 × 105 m/s. What is the work function, in eV of the metal?

Answers

Answer:

12.45eV

Explanation:

Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.

Mathematically, KE =  hf - Ф where;

h is the Planck constant

f is the frequency = c/λ

c is the speed of light

λ is the wavelength

Ф is the work function

The formula will become KE =  hc/λ - Ф. Making the work function the subject of the formula we have;

Ф = hc/λ - KE

Ф = hc/λ - 1/2mv²

Given parameters

c = 3*10⁸m/s

λ = 97*10⁻⁹m

velocity of the electron v = 3.48*10⁵m/s

h = 6.62607015 × 10⁻³⁴

m is the mass of the electron = 9.10938356 × 10⁻³¹kg

Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²

Ф =  6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ -  1/2*9.11*10⁻³¹(3.48*10⁵)²

Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰

Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹

Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷

Ф = 0.1995*10⁻¹⁷Joules

Since 1eV = 1.60218*10⁻¹⁹J

x = 0.1995*10⁻¹⁷Joules

cross multiply

x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹

x = 0.1245*10²

x = 12.45eV

Hence the work function of the metal in eV is 12.45eV

The speed of sound through air is 340 m/s. If a person hears the clap of thunder 9.6 s after seeing the bolt of lightning, how far away is the lightning?

Answers

Explanation:

Distance = speed × time

d = (340 m/s) (9.6 s)

d = 3264 m

At what minimum angle will you get total internal reflection of light traveling in diamond and reflected from ethanol? °

Answers

Answer:

34°

Using the relation

θᶜ = sin^-1(n₂/n₁),

where n1= the refractive index of light is propagating from a medium

And n2 = refractive index of medium into which light is entering

So we know that

refractive index of diamond at 589nm = 2.41= n₁

refractive index of ethanol at 589nm and 20°C = 1.36= n₂

Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°

Explanation:

1.) When the acceleration is zero, what can you say about the velocity of an object?

Answers

Answer:

it is either constant or zero

Explanation:

Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.

Answers

Answer:

A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.

Figure (3) shows a car travelling along the route PQRST in 30 minutes. What is the average speed of the car in km/hour?

Answers

Answer:

60 km/hour.

Explanation:

We'll begin by calculating the total distance traveled by the car. This is illustrated below:

Total distance traveled = sum of distance between PQRST

Total distance = 10 + 5 + 10 + 5

Total distance = 30 km

Next, we shall convert 30 mins to hour. This can obtained as follow:

Recall:

60 mins = 1 hour

Therefore,

30 mins = 30/60 = 0.5 hour.

Finally, we shall determine the average speed of the car as follow:

Distance = 30 km

Time = 0.5 hour

Speed =?

Speed = distance /time

Speed = 30/0.5

Speed = 60 km/hour

Therefore, the speed of the car is 60 km/hour.

What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round window

Answers

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

A convex spherical mirror has a radius of curvature of magnitude36.0cm.

(a) Determine the position of the virtual image and the magnification for object distances of25.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(b) Determine the position of the virtual image and the magnification for object distances of47.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(c) Are the images in parts (a) and (b) upright or inverted?
The image in part (a) is---Select---uprightinverted
The image in part (b) is---Select---uprightinverted

Answers

Answer:

Explanation:

a )

focal length of convex spherical mirror

f = 36/2  cm = 18 cm

object distance u = - 25 cm

mirror formula

[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} + \frac{1}{- 25} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{18}[/tex]

v = 6.28  cm .

It is positive hence the image will be erect / upright  and formed on the back of the mirror.

For object distance of 47 cm

u = - 47 cm

Putting the values in the mirror formula

[tex]\frac{1}{v} + \frac{1}{- 47} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{ 47} + \frac{1}{18}[/tex]

v = 13  cm

It is positive hence the image will be erect / upright and formed on the back of the mirror.

The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.

