What causes irregular galaxies to be irregular

The average angular acceleration of the fan is

[tex]\alpha_{\rm avg}=\dfrac{\Delta\omega}{\Delta t}=\dfrac{200\frac{\rm rev}{\rm min}-1000\frac{\rm rev}{\rm min}}{2\,\rm s}=-6.67\dfrac{\rm rev}{\mathrm s^2}[/tex]

The number of revolutions after time [tex]t[/tex] is given by

[tex]\theta=\omega_0t+\dfrac{\alpha}2t^2[/tex]

Acceleration is assumed to be constant, so [tex]\alpha=\alpha_{\rm avg}[/tex] and over the 2 second interval we have

[tex]\theta=\left(1000\dfrac{\rm rev}{\rm min}\right)(2\mathrm s)+\dfrac{\alpha_{\rm avg}}2(2\,\mathrm s)^2=20\,\mathrm{rev}[/tex]

so the answer is A.

No of revolution is defined as the no of loops taken from the starting to the end to the end from the same point started. The number of the revolution blade experiences will be 20.

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}[/tex]

given ,

initial angular velocity = 1000 rev/min = 16.67 rev /sec

final angular velocity =  200 rev/min = 3.33 rev/sec

time taken = 2 second

[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}\\\\\\\rm{\alpha _{avg}= \frac{3.33-16.67}{0.3}}\\\\\\\rm{\alpha _{avg}= -6.67 rev/ sec^2[/tex]

According to newtons second law of motion,

No of revolution after time t

[tex]\rm{\theta= \omega_it+\frac{1}{2} \alpha t^{2} }\\\\\\\rm{\theta= \ 16.67\times2+\frac{1}{2}(-6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 33.3- 13.3}\\\\\\\rm{\theta= \ 20rev }\\\\\\[/tex]

Therefore the number of revolutions the blade undergoes will be 20 revolutions.

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electric fields and magnetic fields both exhibit similar properties. where magnetic fields are natural electric fields are not but man made and can do the same things as a magnetic field.

this is your teacher get off your phone

[tex](a) \( v_0 = 41 \, \text{m/s} \)(b) \( h = 45 \, \text{m} \)[/tex]

To find the initial speed of the rock, we can use the kinematic equation that relates final velocity [tex](\( v_f \)),[/tex]  initial velocity [tex](\( v_i \)),[/tex] acceleration [tex](\( a \))[/tex], and displacement [tex](\( d \))[/tex]. The equation is:

[tex]\[ v_f^2 = v_i^2 + 2ad \][/tex]

Given:

- Final velocity [tex](\( v_f \))[/tex]= 27 m/s (when it hits the ground)

- Displacement [tex](\( d \))[/tex]= 33 m (height of the cliff, since the rock starts from the top)

Substitute the known values into the equation:

[tex]\[ 27^2 = v_i^2 + 2(-9.8)(33) \]\[ 729 = v_i^2 - 646.8 \]\[ v_i^2 = 729 + 646.8 \]\[ v_i^2 = 1375.8 \]\[ v_i = \sqrt{1375.8} \]\[ v_i ≈ 37.1 \, \text{m/s} \][/tex]

The initial speed of the rock is approximately 37.1 m/s.

Next, to find the greatest height of the rock, we can use the same kinematic equation, but this time solving for the displacement [tex](\( d \))[/tex], with [tex]\( v_f = 0 \)[/tex] since the rock reaches its highest point momentarily before falling back down.

[tex]\[ v_f^2= v_i^2 + 2ad \]\[ 0^2 = 37.1^2 + 2(-9.8)d \]\[ 0 = 1374.41 - 19.6d \]\[ 19.6d = 1374.41 \]\[ d = \frac{1374.41}{19.6} \]\[ d ≈ 70.15 \, \text{m} \][/tex]

However, since the height is measured from the base of the cliff, we need to subtract the initial height of the cliff:

[tex]\[ h = 70.15 - 33 \]\[ h = 37.15 \, \text{m} \][/tex]

So, the greatest height of the rock, measured from the base of the cliff, is approximately 37.15 meters.

