A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displacement of the helicopter?
Express as components of a vector.
x-component_____________________
y-component_____________________

b) What is the displacement of the helicopter? Express as a vector (magnitude and direction).

Answer_____________________

Answers

Answer 1

Answer:

a

  x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

b

 Magnitude [tex]d = 52 \ m[/tex]

direction is  [tex]\theta = 67.4^o[/tex]

Explanation:

From the question we are told that

   The first  vertical distance is  [tex]y_1 = 500 \ m[/tex]

    The  first horizontal distance  is  [tex]x = 20 \ m[/tex]

    The  second vertical distance is  [tex]y_2 = 452 \ m[/tex]

Generally the displacement is  

x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

Generally the helicopters displacement is mathematically evaluated as  

       [tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]

      [tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]

      [tex]d = 52 \ m[/tex]

The  direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as

         [tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]

=>       [tex]\theta = tan ^{-1}[ 2.4 ][/tex]

=>      [tex]\theta = 67.4^o[/tex]

   


Related Questions

The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down

Answers

Answer:

So the minimum force is

32.2Newton

Explanation:

To solve for the minimum force, let us assume it to be F (N)

So

F=mgsinA

But

=>>>> coefficient of static friction x (F + mgcosA

=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)

So making F subject of formula

F + 24.0 = 56.2

F = 32.2N

Remove the wood block from the water. Place the brick block into the water and let it sink. How much water does the brick block displace? Hint: This is the liquid level with the block minus the liquid level before the block is added.

Answers

Answer:

the volume desalinated  that is placed is equal to the volume of the block.

Explanation:

For this exercise we use the Archimedean principle which states that the thrust is equal to the weight of the desalted liquid.

                      B = ρ g V

When we have the block of wood the volume of water desalinated in the volume of the block under water, this is because the wood floats in the water

When placing the block of clay (brick), it sinks, so the volume desalinated  that is placed is equal to the volume of the block.

          V_body = V_waer = l to h

What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?

Answers

Answer:

λ = hc/(eV + h[tex]f_{0}[/tex])

Explanation:

Let the work function of the metal = ∅

the kinetic energy with which the electrons are ejected = E

the energy of the incident electromagnetic wave = hf

Then, we know that the kinetic energy of the emitted electron will be

E = hf - ∅

because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.

This means that the energy of the incident e-m wave can be written as

hf = E + ∅

also, we know that the kinetic energy of the emitted electron E = eV

and the work function ∅ = h[tex]f_{0}[/tex]

we can they combine all equations to give

hf = eV + h[tex]f_{0}[/tex]

we know that f = c/λ

substituting, we have

hc/λ = eV + h[tex]f_{0}[/tex]

λ = hc/(eV + h[tex]f_{0}[/tex])    This is the wavelength of the e-m radiation needed to eject electrons from a metal.

where

λ is the wavelength of the e-m radiation

h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s

c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s

e is the charge on an electron

V is the voltage potential on the electron

[tex]f_{0}[/tex] is the threshold frequency of the metal

A satellite travels around the earth at 37,000 km/hr. How far will it travel after 13 hours?

Answers

Explanation:

Distance = speed × time

d = (37,000 km/hr) (13 hr)

d = 481,000 km

A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered to the string is tripled, by what factor does the amplitude change

Answers

Answer:

n = 1,732    the amplitude must be increased by a factor of 1,732

Explanation:

The power delivered by a wave is given by

                  P = E / t

                  P = ½ μ w² v A²

let's apply this expression to our case the power tripled

                3P₀ = ½ μ w² v A’²

                 

let's write the amplitude function of a initial amplitude

               A ’= n A₀

where n is a number

               3 P₀ = (½ μy w² v  A₀²) n²

               3P₀ = P₀ n²

               n = √ 3

               n = 1,732

therefore the amplitude must be increased by a factor of 1,732

A westward-moving bicycle increased its speed from 2.0m/s to 12.0m/s in 4.0 seconds. What is the magnitude and direction of the acceleration?

