The headlights of a car are 1.4 m apart. What is the maximum distance (in km) at which the eye can resolve these two headlights? Take the pupil diameter to be 0.30 cm. (Assume the average wavelength of visible light is 555 nm.)

Answers

Answer 1

Answer:

5.4x10^4km

Explanation:

See attached file

The Headlights Of A Car Are 1.4 M Apart. What Is The Maximum Distance (in Km) At Which The Eye Can Resolve

Related Questions

A man has vocal cords of length 22 mm, with a mass per length of 0.0042 kg/m. What tension is required in the vocal cords in order to produce a tone of middle C (261.62 Hz)?

Answers

Answer:

Tension, T = 0.556 N

Explanation:

It is given that,

Length of vocal cords, l = 22 mm = 0.022 m

Mass per unit length, [tex]\mu=0.0042\ kg/m[/tex]

We need to find the tension is required in the vocal cords in order to produce a tone of middle C of frequency 261.62 Hz. The frequency in terms if tension is given by :

[tex]f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/tex]

T = tension in the vocal cords

[tex]f^2=\dfrac{1}{4l^2}\times \dfrac{T}{\mu}\\\\T=4l^2\mu f^2\\\\T=4\times (0.022)^2\times 0.0042 \times (261.62 )^2\\\\T=0.556\ N[/tex]

So, the tension in the vocal cords is 0.556 N.

You stand near the edge of a swimming pooland observe through the water an object lying on the bottom of thepool. Which of the following statements correctly describes whatyou see?
a. The apparent depth of the object is less than thereal depth.
b. The apparent depth of the object is greater thanthe real depth.
c. There is no difference between the apparent depth and the actual depth of the object.

Answers

Answer:

a

Explanation:

The correct answer would be that the apparent depth of the object is less than the real depth.

The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.

Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.

The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.

Correct option: a

A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find:

a. The speed of the skater after throwing the stone.
b. The distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between his skates and the ice is 0.03.

Answers

Answer:

Explanation:

know that there is no external force on skater and the stone so the total momentum of the system will remains constant

so we will have

here we have

so the skater will move back with above speed

now the deceleration of the skater is due to friction given as

Answer:

(a) 3.553 m/s

(b) 21.46 m

Explanation:

(a) Applying the law of of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u'  = mv+m'v'.................. Equation 1

Where m and m' are the mass of skater and stone respectively,  u and u' are the initial velocity of skater and stone respectively, v and v' are the final velocity of the skater and the stone respectively.

Note, u = 0 m/s, u' = 0 m/s

Therefore,

0 = mv+m'v'

-mv = m'v'................ Equation 2

make v the subject of the equation

v = -m'v'/m............. Equation 3

Given: m = 45 kg, m' = 7.65 kg, v' = 20.9 m/s

Substitute into equation 3

v = 7.65(20.9)/45

v = -3.553 m/s

Hence the speed of the skater = 3.553 m/s

(b) F = mgμ..............Equation 4

But F = ma

Therefore,

ma = mgμ

a = gμ............... Equation 5

Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction

Given: μ = 0.03, g = 9.8 m/s²

Substitute into equation 5

a = 0.03(9.8)

a = 0.294 m/s²

Using the equation of motion,

v² = u²+2as............. Equation 6

Where s = distance moved by the skater.

note that u = 0 m/s.

therefore,

v² = 2as

s = v²/2a................ Equation 7

Given: v = 3.553 m/s, a = 0.294

Substitute into equation 7

s = 3.553²/(2×0.294)

s = 12.62/0.588

s = 21.46 m

Question 8
A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The
damping factor for the spring is determined to be 0.4 and the spring oscillates 12 times each
second. Find an equation for the displacement, D(t), of the spring from equilibrium in terms of
seconds, t.
D(t) =

Can someone please help me ASAP?!!!!