The calculations used basic kinematic equations of motion to determine the initial speed and greatest height of the rock. The equations take into account the acceleration due to gravity [tex](\( -9.8 \, \text{m/s²} \))[/tex] )and the given displacements and final velocities.

Complete Question:
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 33 m high. When it hits the ground at the base of the cliff the rock has a speed of 27 m/s. (a) Assuming that air resistance can be ignored, find the initial speed of the rock. m/s (b) Find the greatest height of the rock as measured from the base of the cliff. m

(a) 373.4 J

The displacement of the book is

[tex]d=14.0 m - 1.30 m=12.7 m[/tex]

While the gravitational force acting on the book is

[tex]F=mg=(3.00 kg)(9.8 m/s^2)=29.4 N[/tex]

And the force (downward) is parallel to the displacement (downward), so the work done by gravity is

[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]

(b) -373.4 J

The work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground). SInce the total mechanical energy of the Earth-book system must be conserved, this means that potential energy U has been converted into kinetic energy K: therefore, the loss in potential energy U is exactly -373.4 J.

(c) 411.6 J

The gravitational potential energy U is given by:

[tex]U=mgh[/tex]

where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground). Substituting,

[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)=411.6 J[/tex]

(d) 38.2 J

When the book reaches the hand, its height above the ground level is

h = 1.30 m

therefore, the gravitational potential energy this time is

[tex]U=mgh=(3.00 kg)(9.8 m/s^2)(1.30 m)=38.2 J[/tex]

(e) 373.4 J

The work done by gravity does not change if we change the value of the potential energy at ground level. In fact, the work done by gravity is still calculated as before:

[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]

(f) -373.4 J

As we did in point b), the work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground), so the loss in potential energy is equal to the work done by gravity, and this value does not depend on the value of the potential energy at ground level, so it is still -373.4 J.

(g) 511.6 J

The gravitational potential energy U is given by:

[tex]U=mgh+U_0[/tex]

where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground), and [tex]U_0 = 100 J[/tex] is the potential energy at ground level, which must be added into the formula. Substituting,

[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)+100 J=511.6 J[/tex]

(h) 138.2 J

As before, we can calculate the potential energy of the book at a height of h=1.30 m, adding 100 J of energy which is the value of the potential energy at ground level. we find:

[tex]U=mgh+U_0=(3.00 kg)(9.8 m/s^2)(1.30 m)+100 J=138.2 J[/tex]

The work done by gravity, Wg, is calculated to be 372.66 J. The gravitational potential energy, U, decreases by this amount during the fall, from 411.60 J at the release point to 38.94 J at the friend's hands. If the ground level has an arbitrary gravitational potential energy of 100 J, the final values are adjusted accordingly to 511.60 J and 138.94 J.

To calculate the work Wg done by the gravitational force as the book drops to the friend's hands, we use the formula W = mgh, where m is the mass of the book, g is the acceleration due to gravity (9.8 m/s²), and h is the height through which the book falls. The change in height is the total distance D minus the distance d, which the friend's hands are above the ground. This gives us the effective height the book falls through.

(a) The work done by gravity, Wg, is:

Wg = mgh
Wg = (3.00 kg)(9.8 m/s²)(14.0 m - 1.30 m)
Wg = (3.00 kg)(9.8 m/s²)(12.7 m)
Wg = 372.66 J

(b) The change in the gravitational potential energy ΔU is equal to the work done by gravity, so ΔU = 372.66 J.

