Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.

Answers

Answer 1

Answer:

a)0.51°

b)1.47×10^-4m

Explanation:

a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.

bsin(θ)= mλ.........................(*)

Where b= width of the slit

The distance on the screen from Central angle can be expressed as

Sin(θ)= y/d............. (**)

d and y is the horizontal distance between slit and screen

If we input eqn(**) into equation (*) we have

y= mλd/b................(z)

In order to find angle (θ) we have

(θ)= sin-(1.99×10^-2)/2.25

= 0.51°

Therefore, angle of diffraction θ of the second minimum is 0.51°

b)to find the width of the sloth using eqn(z) by substitute the values, we have

b= (2)(649×10^-9)(2.25)/1.99×10^-2

b= 1.47×10^-4m

Therefore, the width of the slit is 1.47×10^-4m


Related Questions

A metal disk lies in the xy-plane, centered about the origin, and rotates with a constant angular velocity about the z-axis. There is a uniform 0.0314 T magnetic field parallel to the z-axis. The radius of the disk is 1.56 m. At what rate is the disk turning if an E of 3.86 V develops between the center of the disk and a point on its rim

Answers

Answer:

 w = -101 rad / s

Explanation:

For this exercise we will use Faraday's law

          E = - dФ / dt

where the magnetic flux is

          Ф = B. A = B A cos θ

In this case, the angle between the magnetic field and the normal to the disk is zero, cos 0 = 1, they indicate that the field is constant, let's find the area

               

The area rotated by the disk is

         A = ½ r s

if we express the angles in radians

         θ = s / r

         s = r θ

where is the arc supported

         A = ½ r (r θ)

let us substitute in the Faraday equation

          E = - d (B ½ r² θ) / dt

          E = - ½ B r² dθ/dt

the definition of angular velocity is

          w = dθ/dt

          E = - ½ B r² w

           w = - 2E / B r²

let's calculate

         w = - 2 3.86 / (0.0314 1.56²)

         w = -101 rad / s

An atom of unknown element Z has mass number of 39 and an atomic number of 18. How many protons and how many neutrons are in this atom? Show your work

Answers

Answer:

proton:18

neutron:39-18=21

Which of the following is the correct equation to calculate electrical energy? P = I/V V = P × I P = I × V E = P × t

Answers

You haven't told us what any of those letters represents.

I'm going to assume that ...

-- ' P ' = power

-- ' I ' = circuit current

-- ' V ' = circuit voltage

-- ' E ' = energy

-- ' t ' = time.

Now I'm ready to answer the question that I just invented.

P = I / V

V = P x I

Both of these are false statements, and can't be used to calculate anything.

P = I x V   is the correct equation to calculate electrical power.

E = P x t   is the correct equation to calculate electrical energy.

What is a hypothesis?
A. A piece of information gathered during an experiment
OB. A source that summarizes a primary source
O C. Something that a scientist intentionally changes during an
experiment
O D. An educated guess about the outcome of an experiment

Answers

Answer:

D!!!!!

Explanation:

A hypothesis is a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

Answer:

D. An educated guess about the outcome of an experiment.

Explanation:

Hypothesis is an educated guess about the outcome of an experiment.

Hope this helps ;) ❤❤❤

A body decelerates uniformly to a constant speed and after some time it accelerates uniformly Draw the shape of speed time graph for such a motion . Label the three sections of this graph. What is the quantity which is measured by the area occupied under speed time graph ?

Answers

Answer:

here I am just giving an idea of how the graph will be like ...

In the pic..

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:)

THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period of the pendulum with uncertainty in it

Answers

Answer:

  2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

  T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

  Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

  Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

  T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

  T ≈ 2.4583 ± 0.0207 . . . seconds

__

We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

PLZ HELP ASAP!!!! THANK YOU The disturbance that occurs as longitudional waves travel through a medium can be described as a series of A:oscillations and refractions B:propagations and compressions C:destructions and constructions D:rarefactions and compressions

Answers

Answer:

D:rarefactions and compressions

Explanation:

Longitudinal waves are readily formed in materials such as a stretched spring. Longitudinal waves are waves which travel in a direction parallel to the vibrations of the medium.