Answers

Answer:

2.5 m/s^2

Explanation:

Acceleration = (Final Velocity - Initial Velocity) / TIme

Acceleration = 12 - 2 / 4

=> Acceleration = 10 /4

=> Accelration = 5/2

=> Acceleration = 2.5 m/s^2

So, Acceleration is 2.5 m/s^2

The acceleration of the bicycle which is the change in the velocity of the bike with change in time will be 2.5m/s in the Westward direction

Initial speed = 2.0 m/s Final speed = 12.0 m/s

Change in time, t = 4.0 seconds

We define acceleration thus :

Acceleration = (△velocity / △time)

Change in velocity = (12 - 2) = 10 m/s

Acceleration = 10 / 4 = 2.5 m/s²

Therefore, the acceleration of the bike will be 2.5m/s² in the West direction

Learn more :https://brainly.com/question/11979832

A volleyball is released three times from three different heights: 5 m, 10 m, and 20 m. Which statement is correct about the volleyball?

The volleyball accelerates more quickly toward the ground when released from a higher height.

The volleyball accelerates the same amount toward the ground each time.

The volleyball accelerates more slowly toward the ground when released from a higher height.

The volleyball may accelerate more quickly or slowly depending on its mass.

A student is setting up an experiment with 10 different balls. The student wants to measure which ball hits the ground with the largest force when released from a box 5 feet above the ground. Which quantity should be measured to correctly calculate the force with which each ball hits the ground?

the initial velocity of each ball
the mass of each ball
the rate at which the velocity of each ball increases
the final velocity of each ball right before hitting the ground

Answers

Explanation:

1)

The answer is

The volleyball accelerates the same amount toward the ground each time.

Because gravitational acceleration is constant irrespective of altitude(height) or mass of the object.

2)

The answer is

the mass of each ball

because gravitational force is given by

F = mg

since g is a constant, he only needs to measure the mass if each ball

The concepts of free fall and Newton's second law allow to find the correct answers

1) Acceleration is constant

2) The mass of the body

The game of volleyball consists of throwing the balls from the serve to pass it to the other court, this is a two-dimensional movement type projectile launch, in this type of movement the acceleration is the constant on the y axis, it is called acceleration of the gravity and vale (g = 9.8 m / s²)

1) Let's review the different statements of the first part regarding acceleration.

a) The ball accelerates faster as it goes down

   False. Acceleration is constant

b) The ball accelerates the same amount

   True. Acceleration is constant and directed on the vertical axis

c) The ball accelerates slower or faster

     False. Acceleration is constant

The vertical launch of kinematics, establishes that the ball as it descends it goes faster with an acceleration equal to the acceleration of gravity, in this part it is asked to look for the force of the ball when it reaches the ground, for this the second is used Newton law

           F = m a

where in this case the acceleration is the acceleration of gravity

          a = g

          F = m g

Let's review the different claims

a) Initial velocity

    False. To calculate the force you only need the mass since acceleration is constant

b) The mass

    True. The product of the mass and the acceleration of gravity gives the force with which the ball hits the ground

c) Increasing speed

    False. Increasing speed allows calculating acceleration not force

d) The final speed

     False. Force does not depend on the speed of the body

In conclusion using the concepts of free fall and Newton's second law we can find the correct answers:

1)  Acceleration is constant

2) Body mass

Learn more about free fall and Newton's second law here:

https://brainly.com/question/19897308

Water flows through a 0.5 cm diameter pipe connected to a 1 cm diameter pipe. Compared to the speed of the water in the 0.5 cm pipe, the speed in the 1 cm pipe is

Answers

A

Answer:

Speed of water in the 0.5cm diameter will be faster because it has a smaller area

Since area x radius ² so if radius is reduced by 0.5 speed is increased by 4times in the 0.5 diameter pipe

The measured value of the latent heat of vaporization of helium (at 1 atm) is
84 J/mol. Use this to estimate the binding energy of helium atoms in the liquid.
a. 2.6 x 10-6 eV
b. 3.6 x 10-4 eV
c. 5.1 x 10-4 eV
d. 8.6 x 10-3 eV
e. 3.6 x 10-2 eV

Answers

Answer:

d. 8.6 x 10-3 eV

Explanation:

Binding energy is energy that binds two atoms together . Energy is required to separate them . In the process of evaporation , energy is required to separate each molecules/ atoms  of liquid so that they are separated and then get evaporated . That is why latent heat of evaporation is required .

1 mole of helium will contain 6.02 x 10²² atoms of it .

binding energy per atom  = 84 / 6.02 x 10²² J

= 13.953 x 10⁻²² J

= 13.953 x 10⁻²² / 1.6 x 10⁻¹⁹ eV .

= 8.6 x 10⁻³ eV

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a picture of an object that is 30 cm away.
How far must the lens move to focus on this second object?

