In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration

Answers

Answer 1

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what


Related Questions

Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago

Answers

Answer:

Your answer is( D) - Arago

What is the inductance of a coil if the coil produces an emf of 2.40 V when the current in it changes from -27.0 mA to 33.0 mA in 11.0 ms

Answers

Answer:

Inductance of a coil(L) = 0.44 H (Approx)

Explanation:

Given:

coil produces emf = 2.40 V

Old current = -27 mA

New current = 33 mA

Time taken = 11 mS

Find:

Inductance of a coil(L)

Computation:

Inductance of a coil(L) = -emf / [Δi / Δt]

Inductance of a coil(L) = -2.4 / [(-33 - 27) / 11]

Inductance of a coil(L) = -2.4 / [-5.4545]

Inductance of a coil(L) = 0.44 H (Approx)

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to:__________.
a) 4 Q.
b) 2 Q.
c) Q.
d) Q/2.
e) Q/4.

Answers

Answer:

D. Q/2

Explanation:

See attached file

I swing a ball around my head at constant speed in a circle with circumference 3 m. What is the work done on the ball by the 10 N tension force in the string during one revolution of the ball

Answers

Answer:

The work done on the ball by the tension force is 0 J.

Explanation:

The work can be calculated as follows:

[tex]W = |F|\cdot |d|cos(\theta)[/tex]

Where:

F: is the tension force = 10 N

d: is the displacement = ball's circumference = 3 m

θ: is the angle between the force and the distance = 90°

Hence, the work is:

[tex]W = |10| \cdot |3| cos(90) = 0 J[/tex]

Since the tension force and the displacement vector are orthogonal, the work done on the ball is zero.

                             

Therefore, the work done on the ball by the tension force is 0 J.

I hope it helps you!              

The work done on the ball by the 10 N tension force is zero ( 0 Joules).

Given that:

the circumference(displacement d) of the ball = 3 mthe tension force of the ball = 10 Nthe angle θ between the tension force and the displacement =90°

Using the work equation;

W = F × d cos θ

W = 10×3× cos (90)

W = 10 × 3 × 0

W = 0 Joules

Learn more about work done here:

https://brainly.com/question/13662169?referrer=searchResults

g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be

Answers

Question:

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

[tex]\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

=> [tex]R_{X} = \frac{R_1 * R_2}{R_1 + R_2}[/tex]             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

[tex]R_{X} = \frac{3 * 7}{3 + 7}[/tex]

[tex]R_{X} = \frac{21}{10}[/tex]

[tex]R_{X} = 2.1[/tex]Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.

Answers

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?

Answers

Answer:

Torque = 8.38Nm

Explanation:

Time= 8.00s

angular speed (w) =400 rpm

Moment of inertia (I)= 1.60kg.m2 about its rotation axis

We need to convert the angular speed from rpm to rad/ sec for consistency

2PI/60*n = 0.1047*409 = 41.8876 rad/sec

What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

Then we need to use the formula below for our torque calculation

from basic equation T = J*dω/dt ...we get

Where : t= time in seconds

W= angular velocity

T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

Therefore, constant torque that is required is 8.38 Nm

Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.

Given-

Inertia of the flywheel is 1.60 kg m squared.

Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,

[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]

[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]

[tex]\omega = 41.89[/tex]

Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.

When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,

[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]

[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]

[tex]\tau=8.38[/tex]

Hence, the required constant torque is 8.38 N-m.

to know more about the torque, follow the link below-

https://brainly.com/question/6855614

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]

1. A coil is formed by winding 250 turns of insulated 16-gauge copper wire, that has a diameter d = 1.3 mm, in a single layer on a cylindrical form of radius 12 cm. What is the resistance of the coil? Neglect the thickness of the insulation and the resistivity of copper is ???? = 1.69 × 10−8 Ω ∙ m.

