Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distance of 5.2 mm from the objective.
1. What is the size and location of the image from the objective? What is the linear magnification of this objective?
2. Treat the image from the objective as an object for the eyepiece. If the eyepiece creates an image at infinity, how far apart are the two lenses?
3. What is the angular magnification of the pair of lenses?

Answer:

speed of the alpha particle is 2 x 10^7 m/s.

Explanation:

energy of alpha particle = 8.3 Mev

1 Mev = 1.602 x 10^-13 J

8.3 Mev = [tex]x[/tex]

solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J

mass of a alpha particle = 6.645 x 10^−27 kg

The energy of the alpha particle is the kinetic energy KE of the alpha particle

KE = [tex]\frac{1}{2}mv^{2}[/tex]

where m is the mass of the alpha particle

v is  the velocity of the alpha particle

substituting values, we have

1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]

[tex]v^{2}[/tex] = 4 x 10^14

[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s

Complete Question

The image of this question  is shown on the first uploaded image

Answer:

a

   [tex]d =0.161 \ m[/tex]

b

  [tex]v = - 0.054 \ m/s[/tex]

c

  [tex]a = 6.12 \ m/s^2[/tex]

Explanation:

From the question we are told that

      The maximum  displacement is  A =  0.17  m  

      The  time considered is  [tex]t = 3.1 \ s[/tex]

     The spring constant is  [tex]k = 137 \ N \cdot m[/tex]

      The mass is  [tex]m = 3.8 \ kg[/tex]

Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as

             [tex]d = A cos (w t )[/tex]

Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as

        [tex]w = \sqrt{\frac{k}{m} }[/tex]

substituting values

       [tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]

        [tex]w =6[/tex]

So the displacement is at  t

      [tex]d = 0.17 cos (6 * 3.1 )[/tex]

       [tex]d =0.161 \ m[/tex]

Generally the velocity of a  SHM(simple harmonic motion) is mathematically represented as

         [tex]v = - Asin (wt)[/tex]

substituting values

         [tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]

          [tex]v = - 0.054 \ m/s[/tex]

Generally the maximum acceleration is  mathematically represented as

         [tex]a = w^2 * A[/tex]

substituting values

         [tex]a_{max} = 6^2 * (0.17)[/tex]

substituting values

         [tex]a = 6^2 * (0.17)[/tex]

        [tex]a = 6.12 \ m/s^2[/tex]

Answer:

SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.

Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

Answer:

b. the particle must be moving parallel to the magnetic field.

Explanation:

The magnetic force on a moving charged particle is given by;

F = qvBsinθ

where;

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field

θ is the angle between the magnetic field and velocity of the moving particle.

When is the charge is stationary the magnetic force on the charge is zero.

Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.

Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.

b. the particle must be moving parallel to the magnetic field.

Answer:

x = A sin (wt + theta)        where w = angular frequency - basic SHM equation

w = 8 pi = 2 pi f

f = 4         basic frequency

N = 8     number of times thru origin

Each cycle the particle will pass thru the origin +x and -x    twice

Answer:

The complete question is

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26  W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.

For totally reflective, F = (2I/c)A    ....1

for totally reflective, F = (I/c)A       ....2

where I is the intensity of the light

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

A is the area the sail

b) The intensity of the light from the sun = power/area

==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]

where r is the distance from the sun and the sail

The Force from the sail from equation 1  is therefore

[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]

gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]

where

G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.

m is the mass of the sail = 10000 kg

M is the mass of the sun = 1.99 x 10^30 kg.

==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

Equating the forces, we have

[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]  =  [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

the distance cancels out

A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2

==> 6428.2 km^2

c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation  is inversely proportional to the square of the distance, so they both cancel out.

Answer;

26.45cm

See attached file for explanation

Answer:

Explanation:

Convert all lengths to meters:

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

Answer:

27yrs

Explanation:

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Potential Energy = 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

Explanation:

F = ma

[tex]a = \frac{f}{m} [/tex]

[tex]a = \frac{1500}{500} = 3[/tex]

[tex]a = \frac{v2 - v1}{t} [/tex]

[tex]3 = \frac{v2 - 10}{10} [/tex]

v2 (final) = 40 m/s to the right direction

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

Where:

Make D' The subject of the equation.

From the question,

Given:

Substitute these values into equation 2

Hence, the apparent depth of the fish is 60 cm

Learn more about apparent depth here: https://brainly.com/question/24319677

Answer:

Answer A : Fusion followed by beta decay (electron emission)

Explanation:

Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).

Therefore, answer A is the correct one.

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

Answer:

C

Explanation:

intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)

I = P/A

A = πr²

r = is the distance between you and the cricket.

so in other form we can get

I = P/πr²

let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.

I(1) / I(2) = r(2)² / r(1)²

1/4 = r(2)²/r(1)²

1/2 = r(2) / r(1)

r(2) = ½ r(1)

or r(2) is decreaases by a factor of 2.

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

Answer:

B 5s

Explanation:

Because of the Displacement in the nth second of the free fall is 

Snth=21g(t12−t22)

Given that (tn−tn−1)=1

Displacement in 3 seconds of the free fall 

S=21gt2

S=21×10×32

S=45m

Given that: Snth=45

On solving that we get:

t1=5sec

Answer:

At the end of the handle farthest from the head of the hammer.