Answers

Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]

where:

|a| is initil displacement

[tex]\frac{2.\pi}{\omega}[/tex] is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]

period = [tex]\frac{2.\pi}{\omega}[/tex]

12 = [tex]\frac{2.\pi}{\omega}[/tex]

ω = [tex]\frac{\pi}{6}[/tex]

Replacing values:

[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]

The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].

4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?

Answers

Answer:

64 times

Explanation:

if increase of 1 gives you 32

then increase of 2 will give you its double

64

If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.

What is Earthquake?

An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.

The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.

The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.

To know more about Earthquake:

https://brainly.com/question/1296104

#SPJ5

A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?

Answers

Answer:

The current is  [tex]I = 2042\ A[/tex]

Explanation:

From the question we are told that

    The length of the solenoid is  [tex]l = 2.2 \ m[/tex]

    The  radius is  [tex]r_i = 30 \ cm = 0.30 \ m[/tex]

    The number of turn is [tex]N = 1200 \ turns[/tex]

    The  magnetic field is  [tex]B = 1.4 \ T[/tex]

The  magnetic field produced  is mathematically represented as

         [tex]B = \frac{\mu_o * N * I }{l }[/tex]

making [tex]I[/tex] the subject

       [tex]I = \frac{B * l}{\mu_o * N }[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]

 substituting values

        [tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]

        [tex]I = 2042\ A[/tex]

How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.

Answers

Answer:

2.5×10¹⁹

Explanation:

4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second

A particle with a charge of 4.0 μC has a mass of 5.0 × 10 -3 kg. What electric field directed upward will exactly balance the weight of the particle?

Answers

Answer:

E = 12.25 x 10³ N/C = 12.25 KN/C

Explanation:

In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,

Electrostatic Force = Weight

E q = mg

where,

E = Electric Field = ?

m = Mass of the Charge = 5 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C

Therefore,

E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)

E = 0.049 N/4 x 10⁻⁶ C

E = 12.25 x 10³ N/C = 12.25 KN/C

(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)

Answers

Answer:

a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c)  λ = 0.486 m , d)     λ = 0.35 m

Explanation:

a) The speed of a wave on a string is

          v = √T /μ

also all the waves fulfill the relationship

          v = λ f

they indicate that the fundamental frequency is f = 980 Hz.

The wavelength that is fixed at its ends and has a maximum in the center

          L = λ / 2

          λ = 2L

we substitute

           v = 2 L f

let's calculate

           v = 2  0.243  980

           v = 476.28 m / s

b) The tension of the rope

             T = v² μ

the density of the string is

            μ = m / L

            T = v² m / L

            T = 476.28²   0.717 / 0.243

            T = 6.69 10⁵ N

           

c)          λ = 2L

            λ = 2  0.243

            λ = 0.486 m

d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air

          v = λ f

          λ= v / f

          λ = 343/980

          λ = 0.35 m

Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. There are several measurements you can make.

Answers

Answer:

The quantities to measure are:

* the distance to the screen

* The distance from the central maximum to each interference

* in order of interference

* wavelength

Explanation:

To determine the gap spacing we must use the constructive interference equation

            d sin θ = m λ

as the angles are small

          tan θ = sin θ / cos θ

          tan θ = sin θ

and the definition of tangent is

          tan θ = y / L

Thus

         sin θ = y / L

when replacing

          d y / L = m λ

          d = m λ L / y

with this equation we can know what parameter should be measured.

The quantities to measure are:

* the distance to the screen

* The distance from the central maximum to each interference

* in order of interference

* wavelength

Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.


Work done on the skateboard


a. Rodger Work= 0J


b. Rodger work= 1300J


c. sally work= 1300J


d. sally work= 5850J


e. rodger work= 5850J

Answers

Answer:

b. Rodger work = 1300 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of the force.

From the question,

Work is done by Rodger using a force of 100 N  in pushing the skateboard through a distance of 13.0 m.

W = F×d............. Equation 1

Where W = work done, F = force, d = distance.

Given: F = 100 N, d = 13 m

Substitute these values into equation 1

W = 100(13)

W = 1300 J.