(c) Initially, the gravitational potential energy U at the release point is U = mgh, which is:

U = (3.00 kg)(9.8 m/s²)(14.0 m)
U = 411.60 J

(d) When the book reaches the friend's hands, the gravitational potential energy is:

U = (3.00 kg)(9.8 m/s²)(1.30 m)
U = 38.94 J

If the gravitational potential energy at ground level is 100 J instead of 0 J, the values for Wg and ΔU do not change because these are quantities that depend only on the change in height. However, the absolute values of U at the initial and final points will be 100 J more than previously calculated:

(e) Wg remains 372.66 J

(f) ΔU remains 372.66 J

(g) U at the release point now is U + 100 J:
U = 411.60 J + 100 J
U = 511.60 J

(h) U at the friend's hands now is U + 100 J:
U = 38.94 J + 100 J
U = 138.94 J

Answer:

1. When light passes from a more dense to a less dense substance the light is refracted.

2. The diamond would act like a prism and make rainbows.

3. You observe all of these pretty much every day. Like if you notice how on a glass building it reflects the sunlight and some other things around it. Then another example could be just shining a flash light in a room.

I do FLVS and I got a 100 on this assignment so I hope it helped a little :)

Answer:

both energy will remain constant and potential energy must be more than the kinetic energy of the block

So correct Graph will be

A) option

Explanation:

As we know that kinetic energy of the block is given as

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

potential energy is given as

[tex]U = mgh[/tex]

so when a block is stationary at the surface of table so due to height of the table the potential energy of the block is given as

[tex]U = mgh[/tex]

but as we know that the speed of the block is zero

so kinetic energy must be

[tex]K = 0[/tex]

so here in the graph both energy will remain constant and potential energy must be more than the kinetic energy of the block

In order of decreasing wavelength:

Radio Waves

Microwaves

Infrared light

Red light

Violet light

Ultraviolet light

X-rays

Gamma rays

Explanation:

The electromagnetic spectrum is a classification of all the electromagnetic waves depending on their wavelength and their frequency. The waves with longest wavelength are radio waves, which have wavelenght that spans from a few mm up to several km. The waves with shortes wavelength are gamma rays, which have wavelength shorter than 10 picometers.

More or less at the centre of the electromagnetic spectrum light the visible part of the spectrum, which is usually classified into 7 colors: red, orange, yellow, green, blue, indigo and violet (decreasing wavelength). Red light is the color with longer wavelenght (approx. 750 nanometers), while violet light has the shortest wavelength (approx. 380 nanometers).

Electromagnetic radiation can take multiple forms. The electromagnetic radiation with the longest wavelength is of radio while the shortest is of gamma rays.

Electromagnetic radiation can be defined as the flow of energy at the universal speed of light through a material medium or free space.

Some examples are radio waves, gamma-ray, etc.

The ranking of the following kinds of electromagnetic radiation in order of decreasing wavelength is shown below,

The topmost is having a low frequency, long wavelength, and low quantum energy, while the radiation at the bottom is having a high frequency, short wavelength, and high quantum energy.

Hence, the electromagnetic radiation with the longest wavelength is of radio while the shortest is of gamma rays.

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There are four wavelengths seen in the image of the sound wave.

Four wavelengths are seen in this image of a sound wave.

The distance between the corresponding spots of two successive waves is known as wavelength. Two points or particles that are in the same phase—i.e., points that have completed identical fractions of their periodic motion—are referred to as "corresponding points."

The wavelength is typically measured from crest to crest or from trough to trough in transverse waves (waves with points vibrating at right angles to the direction of their advance); from compression to compression or from rarefaction to rarefaction in longitudinal waves (waves with points vibrating in the same direction as their advance).

The wavelength also describes the pattern of disturbance brought about by the energy moving away from the sound source. Longitudinal waves are what make up sound. This indicates that the direction of energy wave propagation is parallel to the direction of particle vibration propagation.

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electric potential

Electric potential is the electric potential energy per unit charge.

Mathematically; V =PE/q

Where; PE is the electric potential energy, V is the electric potential and q is the charge.

Electric potential is more commonly known as voltage.  If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.