Answers

Answer:

2.39 Ω

Explanation:

Given that

Number of winnings on the coil, = 250 turns

Radius if the copper wire, r(c) = 1.3/2 = 0.65 mm

Radius of single cylinder layer, R = 12 cm

Length of the cylinderical coil, L = 250 * 2π * 12 = 188.4 m

Resistivity of copper, ρ = 1.69*10^-8 Ωm

Area is πr(c)², which is

A = 3.142 * (0.65*10^-3)²

A = 3.142 * 4.225*10^-7

A = 1.33*10^-6 m²

The formula for resistance is given as

R = ρ.L/A, if we substitute, we have

R = (1.69*10^-8 * 188.4) / 1.33*10^-6

R = 3.18*10^-6 / 1.33*10^-6

R = 2.39 Ω.

Therefore, the resistance is 2.39 Ω

Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?

Answers

Answer:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.10 m. A concrete foundation wall is built all the way across the 8.90 m width of the excavation. This foundation wall is 0.189 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by 2.10 m ✕ 8.90 m ✕ 0.189 m ✕ 1000 kg/m3 ✕ 9.80 m/s2 = 34.6 kN.

Answers

Answer:

The  force on the foundation wall is   [tex]F_f = 191394 \ N[/tex]

Explanation:

From the question we are told that

     The  depth of the hole's vertical side is  [tex]d = 2.10 \ m[/tex]

      The  width of the hole is  [tex]b = 8.90 \ m[/tex]

      The distance of the concrete wall from the front of the cellar is  [tex]c = 0.189 \ m[/tex]

Generally the area which the water from the drainage covers is mathematically represented as

        [tex]A = d * b[/tex]

substituting values

        [tex]A = 2.10 * 8.90[/tex]

       [tex]A = 18.69 \ m^2[/tex]

Now the gauge pressure exerted on the foundation wall is mathematically evaluated as

          [tex]P_g = \rho * d_{avg} * g[/tex]  

Here  is the average height foundation wall where the pressure of the water is felt and it is evaluated as

         [tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]

where [tex]h_1[/tex] at the height at bottom of the hole which is equal to  [tex]h_1 = 0[/tex]

and  [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]

        [tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]

       [tex]d_{avg} = 1.05[/tex]

Where  [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]

substituting values

          [tex]P_g = 1000 * 1.05 * 9.8[/tex]

         [tex]P_g = 10290 \ Pa[/tex]

Then the force exerted by the water on the foundation wall mathematically represented as

      [tex]F_f = P_g * A[/tex]

substituting values

      [tex]F_f = 10290 * 18.69[/tex]

       [tex]F_f = 191394 \ N[/tex]

A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t

Answers

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same

Answers

Answer:

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70

Answers

Answer:

The  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

      [tex]I = \frac{\epsilon}{ R }[/tex]

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

The fractional change of reacting mass to energy in a fission reactor is about 0.1%, or 1 part in a thousand. For each kilogram of uranium that is totally fissioned, how much energy is released

Answers

Answer:

9*10^13 J

Explanation:

Given that

mass of the material, m = 0.001 kg

speed of light, c = 3*10^8 m/s

To solve this, we would be adopting Einstein's mass - energy relation...

E = mc²

Where

E = Energy, in joules

m = mass, in kg

c = speed of light, in m/s

E = 0.001 * (3*10^8)²

E = 0.001 * 9*10^16

E = 9*10^13 J

Thus, the total energy released would be 9*10^13 J

A telewision weighs 8.50 pounds. How many grams is this? (Hint: You need to
use two unit conversion fractions. 1 pound equals about 0.454 kg.)​

Answers

Answer:

3859 g

Explanation:

1 pound = 0.454 kg

therefore, 8.50 ponds = 0.454*8.50 = 3.859

to covert kilograms into grams you need to multiply it by 1000

=3.859*1000

= 3859 grams

A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from the star and is moving at 55 km/s. What is the semimajor axis of the planet's orbit

Answers

Answer:

32

Explanation:

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]

A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficient of kinetic friction between the rod and rails is 0.18 and the kinetic friction force is 0.212 N , what vertical magnetic field is required to keep the rod moving at a constant speed of 5.1 m/s

Answers

Answer:

The magnetic field is [tex]B = 8.20 *10^{-3} \ T[/tex]

Explanation:

From the question we are told that

   The  mass of the metal rod is  [tex]m = 0.12 \ kg[/tex]