Longitudinal waves are characterized by a series of compressions and rarefactions. The compressions are areas of clusters while rarefactions are areas of expansions. The same can be observed in a sound wave.

A group of Advanced Open Water Divers plans to make two dives. The first dive is on a reef in 90 feet of water for 20 minutes. The group then remains on the surface for 1 hour. The second dive is on a wreck in 60 feet of water, with a planned bottom time of 30 minutes. What will be the ending pressure group after the second dive

Answers

Answer:

The ending pressure is R.

Explanation:

This is Recreational scuba diving (R). According to standards, this diving has prescribed depth limit of 130 ft. This means that the limit must include a depth that is not greater than 130 ft while using only compressed air and not even requiring a decompression stop.

Now, from the question, we can see that that there was a prescribed limit of 60ft of water with a planned bottom time of 30 minutes.

Thus, the ending pressure is R.

You have a cup with 50cm filled with water. How much pressure will the water act on the bottom of the cup? The density of water is 1000kg/m^3 and g = 10N/kg

Answers

Answer:

5000 N/m².

Explanation:

The following data were obtained from the question:

Density (d) = 1000 kg/m³

Acceleration due to gravity (g) = 10 N/kg

Height (h) = 50 cm = 50/100 = 0.5 m

Pressure (P) =.?

Pressure is related to density and height by the following equation:

P = dgh

Where

P is the pressure.

d is the density.

g is the acceleration due to gravity.

h is the height.

With the above formula, we can obtain the pressure at the bottom of the cup as follow:

P = dgh

P = 1000 x 10 x 0.5

P = 5000 N/m²

Therefore, the pressure at bottom of the cup is 5000 N/m².

What form of energy does a block of chocolate have?

Answers

Answer:

Chemical Energy

Explanation:

One glass microscope slide is placed on top of another with their left edges in con- tact and a human hair under the right edge of the upper slide. As a result, a wedge of air exists between the slides. An interference pattern results when monochromatic light is incident on the wedge. What is observed at the left edge of the slides? a. A dark fringe b. A bright fringe c. Impossible to determine

Answers

Answer:

A dark fringe

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?

Answers

Answer:

Rounded to three significant figures:

(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:

A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].Maxima

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].

The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].

Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].

Minima

The dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.

First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].

For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].

A construction worker uses an electrical device to attract fallen nails and sharp objects
from a construction site. What is causing the attraction of the metal objects?
O An electrical wave oscillating perpendicular to the electrical device.
O An electrical charge radiating perpendicular to the wire
O Amagnetic wave radiating perpendicular to an electrical device
O A magnetic wave and electrical current moving in opposite directions

Answers

Answer:

is the last one, a magnetic wave and electrical current moving in opposite directions

Explanation:

opposite directions always attract in magnetic waves and fields

A mass, M, is at rest on a frictionless surface, connected to an ideal horizontal spring that is unstretched. A person extends the spring 30 cm from equilibrium and holds it at this location by applying a 10 N force. The spring is brought back to equilibrium and the mass connected to it is now doubled to 2M. If the spring is extended back 30 cm from equilibrium, what is the necessary force applied by the person to hold the mass stationary there

Answers

Answer:

The necessary force applied by the person to hold the mass stationary there is 10 N

Explanation:

We are told that this person extends the spring 30 cm from equilibrium and holds it at this location by applying a 10 N force.

Thus, based on Hooke's law formula which is F = k Δx, we can say that the mass attached to the spring does not change the spring constant. Thus, the

same resistive spring force will still be in place and in turn, the same stretching force of 10N would still be required.