Answers

Answer:

The distance is [tex]z = 0.008 \ m[/tex]

Explanation:

From the question we are told that

   The  focal length is  [tex]f = 50 \ mm = 50*10^{-3} \ m[/tex]

   

Generally the lens equation is mathematically represented as  

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

At  image  distance  u =  1.5 m

       [tex]\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}[/tex]

=>      [tex]\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}[/tex]

=>[tex]v = 0.052 \ m[/tex]

At  image  distance  [tex]u = 30\ cm = 0.30 \ m[/tex]

        [tex]\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}[/tex]

=>     [tex]\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}[/tex]

=>    [tex]v_1 = 0.06 \ m[/tex]

The distance the lens need to move is evaluate as

   [tex]z = |v - v_1|[/tex]

   [tex]z = |0.052 - 0.06|[/tex]

   [tex]z = 0.009 \ m[/tex]

Maggie completed a 10000-m race at an average speed of 160
m/min. If Tom took 12.5 fewer minutes to complete the race,
what was Tom's average speed?

Answers

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

When stretched beyond it's elastic limit, a metal rod such as steel​

Answers

Answer:

When an elastic material is stretched, depending on the stress, it might reach a point beyond which it will no longer return to its original size and shape. This point is called the elastic limit.

When the material (such as a metal rod (e.g steel, copper)) is stretched beyond its elastic limit, Hooke's law is no longer obeyed. i.e strain is no longer directly proportional to stress.

a pistol fires a bullet towards a target located 175m away. The bullet is traveling at 320 m/s. How long does it take for the bullet to hit the target?

Answers

Time = distance/speed

Time = (175 m) / (320 m/s)

Time = 0.547 second

Goslo and Speedo are twins. They have just celebrated their 20th birthday (Yay!). Speedo travels to the planet Zorth which is 40 light years away, he immediately returns. Neglect any time Speedo spent accelerating. If Speedo travels at 0.997 relative to the Earth, who will measure the proper time?

a. Goslo
b. Speedo
c. Nobody
d. Everybody

Answers

Answer:

a. Goslo

Explanation:

given data

Speedo travels = 40 light year

Speedo travels relative to the Earth = 0.997

solution

Goslo will measure here proper time and this time is expres as

time T =  Speedo travels ÷ Speedo travels relative to the Earth   ..........................1

put here value

time T = [tex]\frac{40}{0.997}[/tex]

time T = 40.1 year

so Goslo will measure the proper time

Scientists pose a ___ to examine cause and effect relationships.

Answers

Answer:

hypothesis

Explanation:

I think this is right I dont remember

What is process in which long thin strips of flexible ribbon are produced by pouring onto a flat surface?

Answers

Answer:

Tape casting

Explanation:

Tape casting is also known as knife coating. This process is used specially in the production of thin surfaces of a ceramic work.

Tape casting is also the process which which involves long thin strips of flexible ribbon being produced by pouring onto a flat surface. It is then left to dry by applying a high temperature and then ready for use.

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?

Answers

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

One train is at the 20 Km position, another is at the 200Km position, both are approaching, the first with a speed of 30Kmh-1 and the second with a speed of 48 Kmh-1. If they leave at 7:12, what time do they cross? Answer: t = 2.30 H Time: 9:30 am.

Answers

Explanation:

The distance between them is 200 km − 20 km = 180 km.

The relative velocity is 30 km/h − (-48 km/h) = 78 km/h.

The time it takes is 180 km / (78 km/h) = 2.31 hours, or 2 hrs 18 min.

Therefore, the trains meet at 9:30 AM.

A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path

Answers

Answer:

The particle will continue in a straight line parallel to the field

Explanation:

If a charged particle moves parallel to a magnetic field, without cutting across the field, then there will be no force on the particle. Since there is no force on the charged particle from the magnetic field, its velocity will continue in its original path, which is parallel to the field.

Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in minutes? Use 3 × 108 m/s for the speed of light.

Answers

Answer:

The value is   [tex]t =  14.129 \  minutes[/tex]    

Explanation:

From the question we are told that

  The distance of planet Tatoone is  [tex]d =  1.7 \ AU  =  1.7 *1.496* 10^{11}=2.543*10^{11} \ m[/tex]

   The  speed of light is  [tex]c =  3.0*10^{8} \  m/  s[/tex]

Generally the time taken is mathematically represented as

     [tex]t =  \frac{d}{c}[/tex]

=> [tex]t =  \frac{2.543*10^{11}}{3.0*10^{8} }[/tex]

=>    [tex]t =  847.7 \  s[/tex]

Now converting to minutes

       [tex]t =  \frac{847.7}{60}[/tex]

   =>     [tex]t =  14.129 \  minutes[/tex]    

J. Henry Alston was the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal. Please select the best answer from the choices provided T F

Answers

Answer:

True

Explanation:

J. Henry Alston was known as a famous African American psychologist. He was known through his detailed study of the sensations of heat and cold which is a necessity for all humans.