    The current on the rod is  [tex]I = 4.1 \ A[/tex]

    The distance of separation(equivalent to length of the rod ) is [tex]L = 6.3 \ m[/tex]

     The coefficient of kinetic friction is [tex]\mu_k = 0.18[/tex]

      The kinetic frictional force is  [tex]F_k = 0.212 \ N[/tex]

     The constant speed is [tex]v = 5.1 \ m/s[/tex]

Generally the magnetic force on the rod is mathematically represented as  

      [tex]F = B * I * L[/tex]

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        [tex]F_ k = B* I * L[/tex]

=>      [tex]B = \frac{F_k}{L * I }[/tex]

=>       [tex]B = \frac{0.212}{ 6.3 * 4.1 }[/tex]

=>       [tex]B = 8.20 *10^{-3} \ T[/tex]

Electromagnetic waves are traveling in the vacuum of space. Calculate the wavelengths of these electromagnetic waves with the following frequencies. (Enter the first wavelength in pm and the second wavelength in cm.)
(a) 2.00 x 1019 Hz
(b) 4.50 x109 Hz

Answers

Answer:

(a) 1.5×10⁻¹¹ m.

(b) 6.7×10⁻² m

Explanation:

Note: All Electromagnetic wave travels in with the same speed, which 3×10⁸ m/s

(a) Give a frequency of 2.00×10¹⁹ Hz.

Using the equation of a wave,

V = λf................ Equation 1

Where V = Speed of electromagnetic wave, λ = wavelength, f = frequency.

make λ the subject of the equation

λ = V/f................. Equation 2

Given: f = 2.00×10¹⁹ Hz.

Constant: v = 3×10⁸ m/s.

Substitute into equation 2

λ = 3×10⁸/2.00×10¹⁹

λ = 1.5×10⁻¹¹ m.

(b) Similarly using

λ = v/f

Given: f = 4.5×10⁹ Hz, and v = 3×10⁸ m/s.

Substitute these values into equation 2 above.

λ = 3×10⁸/4.5×10⁹

λ = 6.7×10⁻² m

Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires

Answers

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

=  10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

=  10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

a uniform ladder of mass 100kg leans at 60° to the horizontal against a frictionless wall, calculate the reaction on the wall.​

Answers

Answer:

[tex]500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Explanation:

The reaction force is the force that is in the perpendicular direction to the wall.

We have an angle and a hypotenuse, we need to find the adjacent angle - so we can just use cos:

[tex]cos(\theta)=\frac{\text{adj}}{\text{hyp}}\\\text{hyp}*cos(\theta)=\text{adj}\\100*cos(60)=100*0.5=50\text{kg}[/tex]

However, we would like a force and not a mass.

[tex]W=mg\\W=50g\\W=500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Answer 1 if you use g as 10, answer 2 if you're studying mechanics in maths, answer 3 if you're studying mechanics in physics.

Which examples are simple machines?
Select all correct answers.
a hammer
an automobile
O a pulley
an inclined plane

Answers

A hammer and a pulley

You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.

a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.

Answers

Answer:

a)   d sin θ = m λ₀ / n

b)   d sin θ = (m + ½) λ₀ / n

c)    d = 2,439 10⁻⁷ m

Explanation:

For the interference these rays of light we must take as for some aspects,

* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2

* when the ray penetrates the lens the donut length changes by the refractive index

            λ = λ₀ / n

now let's write the destructive interference equation for these lightning bolts

           d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n

           d sin θ = m λ₀ / n

b) now nc> np

in this case there is no phase change in the reflected ray and the equation for destructive interference remains

             d sin θ = (m + ½) λ₀ / n

c) select the value of nc = 2.05 of the ZnO

we calculate the thickness of the film (d)

            d = m λ / (n sin 90)

in this type of interference the observation is normal, that is, the angle is 90º)

           d = 1 500 10-9 / (2.05 1)

           d = 2,439 10⁻⁷ m

d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference

A sample of lead has a mass of 26.00 kg and a density of 1.130 104 kg/m3 at 0°C. (Assume the average linear expansion coefficient for lead is 2.900 10-5(°C-1).)
(a) What is the density of lead at 82.00°C? (Give your answer to four significant figures.)
____ kg/m3
(b) What is the mass of the sample of lead at 82.00°C?
_____ kg