Hence the right option is b. Rodger work = 1300 J

Huygens claimed that near the surface of the Earth the velocity downwards of an object released from rest, vy, was directly proportional to the square root of the distance it had fallen, . This is true if c is equal to

Answers

Answer:

the expression is False

Explanation:

From the kinematics equations we can find the speed of a body in a clean fall

        v = v₀ - g t

         v² = V₀² - 2 g y

If the body starts from rest, the initial speed is zero (vo = 0)

            v= √ (2g y)

In the first equation it gives us the relationship between speed and time.

With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.

Therefore the expression is False

A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s
With what angular speed is the stick spinning after the collision?

Answers

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s

final velocity of bullet [tex]v_{2}[/tex] = 140 m/s

loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]

==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

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

70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]

566.28 = [tex]v^{2}[/tex]

[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
O A. Ferromagnetic
O B. Non-magnetic
C. Diamagnetic
O D. Paramagnetic​

Answers

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

Answer:

its paramagnetic

Explanation:

i took this quiz

Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center

Answers

Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m  

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I =  3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"

Answers

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = 14691.456 J   is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

[tex]P = \frac{w}{t}[/tex]

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

Solve 3* +5-220t = 0​

Answers

Answer:

t = 27.5

Explanation:

[tex]3 + 5 -220t = 0[/tex]

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

A horizontal uniform meter stick is supported at the 50.0 cm mark. It has a mass of 0.52 kg, hanging from it at the 20.0 cm mark and a mass of 0.31 kg mass hanging from the 60.0 cm mark. Determine the position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance. Group of answer choices

Answers

Answer: 70.5 cm

Explanation:

The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.

You will use the moment techniques.

That is,

Sum of the clockwise moment = sum of anticlockwise moments

Please find the attached file for the remaining explanation and solution.

If the direction of the position is north and the direction of the velocity is up, then what is the direction of the angular momentum

Answers

Answer:

the direction of angular momentum = EAST

Explanation:

given

Direction of position = r = north

Direction of velocity = v = up

angular momentum = L = m(r x v)

where m is the mass, r is the radius, v is the velocity

utilizing the right hand rule, the right finger heading towards the course of position vector and curl them toward direction of velocity, at that point stretch thumb will show the bearing of the angular momentum.

then L = north x up = East

EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4​πrhoa4 rhoπa2 = a2 4​ .

Answers

Answer:

Radius of gyration = a/2.

Explanation:

So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;

"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."

(NB: I changed the "rho" word to its symbol).

Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.

Therefore, the Radius of gyration = a/2.

A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 2.60 cm thick flat piece of crown glass and back to air again. The beam strikes the glass at a 28.0° incident angle.
A) At what angles do the two colors emerge?
B) By what distance are the red and blue separated when they emerge?

Answers

Answer:

A: 28°

B. 1x10^-3M

Explanation:

See attached file

Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.

Answers

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = [tex]m_{p}[/tex] = 117.5 kg

speed of player = [tex]v_{p}[/tex] = 6.5 m/s

mass of ball = [tex]m_{b}[/tex] = 0.43 kg

velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] =  2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J  gained

c) if they travel in opposite direction, equation becomes

[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex]  = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J  lost

In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum

Answers

Complete  Question

In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?

Answer:

a

  [tex]\theta = 0.3819^o[/tex]

b

  [tex]y = 0.00386 \ m[/tex]

Explanation:

From the question we are told that

    The slit separation is  [tex]d = 150 \lambda[/tex]

    The  distance from the screen is  [tex]D = 57.9 \ cm = 0.579 \ m[/tex]

 

Generally the condition for constructive interference is mathematically represented as

            [tex]dsin (\theta ) = n * \lambda[/tex]

=>        [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]

where  n is the order of the maxima  and value is 1 because we are considering the central maximum and an adjacent maximum

     and  [tex]\lambda[/tex] is the wavelength of the light

So

       [tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]

       [tex]\theta = 0.3819^o[/tex]

Generally the distance between the maxima is mathematically represented as

       [tex]y = D tan (\theta )[/tex]

=>    [tex]y = 0.579 tan (0.3819 )[/tex]

=>    [tex]y = 0.00386 \ m[/tex]

An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsional spring with spring constant k. If a concentrated torque of magnitude Ta is applied in the center of the bar, what is the rotation at the end of the bar, φ(L)? Assume a constant shear modulus G and polar moment of inertia J.