Answer:

c. electric potential

Explanation:

on edg

Answer:

All the 4 options

Explanation:

Electromagnetic induction occurs when the magnetic flux through a coil of wire is changing over time:

[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]

where

[tex]\epsilon[/tex] is the emf induced in the coil

[tex]\Delta \Phi[/tex] is the variation of magnetic flux

[tex]\Delta t[/tex] is the variation of time

The presence of an emf in the coil will generated an induced current.

The magnetic flux through the coil is given by

[tex]\Phi = BA cos \theta[/tex]

where

B is the intensity of the magnetic field

A is the area of the coil

[tex]\theta[/tex] is the angle between the direction of the field and the axis of the coil

We see that any actions that changes one of these 3 variables will change the magnetic flux through the coil, so it will also induce a current.

The 4 options are:

1.Moving the loop outside of the magnetic field region. --> this will decrease the intensity of the magnetic field, B, therefore it will change the flux, and it will induce a current

2.Spin the loop such that its axis does not consistently line up with the magnetic field direction. --> this will change the angle between the direction of the coil's axis and the field B, so this will also change the flux, and therefore will induce a current

3.Change the magnitude of the magnetic field. --> this will change the magnitude of B, so this will also change the flux, and therefore will induce a current

4.Change the diameter of the loop --> this will change the area of the coil A, so this will also change the flux, and therefore will induce a current

Therefore, all 4 options are correct.

An induced emf in a loop of wire can result from moving the loop out of the magnetic field region, spinning the loop to change its orientation, or changing the magnitude of the magnetic field itself.

An induced electromotive force (emf) occurs in a loop of wire when the magnetic flux through the loop changes. This can happen in several ways, such as:

Changing the diameter of the loop does not induce an EMF unless the magnetic field or the loop's orientation relative to the field changes at the same time.

Kinetic (mechanical) energy

Protons are about 1836 times bigger than electrons.

Hope this helps :)

Please mark brainliest xo

I see TWO statements here:

#1).   Earth is much farther away from the Sun in Winter.

#2).  That's what makes it cold in the Winter.

Let's use our heads on #1 for just a little moment:

--  When it's Winter in Canada, the USA, and France, it's Summer in Australia, Paraguay, and Namibia.  This is the end of December, all of January and February, and most of March.

-- When it's Winter in Tasmania, South Africa, and Botswana, it's Summer in Germany, Israel, and Mexico.  This is the end of June, all of July and August, and most of September.

So WHOSE Winter is the question asking about ? ?  It doesn't say.

Earth is FARTHEST away from the Sun during the first week in July, and it's NEAREST to the Sun during the first week in January.

The difference between the nearest distance and the farthest distance is about 3%.  It has almost ZERO effect on how hot or cold the days are.  Anywhere !

So BOTH of the statements in the question are FALSE.

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

[tex]\tau = Fdsin \theta[/tex]

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

[tex]\theta[/tex] is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

Mechanical advantage

Answer:

Mechanical Advantage

Explanation:

5250/750 m³ or 7 m³

formula is density x volume = mass

Answer:

Volume of wood, [tex]V=7\ m^3[/tex]

Explanation:

It is given that,

Mass of wood, m = 5250 kg

The density of wood, [tex]d=750\ kg/m^3[/tex]

We need to calculate the volume of wood. It can be calculate using the formula of density. It is given by :

[tex]d=\dfrac{m}{V}[/tex]

[tex]V=\dfrac{m}{d}[/tex]

[tex]V=\dfrac{5250\ kg}{750\ kg/m^3}[/tex]

[tex]V=7\ m^3[/tex]

So, the volume of wood is 7 cubic meter. Hence, this is the required solution.

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

[tex]r'=\frac{2}{3}r[/tex]

So the new force will be

[tex]F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F[/tex]

So, the force will be 2.25 times the previous value.