Thus;

The necessary force applied by the person to hold the mass stationary there is 10 N

The transmutation of a radioactive uranium isotope, 234/92U , into a radon isotope, 222/86Rn , involves a series of three nuclear reactions. At the end of the first reaction, a thorium isotope,230/90Th , is formed and at the end of the second reaction, a radium isotope, 226/88Ra , is formed. In both the reactions, an alpha particle is emitted. Write the balanced equations for the three successive nuclear reactions.

Answers

Answer:

See explanation below

Explanation:

234/92U--------> 230/90Th + 4/2He

230/90Th----->226/88Ra + 4/2He

There are various modes of radioactive decay. One of them is alpha decay.

Alpha decay involves the loss of an alpha particle from a nucleus. When this occurs, the mass number of the daughter nucleus is four units less than that of the parent while the atomic number of the daughter nucleus is two units less than that of its parent as shown in the equations above.

An alpha particle is akin to a helium nucleus in terms of mass and charge

obesity occurs due to (a) overeating of carbohydrates and fats(b) not eating enough carbohydrates and fats (c) overeating of vitamins and minerals (d) not eating enough vitamins and minerals

Answers

Answer:

B for BIG

Explanation:

carbs and fats is energy, if not being use or burned off, it'll be becomes body fat.

2 Which invention was crucial to the development of cell theory?

Answers

Answer:

I guess i would say the microscope

Explanation:

Because of the microscope, we can see the cells.

A parallel-plate capacitor has square plates that are 7.40 cm on each side and 3.20 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.40 cm on a side and 1.60 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 84.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.

Answers

Answer:

The energy that can be stored in the capacitor is 239 nJ

Explanation:

We first calculate the capacitance of each material. Let C₁ be the capacitance of pyrex glass and C₂ be the capacitance of polystyrene.

C₁ = κ₁ε₀A/d where κ₁ = dielectric constant of pyrex glass = 5, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of pyrex slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 5 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 2424.2252/1.60 × 10⁻¹¹ F = 1515.14 × 10⁻¹¹ F = 15.2 × 10⁻⁹ F = 15.2 nF

C₂ = κ₂ε₀A/d where κ₂ = dielectric constant of polystyrene = 3, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of polystyrene slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 3 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 1454.5351/1.60 × 10⁻¹¹ F = 909.08 × 10⁻¹¹ F = 9.09 × 10⁻⁹ F = 9.09 nF

Since the capacitors are in series, we find their effective capacitance C from

1/C = 1/C₁ + 1/C₂

C = C₁C₂/(C₁ + C₂)

= 15.2 × 10⁻⁹ F × 9.09 × 10⁻⁹ F/(15.2 × 10⁻⁹ F + 9.09 × 10⁻⁹ F)

= 138.168 × 10⁻¹⁸/24.29 × 10⁻⁹ F

= 5.69 × 10⁻⁹ F

The amount of electrical energy stored in a capacitor is given by W = 1/2CV² where C = capacitance and v = voltage applied. Now C = 5.69 × 10⁻⁹ F and V = 84.0 V for this capacitor

So W = 1/2 × 5.69 × 10⁻⁹ F × 84.0 V

= 238.98 × 10⁻⁹ J

≅ 239 × 10⁻⁹ J

= 239 nJ

So the energy that can be stored in the capacitor is 239 nJ

What did Bohr's model of the atom include that Rutherford's model did not have?
a nucleus
energy levels
electron clouds
smaller particles

Answers

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

Answer:

energy levels

hope this helped!

Explanation:

Write the relation between:
1) applied force and pressure.
2) surface area of contact and pressure.

Answers

realtion between applied force.and pressure is more force exerts more pressure whereas less force exerts less pressure

confused in another one

Answer:

See below

Explanation:

1) Applied Force and Pressure

Pressure = Force / Area

This shows that Applied force and pressure are in direct relationship. This means that If the Applied force is more, the Pressure is also More and vice versa.

2) Surface Area of Contact and Pressure

Pressure = Force / Surface Area of Contact

This shows that Pressure and Surface area of contact are inversely related. This means that if pressure is increased on an object, the surface area of contact decreases and vice versa.