Answers

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

             

When an LRC series circuit is driven at resonance, which of the following statements about the circuit are correct? (There may be more than one correct choice.)a) The impedance of the circuit has its minimum value.b) The inductive reactance and the capacitive reactance are exactly equal to each other.c) The impedance of the circuit is zero.d) The inductive reactance and the capacitive reactance are both zero.e) The impedance of the circuit has its maximum value.

Answers

Answer:

Answers a) and b) should be marked as correct.

Explanation:

Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance  (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:

[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]

as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.

Therefore, answers:

a) "The impedance of the circuit has its minimum value."

and

b) "The inductive reactance and the capacitive reactance are exactly equal to each other."

are correct answers.

(a) The impedance of the circuit has its minimum value.

(b) The inductive reactance and the capacitive reactance are exactly equal to each other

LRC series circuit consists of inductor, resistor and capacitor is series.

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]

where;

R is the resistance[tex]X_C[/tex] is the capacitive reactance[tex]X_L[/tex] is the inductive reactance

The impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.

[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]

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The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm

Answers

Answer:

The maximum force that can be lifted by this system is  51,478.4 kg

Explanation:

Given;

maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²

diameter of the piston, d = 65 cm = 0.65 m

The maximum gauge pressure of the piston is given as;

[tex]Hp = \frac{F}{A}[/tex]

Where;

F is the maximum force of the piston

A is the area of the piston

[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]

F = Hp x A

F = 1.52 x 10⁶N/m² x 0.3319m²

F = 504488 N

Force is given as;

F = mg

m = F/g

m = 504488/9.8

m = 51,478.4 kg

Therefore, the maximum force that can be lifted by this system is  51,478.4 kg

If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength

Answers

Answer:

0.09 x10^-10m

Explanation:

Using wavelength=( 12.27 A)/√V

= 12.27 x 10^-10/ √1.6x10^2

= 0.09x10^-10m

1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?

Answers

Explanation:

1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.

2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.

Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.

We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.

These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.

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: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0

Answers

Answer:

145060km

Explanation: Given that

speed = dx/dt = 2t^2 +1

integrate

x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)

given that at t=0, x = 1000

so 1000 = 2/3 X (0)^3 + 0 + c

or c = 1000

So x = 2/3t^3 + t + 1000

for t = 1 min = 60s

x = 2/3 X 60^3 + 60 + 1000

x = 2/3×216000+ 1060

x = 144000+1060

= 145060km

At one minute, it will be 145060km far from the earth

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.


Required:

What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Answers

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is [tex]v = 80.82 \ m/s[/tex]

Explanation:

From  the question we are told that

   The magnitude of charge one is  [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]

   The  magnitude of charge two  [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]

   The distance of separation is   [tex]d = 58.0 \ cm = 0.58 \ m[/tex]

 

Generally the electric potential of the electron at the midway point is mathematically represented as

         [tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]

substituting values

         [tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]

          [tex]V = 1.8103 *10^{-8} \ V[/tex]

Now when the electron is 10 cm   =  0.10 m  from charge 1 , it is  (0.58 - 0.10 =  0.48 m ) m from charge two

Now the electric potential  at that point is mathematically represented as

       [tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]

 substituting values

      [tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]

      [tex]V_1 = 3.67*10^{-8} \ V[/tex]

Now the law of energy conservation ,

   The  kinetic energy of the electron  =  potential energy of the electron

i.e     [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]

where q is the magnitude of the charge on the electron with value

      [tex]q = 1.60 *10^{-19} \ C[/tex]

While m is the mass of the electron with value  [tex]m = 9.11*10^{-31} \ kg[/tex]

        [tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]

         [tex]v = \sqrt{6532.4}[/tex]

        [tex]v = 80.82 \ m/s[/tex]

 

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?​

Answers

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

a body accelerate uniformly from rest at the rate of 3meters per seconds for 8 sec . calculate the distance covered by the body during the acceleration
SOL​

Answers

Answer:

96 m

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²

Δx = 96 m

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