Answers

Answer:

Explanation:

coefficient of linear expansion α = 2.9 x 10⁻⁵

coefficient of volume expansion γ = 3 x 2.9 x 10⁻⁵ = 8.7 x 10⁻⁵

[tex]d_t = d_0( 1 - \gamma t )[/tex]

[tex]d_{82} = 1.13\times 10^4( 1 - 8.7\times 10^{-5}\times82 )[/tex]

= 1.13 x 10⁴ - 806.14 x 10⁻¹

= 1.13 x 10⁴ - 0.00806 x 10⁴

= 1.1219 x 10⁴ kg / m³

b ) mass of the sample will remain the same as mass does not increase or decrease with temperature .

Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N).
A) What is the magnitude of the component of Tom's weight parallel to the ladder?
B) What is the magnitude of the component of Tom's weight perpendicular to the ladder?

Answers

Answer: A) [tex]P_{x}[/tex] = 564.4 N

              B) [tex]P_{y}[/tex] = 374 N

Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:

[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]

[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]

side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]

side = 1.65

Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.

The image in the attachment illustrates the described above.

A) [tex]P_{x}[/tex] = P sen θ

[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]

[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]

[tex]P_{x}[/tex] = 564.4 N

B) [tex]P_{y}[/tex] = P cos θ

[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]

[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]

[tex]P_{y}[/tex] = 374 N

You want the output current from the secondary coil of a transformer to be 10 times the input current to the primary coil. The ratio of the number of turns N2/N1 must be:_____________.
A. 100
B. 10
C. 1
D. 0.1

Answers

Answer:

D. 0.1

Explanation:

Using transformer equation,

N2/N1 = I1/I2................... Equation 1

Where N2 = secondary coil, N1 = primary coil, I1 = input current, I2 = output current.

make I2  the subject of the equation

I2 = I1/(N2/N1)............ Equation 2

From equation 2 above, For the output current of the secondary coil to be 10 times the input current, N2/N1 = 0.1

Hence the right option is D. 0.1

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with a mass of 300 kg , is spinning at 23 rpm. John runs tangent to the merry-go-round at 4.4 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.

Required:
What is the merry-go-round's angular velocity, in rpm, after John jumps on?

Answers

Answer:

The merry-go-round's angular velocity 23.84 RPM

Explanation:

Given;

diameter of merry go round, d = 3 m

radius of the merry go round, R = 1.5 m

mass of the merry go round, m = 300 kg

angular velocity = 23 rpm

velocity of John, v = 4.4 m/s

mass of John, m = 30 kg

Apply conservation of angular momentum;

[tex]L_i = L_f[/tex]

[tex]I \omega_i + mvR = (I + mR^2)\omega _f[/tex]

where;

I is moment of inertia of disk

[tex]I = \frac{1}{2} mR^2\\\\I = \frac{1}{2} *300*1.5^2\\\\I = 337.5 \ kg.m^2[/tex]

Substitute in this value in the above equation;

[tex]337.5(2\pi \frac{23}{60} ) + (30*4.4*1.5) = (337.5 + 30*1.5^2) \omega_f\\\\812.9925 \ + \ 198 = 405 \omega _f\\\\1010.9925 = 405 \omega _f\\\\\omega _f = \frac{1010.9925}{405} \\\\\omega _f = 2.496 \ rad/s[/tex]

1 rad/s = 9.5493 rpm

2.496 rad/s = 23.84 RPM

Therefore, the merry-go-round's angular velocity 23.84 RPM

(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading is 37.25 cm). (Round your answer to one decimal place.)

Answers

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

[tex] M = log(\frac{I}{S}) [/tex]

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

Unpolarized light is incident upon two polarization filters in sequence. The two filters transmission axes are not aligned. If 18% of the incident light passes through this combination of filters, what is the angle between the transmission axes of the filters

Answers

Answer:

53°

Explanation:

I/Io*2= 0.18

0.18= cos²theta

Cos^-1(0.36) = 53°

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