He however became the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal which gave him recognition in his field.

Answer:

True        YOUR WELCOME        

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into the water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius r_a, outer radius r_b, and length L much larger than r_b. The scientist applies a potential difference Delta V between the inner and outer surfaces, producing an outward radial current I. Let rho represent the resistivity of the water.
(a) Find the resistance of the water between the cylinders in terms of L, rho, r_a, and r_b.
(b) Express the resistivity of the water in terms of the measured quantities L, r_a, r_b, Delta V and I.

Answers

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

           A = π r_b² - π r_a²

           A = π (r_b² - r_a²)

we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

Neutrons travelling at 0.400 m/s are directed through a pair of slits having 1.00 m separation. an array of detectors is placed 10.0 m from the slits.
a) What is the de Broglie wavelength of the neutrons?
b) How far off axis if the first zero-intensity point on the detector array?
c) When a neutron reaches a detector, can we say which slit the neutrons passed through? Explain.

Answers

Answer:

9.91*10^-7 m

4.955*10^-6 m

Explanation:

Given that

v = 0.4 m/s

d = 1 m

L = 10 m

h = 6.62*10^-34 Js

m(neutron) = 1.67*10^-27 kg

To find the debroglie wavelength of the neutron, we use the formula

λ = h/mv

Now, we plug in the values we have listed.

λ = 6.62*10^-34 / (1.67*10^-27 * 0.4)

λ = 6.62*10^-34 / 6.68*10^-28

λ = 9.91*10^-7 m

b)

y1 = L (m + ½) λ/d, where m = 0

y1 = L (0 + ½) λ/d

y1 = L (½) λ/d

y1 = L/2 * λ/d or

y1 = Lλ/2d

now, we substitute the values for each of them, we have

y1 = (10 * 9.91*10^-7) / (2 * 1)

y1 = 9.91*10^-6 / 2

y1 = 4.955*10^-6 m

c) no, we can not say the neuron passed through one slit

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m

Answers

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m[/tex]

(a) Magnification,

[tex]m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2[/tex]

(b) Magnification, [tex]m=\dfrac{h'}{h}[/tex]

h' is image height

[tex]-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m[/tex]

Hence, this is the required solution.

It is found that an engine rejects 100.0 J while absorbing 125.0 J each cycle of operation.
(a) What is the efficiency of the engine?
(b) How much work does it perform per cycle?

Answers

a) 0.2
b) 25 J
.........

A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. What is the cheetah's average speed for the first 3.11 s of its sprint

Answers

Answer:

     18.75m/s

Explanation:

Given data

initial speed ,u= o m/s

final speed v= 25 m/s

time t= 3.11 seconds

Applying the first equation of motion we have

[tex]v= u+at\\\\v-u=at\\\\a= \frac{v-u}{t} \\\\a= \frac{25-0}{6.22} \\\\a= 4.02 m/s^2[/tex]

Also applying [tex]v=u+at[/tex] with initial speed set at 0, we can also find speed at 3.11 seconds

[tex]v=0+4.02*3.11\\v= 12.50 m/s[/tex]

Hence the average speed in the first 3.11 seconds is

[tex]\bar v= \frac{v}{2} \\\\\bar v=\frac{ 12.50}{2} \\\\ \bar v= 6.25 m/s[/tex]

Applying the expression [tex]\bar v=\bar u+at[/tex]  we can now fing the average speed in the second 3.11 second

[tex]\bar v=\bar u+at\\\\ \bar v=6.25+4.02*3.11\\\\ \bar v=18.75m/s[/tex]

Two long wires hang vertically. Wire 1 carries an upward current of 1.20 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.20 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force.
(a) Is this situation possible?
A. Yes
B. No
Is it possible in more than one way?
A. Yes
B. No
(b) Describe the position of wire 3.
distance ________ cm
direction: left of wire 1
(c) Describe the magnitude and direction of the current in wire 3.
magnitude
direction
down

Answers

Answer:

a) NO,   YES that cable 3 should be placed to the left of cable 1

b)  a = 8cm   the current is UP

Explanation:

The force between two cables that carries a current is given through the magnetic field created between them, it has the expression

        F₁ = μ₀ I₁ I₂ / 2π a l

where I₁ and I₂ are the current in each wire, at the separation between them and l the length of wire 1 on which the force is applied.