Choose a heavier bat

Final answer:

To boost momentum, a batter is suggested to choose a heavier bat due to its larger moment of inertia and the greater force it can impart on the ball, according to the principles of moment of inertia, angular momentum, and Newton's third law of motion.

Explanation:

To increase momentum at the plate, a batter can choose a heavier bat. This is because a more massive bat has more inertia and therefore, when swung, will impart a larger force on the ball for the same amount of time compared to a lighter bat. This is evident from the concepts of moment of inertia and angular momentum. When a bat is swung, the moment of inertia plays a role in determining the bat's resistance to changes in its rotational speed. A heavier bat, generally, has a larger moment of inertia, allowing it to maintain its speed through the batting zone and thus create a more forceful impact, enhancing the ball's exit velocity.

Considering the batter swinging a Wiffle ball bat, swinging at the end, furthest from the pivot point, increases the moment of inertia and thus requires more torque to achieve the same angular acceleration as grabbing the middle. However, the increased distance from the pivot allows for a higher linear speed at the end of the bat, creating a larger angular momentum on contact which could result in a more powerful impact if the batter can successfully swing the bat quickly.

In a collision, such as a bat hitting a ball, both objects exert forces on each other, as explained by Newton's third law of motion. This interaction also informs us that not only the ball is affected by the bat but the force exerted by the ball also impacts the bat, as any baseball player who has felt the sting of a ball hit off the end of the bat can attest.

This is a conceptual problem, I’ll try to upload a picture:

Answer :

a = 1 m/s^2

Explanation:

Given information

Net force is a constant

The mass of the truck is [tex]m_{o}[/tex]

Initial acceleration [tex]a_{o}[/tex]= 2 m/[tex]s^{2}[/tex]

The mass of the truck is increased twice as much

[tex]F_{net}= m_{o} a_{o}[/tex] = ma

[tex]F_{net}= m_{o} a_{o}[/tex] = [tex]2m_{o}[/tex]a

a=[tex]\frac{a_{o} } {2}[/tex]

a = 2/2

a = 1 m/[tex]s^{2}[/tex]

Thermal energy also known as 'heat energy' is a form of kinetic energy.

A) the sound waves were pushed closer together

Answer:

The correct choice is

A) the sound waves were pushed closer together.

Explanation:

As the ambulance gets closer, the distance between the ambulance and marge decreases and so does the wavelength of the siren observed by marge. As the wavelength decrease and speed remains the same for the siren, the frequency of the siren increases because wavelength, frequency and speed are related as

Frequency = Speed/wavelength

Clearly, the frequency is inversely related to wavelength.

A= -2 due to is a desacceleration

Answer:

[tex]a = -0.89 m/s^2[/tex]

Explanation:

initial speed of the car is given as

[tex]v_i = 20 miles/hour[/tex]

here we know that

[tex]v_i = 20 \times \frac{1609}{3600}[/tex]

[tex]v_i = 8.94 m/s[/tex]

final speed of the car is given as

[tex]v_f = 10 mile/hour[/tex]

[tex]v_f = 10 \times \frac{1609}{3600}[/tex]

[tex]v_f = 4.47 m/s[/tex]

now we have

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

[tex]a = \frac{4.47 - 8.94}{5}[/tex]

[tex]a = -0.89 m/s^2[/tex]

a. [tex]11.28\Omega[/tex]

The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:

[tex]R_{eq}=R_1 + R_2[/tex]

In this circuit, we have

[tex]R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega[/tex]

Therefore, the equivalent resistance is

[tex]R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega[/tex]

b. 5.8 V, 3.2 V

First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:

[tex]I=\frac{V}{R_{eq}}[/tex]

where V = 9.00 V and [tex]R_{eq}=11.28 \Omega[/tex]. Substituting,

[tex]I=\frac{9.00 V}{11.28 \Omega}=0.8 A[/tex]

Now we can calculate the potential difference across each resistor by using Ohm's law again:

[tex]V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V[/tex]

[tex]V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V[/tex]

The equivalent resistance of the circuit is 11.28 Ω, and the current through the circuit is approximately 0.798 A. The potential differences across the 7.25 Ω resistor and the 4.03 Ω resistor are approximately 5.79 V and 3.21 V, respectively.