A high-voltage powerline operates at 500000 V-rms and carries an rms current of 500 A. If the resistance of the cable is 0.050Ω/km, what is the resistive power loss in 200 km of the powerline?

Answers

Answer:

2,500,000W or 2.5MW

Explanation:

The power lost due to resistance is given by I^2R. We must first obtain R as follows;

Resistance per kilometer= 0.050Ω/km

Distance covered= 200km

R = 200km x 0.050Ω/km = 10Ω

The lost power as a 500A current passes through the powerline is:

P = I²R

P= 500² x 10

P= 2,500,000 W or 2.5MW

The resistive power loss in 200 km of the powerline is of 2.5 MW.

Given data:

The root mean square voltage is, V' = 500000 V.

The magnitude of current through the power line is, I =500 A.

The magnitude of resistance of cable is, R = 0.050 Ω/km.

The length of powerline is, L = 200 km.

Whenever there is a flow of current through the wire, then there are various losses out of which the power loss is a major factor. The mathematical expression for the power loss is given as

P = I²R

Solving as,

P= 500² x 10

P= 2,500,000 W or 2.5MW

Thus, we can conclude that the resistive power loss in 200 km of the powerline is of 2.5 MW.

Learn more about the resistive power loss here:

https://brainly.com/question/15158529

a object 1.5cm high produces a real image 2cm high. placed at a distance of 20cm
from a concave mirror calculate: [a] the position of the image [b] focal lenght of the concave mirror

Answers

Answer:

a. 26.7 cm. b. 11.4 cm.

Explanation:

a. We know h'/h = d'/d where h' = image height = + 2 cm (since it is a real image), h = object height = + 1.5 cm, d' = image distance from mirror and d = object distance from mirror = 20 cm

So, from h'/h = d'/d

d = h'd/h

= 2 cm × 20 cm/1.5 cm

= 40/1.5 cm

=  26.67 cm

≅ 26.7 cm

The position of the image is 26.7 cm from the mirror

b. Using the mirror formula

1/d + 1/d' = 1/f where d = object distance from mirror = + 20 cm, d' = image distance from mirror = + 26.7 cm (its positive since its a real image) and f = focal length of mirror.

So, 1/d + 1/d' = 1/f

⇒ f = dd'/(d + d')

= 20 cm × 26.7 cm/(20 cm + 26.7 cm)

= 534/46.7

= 11.43 cm

≅ 11.4 cm

The focal length of the mirror is 11.4 cm

A spring of initial length 35 cm acquires a length of 55 cm when we hang from it a mass of 3.5 kg. Calculate:

a) The elastic constant of the spring.
b) The length of the spring when we hang a mass of 5 Kg.

Answers

Answer:

the elastic constant of the spring=1.715

the length of the spring=0.28

Explanation:

we know that according to hooks law

F=-k x

F= force

k= elastic constant

x= extension or compression

given

length change from 35cm to 55 cm so delta x = L2-L1= 55-35=20 cm

now to find k we need F and F =ma

M for part a is 3.5 kg

so F=3.5 kg *9.8=34.3

now k=F/x

k=34.3/20=1.715 N/cm=171.5 N/m

now to find length given mass is 5 kg so

F= ma

F=5*9.8=49 N

so x =F/k

x=49/171.5

x=0.28

An average skater averages 11 m/s over the first 5 seconds of a race. find the average speed required over next 10 seconds to average 12 m/s overall. ​

Answers

Answer:

usa

Explanation:

A fish inside the water 12cm below the surface looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 the radius of circle is​

Answers

Answer:

13.6 cm

Explanation:

From Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.

Given n₂ = 4/3:

1 = 4/3 sin θ

sin θ = 3/4

If x is the radius of the circle, then sin θ is:

sin θ = x / √(x² + 12²)

sin θ = x / √(x² + 144)

Substituting:

3/4 = x / √(x² + 144)

9/16 = x² / (x² + 144)

9/16 x² + 81 = x²

81 = 7/16 x²

x ≈ 13.6

explore how archemides principle is applied in building a ship and submarine​

Answers

Answer:

Principle Archimedes is applied in building a ship and submarine using the manipulating that buoyancy, is controlled the ballast tank system.