The direction of the force this given is given by the vector product between the current in wire 1 and the field created by 2, but we can summarize it:

* if the two currents are in the same direction the force is attractive

* if the two currents are contrary, the force is repulsive

With these relationships we can examine the different situations presented

a) Between cable 1 and 3 the force is zero. If cable 1 the current is upward in cable 3 it must be downward, so that the force is repulsive

Between wire 2 and 3. If wire 2 the current is downward, in wire 3 the current must be upward, for a repulsive force.

We see that in the two conditions the current in cable 3 has different directions, which is impossible, therefore this situation cannot occur

the answer is NO

The only way this is possible is that cable 3 should be placed to the left of cable 1

Answer YES

b) and c) Your question is unclear, I interpret that you want the position of cable 3 so that the force is zero on cable 3

we will assume that the current in cables 1 and 2 is ascending and that in cable 3 is descending

We write Newton's second law at the point of cable 3, the acceleration is zero

                    F13 - F23 = 0

                    F13 = F23

Let's calculate the forces

                  F₁₃ = μ₀ I₁ I₃ /2π a    l

                  F₂₃ = μ₀ I₂ I₃ /2π (d + a)    l

we substitute and simplify

                 I₁ / a = I₂ / (d + a)

where we have assumed that the length of all cables is the same

We solve to find

                I₁ (d + a) = I₂ a

                a (I₂ - I₁) = I₁ d₁

                a = I₁ / (I₂ - I₁) d

let's calculate

                a = 1.20 / (4.2 -1.2) 20

                a = 8cm

to the left of wire 1 and the direction is the current is UP

so that it is an attractive force between 1 and 3 and a repulsive force between 2 and 3

light of wavelength 610 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central meximum and the first minimum on the other side is 1.23°. What is the width of the slit?

Answers

Answer:

The width of the slit is    [tex]d = 5.68 *10^{-5} \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 610 \ nm = 610 *10^{-9} \ m[/tex]

     The  angle is  [tex]\theta = 1.23 ^o[/tex]

Generally the angle between the first minimum on one side and that the central maximum is evaluated as

            [tex]\theta _1 = \frac{\theta}{2}[/tex]

=>         [tex]\theta _1 = \frac{1.23}{2}[/tex]

=>         [tex]\theta _1 = 0.615 ^o[/tex]

Generally the condition for minimum diffraction is mathematically represented  as

        [tex]d sin \theta_1 = n\lambda[/tex]

For first minimum n = 1

=>    [tex]d = \frac{n \lambda }{ sin (\theta_1)}[/tex]

=>    [tex]d = \frac{1 * 610 *10^{-9}}{ sin (0.615)}[/tex]

=>    [tex]d = 5.68 *10^{-5} \ m[/tex]

A rectangular trough, 1.6 m long, 0.60 m wide, and 0.35 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge.
When you pull the plug, at what speed does water emerge from the hole?

Answers

Answer:

v = 2.61 m/s

Explanation:

Given that,

Length of a rectangular trough, l = 1.6 m

Breadth of a rectangular trough, b = 0.6 m

Depth of a rectangular trough, h = 0.35 m

We need to find the speed with which water emerge from the hole. It is a concept of efflux. The velocity of efflux is given by :

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.8\times 0.35}\\\\v=2.61\ m/s[/tex]

So, water will emerge from the hole with a speed of 2.61 m/s.

What is the wavelength of the light entering an interferometer if 402 bright fringes are counted when the movable mirror moves 0.115 mm?

Answers

Answer:

The  value is  [tex]\lambda = 572 *10^{-9} \ m[/tex]

Explanation:

From the question we are told that

    The  number of  bright fringes is  n =  402

     The  displacement of the mirror is  [tex]\delta l = 0.115 \ mm = 1.15 *10^{-4} \ m[/tex]

 

Generally the number of fringes is mathematically represented as

       [tex]n = \frac{2 * \delta l }{\lambda }[/tex]

=>    [tex]\lambda = \frac{2 * \delta l }{n }[/tex]

=>     [tex]\lambda = \frac{2 * 1.15*10^{-4} }{402 }[/tex]

=>   [tex]\lambda = 572 *10^{-9} \ m[/tex]

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