Explanation:

The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

Answer:

The power of the eye is 51.64 diopters

Explanation:

The power of the eye is given by;

[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]

where;

P is the power of the eye in diopter

f is the focal length of the eye

[tex]d_o[/tex] is the distance between the eye and the object

[tex]d_i[/tex] is the distance between the eye and the image

Given;

[tex]d_o[/tex] = 61.0 cm = 0.61 m

[tex]d_i[/tex] = 2.0 cm = 0.02 m

[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]

Therefore, the power of the eye is 51.64 diopters.

The power P of the eye when viewing an object 61.0 cm away is 51.639D

The power of a lens is a reciprocal of its focal length and it is expressed as:

[tex]P=\frac{1}{f}[/tex]

According to the mirror formula

[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]

where

[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m

[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m

[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]

Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D

Learn more here: https://brainly.com/question/14870552

Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

To know more about friction:

https://brainly.com/question/28356847

#SPJ5

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]

Thus, the effective dose allowable yearly = 16 times

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

Answer:

It tell you that the velocity is constant, what means that there's no acceleration

Answer:

The values is  [tex]m_{max} = 8001 \ bright \ spots[/tex]

Explanation:

From the question we are told that

    The slit distance is  [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

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

For maximum number of spot [tex]\theta = 90^o[/tex]

So  

       [tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]

        [tex]n =4000[/tex]

Now for the both sides plus the central maxima  we have

      [tex]m_{max} = 2 * n + 1[/tex]

substituting values

       [tex]m_{max} = 2 * 4000 + 1[/tex]

       [tex]m_{max} = 8001 \ bright \ spots[/tex]

Answer:

Water.

Explanation:

To know which of the substance that will absorbed the greatest amount of thermal energy, we'll simply determine the amount of energy absorbed by each substance.

This is illustrated below:

For Air:

Mass (M) = 1 kg

Change in temperature (ΔT) = 15 °C

Specific heat capacity (C) = 1.01 KJ/Kg°C

Heat Absorbed (Q) =..?

Q = MCΔT

Q = 1 x 1.01 x 15

Q = 15.15 KJ

Therefore, the heat absorbed by air is 15.15 KJ.

For Plastic:

Mass (M) = 1 kg

Change in temperature (ΔT) = 15 °C

Specific heat capacity (C) = 2.60 KJ/Kg°C

Heat Absorbed (Q) =..?

Q = MCΔT

Q = 1 x 2.60 x 15

Q = 39 KJ

Therefore, the heat absorbed by plastic is 39 KJ.

For Water:

Mass (M) = 1 kg

Change in temperature (ΔT) = 15 °C

Specific heat capacity (C) = 4.18 KJ/Kg°C

Heat Absorbed (Q) =..?

Q = MCΔT

Q = 1 x 4.18 x 15

Q = 62.7 KJ

Therefore, the heat absorbed by water is 62.7 KJ.

For Wood:

Mass (M) = 1 kg

Change in temperature (ΔT) = 15 °C

Specific heat capacity (C) = 1.68 KJ/Kg°C

Heat Absorbed (Q) =..?

Q = MCΔT

Q = 1 x 1.68 x 15

Q = 25.2 KJ

Therefore, the heat absorbed by Wood is 25.2 KJ.

Summary:

Substance >>>>> Heat Absorbed

Air >>>>>>>>>>>> 15.15 KJ.

Plastic >>>>>>>>> 39 KJ

Water >>>>>>>>>> 62.7 KJ

Wood >>>>>>>>>> 25.2 KJ.

From the above calculations, we can see that water will absorb the greatest amount of thermal energy.

The substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.

The specific heat capacity of each substance can be used to determine the substance with the smallest change in temperature.

Q = mcΔθ

where;

Δθ = Q/mc

Δθ = (250)/(1.01)

Δθ = 247.5 ⁰C

Δθ = (250)/(2.6)

Δθ = 96.15 ⁰C

Δθ = (250)/(4.18)

Δθ = 59.81 ⁰C

Δθ = (250)/(1.68)

Δθ = 148.81 ⁰C

Thus, the substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.

Learn more about heat capacity here: https://brainly.com/question/16559442

Answer:

C

They allow light to pass only if their directions of polarizations are exactly 90° apart.

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.''

The horizontal force applied to the lower block is approximately 1,420.85 Newtons

The known parameters are;

The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N

The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N

The length of the string attached to the block, W₂, l = 6 m

The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m

The coefficient of friction between the surfaces, μ = 0.25

Let T represent the tension in the string

The upward force on W₂ due to the string = T × sin(θ)

The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)

The tension in the string, T = N₂ × μ × cos(θ)

∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)

sin(θ) = √(6² - 5²)/6

cos(θ) = 5/6

∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T ≈ 183.27 N

The normal reaction on W₂, N₂ = T/(μ × cos(θ))

∴ N₂ = 183.27/(0.25 × 5/6) = 879.7

N₂ ≈ 879.7 N

The friction force, [tex]F_{f2}[/tex] = N₂ × μ

∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N

The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂

[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N

The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ

∴  [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N

The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = 1,200.925 N + 219.925 N = 1,420.85 N

The horizontal force applied to the lower block, P ≈ 1,420.85 N

Answer:

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Explanation:

Newton's 3rd law states that every action has and equal but opposite reaction.

If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.

Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.

The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.

The same amount of force used to move/stop something is used to stop/move it.

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.

There are two kinds of friction;

Since more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.

Learn more: https://brainly.com/question/18754989

Answer:

» The image is laterally inverted.

» The image is upright.

» The image geometry is same as object geometry.

» Image distance is same as object distance.

» Image is not real, it's virtual ( not formed on screen ).

[tex].[/tex]