Explanation:

Submarine is rather had they focused on main parts of the submarine,he is complex and long process implementation,the most submarine design like submarine stability.

Submarine stability is complete and the fundamental Archimedes principle to arrive the weight of submarine is equal to buoyancy force.

Submarine into the parts and components of ballast tank the sequence in diving and surfacing,there two vital parts:-  flood parts and air vents

flood parts:- at the bottom position and allow water to enter or leave that tank.

air vents:- air vents at the top of the pressure hall,and that they submarine dive.

this time submarine is most modern system is depth is 300 to 450 meters,high pressure  air is 15 bar is tank air valve.

submarine is basic of the effective volume of all the submarine surfaced condition,submarine minus to the free water flood is equal to the fully pressure hull,submarine is the surfaced condition.

3) A charged particle is moving with velocity of V in a magnetic field of B, which one of the followings is correct: A) The direction of force F on the charge is parallel to magnetic field B B) The direction of force F on the charge is parallel to velocity direction V C) The force is maximized when velocity direction and magnetic field are parallel D) The force F is perpendicular (normal) to both velocity V and magnetic field B E) The direction of force on positive charge or negative charge would be the same

Answers

Answer:

D) The force F is perpendicular (normal) to both velocity V and magnetic field B.

Explanation:

When a charged particle enters a magnetic field, it experiences a force which changes its direction of travel. The direction of motion of the charged particle, the magnetic field direction, and the direction of the force, are all perpendicular to one another. According to Lorentz right hand rule, hold the right hand parallel to the ground, with the palm facing up, and the thumb held out at right angle to the other fingers. If the direction of the other fingers represents the magnetic field line and direction, and the thumb represents the direction of motion of a positively charged particle, then, the palm will push up in the direction of the force. For a negatively charged particle, the force will push down in the direction of the back of the hand.

what is SI unit System ? why has SI system been developed ? Give reasons​

Answers

Explanation:

SI is the international system of units

It was developed to express magnitudes and quantities

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.27 m/s . Suppose the collision is elastic.1. Find the magnitude of the final velocity of the 0.157kg glider.
2. Find the magnitude of the final velocity of the 0.306kg glider.

Answers

Answer:

v1 = −2.201946 m/s ( to the left)

v2 = 0.7780534 m/s ( to the right)

Explanation:

Given the following :

Mass of first glider (m1) = 0.149kg

Initial Speed of first glider (u1) = 0.710 m/s

Mass of second glider (m2) = 0.308kg

Initial Speed of second glider (u2) = 2.27m/s

For elastic collision:

m1u1 + mu2u2 = m1v1 + m2v2

Where V1 and v2 = final velocities if the body after collision.

Taking right as positive ; left as negative

u1 = 0.710m/s ; u2 = - 2.27m/s

u1 - u2 = - (v1 - v2)

0.710 - - 2.27 = - v1 + v2

v2 - v1 = 2.98 - - - - (1)

From:

m1u1 + mu2u2 = m1v1 + m2v2

(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)

0.10579 + (-0.69916) = 0.149 v1 + 0.308v2

−0.59337 = 0.149 v1 + 0.308v2

Dividing both sides by 0.149

v1 + 2.067v2 = −0.59337 - - - - - (2)

From (1)

v2 = 2.98 + v1

v1 + 2.067(2.98 + v1) = −0.59337

v1 + 6.16 + 2.067v1 = −0.59337

3.067v1 = −0.59337 - 6.16

3.067v1 = −6.75337

v1 = −6.75337 / 3.067

v1 = −2.201946 m/s ( to the left)

From v2 = 2.98 + v1

v2 = 2.98 + (-2.201946)

v2 = 0.7780534 m/s ( to the right)

The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU

Answers

Answer:

The answer is 29 AU

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Answer:

29

Explanation:

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