Answers

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.

Answers

Answer:

a) 6738.27 J

b) 61.908 J

c)  [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

[tex]I[/tex] =  [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] =  [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

[tex](I_{1} +I_{2} )w[/tex]

where the subscripts 1 and 2 indicates the values first and second  flywheels

[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

where m is the mass of the car

[tex]v_{car}[/tex] is the velocity of the car

Equating the energy

2246.09 =  [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

making m the subject of the formula

mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

As a wheel turns, the angle through which it has turned varies with time as β(t)=Ct + Bt3 where C=0.400rad/s and B=0.0120rad/s3. Calculate the angular velocity w(t) as a function of time.

Answers

Answer:

ω(t) = 0.4 + 0.036 t²

Explanation:  

The angular displacement of the disk is given as the function of time:

β(t) = Ct + B t³

where,

C = 0.4 rad/s

B = 0.012 rad/s³

Therefore,

β(t) = 0.4 t + 0.012 t³

Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:

ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)

ω(t) = 0.4 + 0.036 t²

A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?

Answers

Answer:

Explanation:

a = F / m

where a is acceleration , F is thrust and m is mass

taking log and differentiating

da / a = dF / F - dm / m

(da / a)x 100 = (dF / F)x100 - (dm / m) x100

percentage increase in a = percentage increase in F - percentage increase in m

= percentage increase in acceleration a   = 39 - 13 = 26 %

required increase = 26 %.

"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."

Answers

Answer:

9.6×10⁹ A

Explanation:

From the question above,

P = VI.................... Equation 1

Where P = Electric power, V = Voltage, I = current.

make I the subject of the equation

I = P/V............. Equation 2

Given: P = 200 kW = 200×10³ W, V = 48000 V.

Substitute these vales into equation 2

I = 200×10³×48000

I = 9.6×10⁹ A.

Hence the current in the line is 9.6×10⁹ A.

What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The wavelength is   [tex]\lambda= \frac{2 \pi }{k}[/tex]

Explanation:

From the question we are told that  

      The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]

       The magnetic field is  [tex]\= B = B_0 sin (kx -wt) \r k[/tex]

From the above equation

and  k is the wave number which is mathematically represented as

        [tex]k = \frac{2 \pi }{\lambda }[/tex]

=>     [tex]\lambda= \frac{2 \pi }{k}[/tex]

Where [tex]\lambda[/tex] is the wavelength

A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home

Answers

Answer:

7.68×10^25kg

Explanation:

The formula for energy used per year is calculated as

Energy used per year =12 x Energy used per month

By substituting Energy used per month in the above formula, we get

Energy used per year =12 x 1600kWh

= 19200kWh

Conversion:

From kWh to J:

1 kWh=3.6 x 10^6 J

Therefore, it is converted to J as

19200 kWh =19200 x 3.6 x 10^6 J

= 6.912×10^10 J

Hence, energy used per year is 6.912×10^10 J

To find the mass that is converted to energy per year.

E = MC^2 ............1

E is the energy used per year

C is the speed of light = 3.0× 10^8m/s

Where E= 6.912×10^10 J

Substituting the values into equation 1

6.912×10^10 J = M × 3.0× 10^8m/s

M = 6.912×10^10 J / (3.0× 10^8m/s)^2

M = 6.912×10^10 J/9×10^16

M = 7.68×10^25kg

Hence the mass to be converted is

7.68×10^25kg

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