To solve the problem, we need to brreak it down in the following steps:

Part a):

Part b):

Thus, the potential differences across the resistors are approximately 5.79 V and 3.21 V, respectively.

Neither source is a renewable one.

Answer:

B. Whales and porpoises signal each other by making sounds

Explanation:

Sound is a type of mechanical wave, which consists of oscillations of the particles in a medium. It is also classified as longitudinal wave, which means that the direction of the oscillation of the particles in the medium is parallel to the direction of propagation of the wave, forming alternating regions of higher particle density (compressions) and lower particle density (rarefactions).

Since sound is a mechanical wave, it just need a medium to propagate through. Therefore, it can travel through solids, liquids and gases.

In particular, the option

B. Whales and porpoises signal each other by making sounds

is an example of this: these animals emit ultrasound waves (a type of sound waves with high frequency), that travel through the water, and then are reflected back to the animal, allowing the animal to understand the distance of the object from which the wave has been reflected back.

Well hydrogen would be the main element, as a process called nuclear fusion with both helium and hydrogen atoms occurs within stars. And planets are the products of dead stars that have burned through their supplies of hydrogen, helium, and carbon. Planets are a product of this.

Stars and planets are mainly composed of hydrogen and helium, making up from 96 to 99% of their total mass. They also contain heavier elements, the concentrations of which vary among different stars. This fundamental composition was first discovered by Cecilia Payne-Gaposchkin in 1925.

The stars and planets in the universe are composed mainly of hydrogen and helium, as determined through analyses of stellar spectra. These elements constitute from 96 to 99% of the total mass of most stars. However, they also contain heavier elements in varying quantities. For instance, Population I stars, like the Sun, contain around 1-4% of heavy elements. However, Population II stars found in the outer galactic halo and in globular clusters have much lower concentrations.

These elements must have been fused somewhere. The stars, having incredibly high temperatures, are speculated to be the source of this element formation. Therefore, stars are believed to play a critical role in the chemical richness that characterizes our world.

The discovery that stars are primarily composed of light gases like hydrogen and helium was first made in a pioneering thesis by Cecilia Payne-Gaposchkin in 1925, making a significant contribution to our understanding of the universe. This composition is also found in the material between the stars, with about 99% of it being a gas composed of individual atoms or molecules of hydrogen and helium.

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I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door [tex]d[/tex])...

The door forms a right triangles that satisfies

[tex]\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}[/tex]

We also have

[tex]\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd[/tex]

so if you happen to know the height of the door, you can solve for [tex]b[/tex] and [tex]a[/tex].

[tex]d[/tex] is fixed, so

[tex]a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}[/tex]

We can solve for the angular velocity [tex]\dfrac{\mathrm d\theta}{\mathrm dt}[/tex]:

[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}[/tex]

At the point when [tex]\theta=40^\circ[/tex] and [tex]\dfrac{\mathrm db}{\mathrm dt}=1.8[/tex] ft/s, we get

[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}[/tex]

The angular velocity of the door is [tex]\rm \( 0.744 \, \text{rad/s} \)[/tex], and the velocity of end D is approximately [tex]\rm \( 3.6 \, \text{ft/s} \)[/tex] in the horizontal direction and [tex]\rm \( 2.6964 \, \text{ft/s} \)[/tex] in the upward direction.

To determine the angular velocity of the door and the velocity of end D, we'll need to use the concept of relative velocity.

Let:

- [tex]\rm \( v_b = 1.8 \, \text{ft/s} \)[/tex] (velocity of wheel B)

- [tex]\( \theta = 40^\circ \)[/tex] (angle between horizontal and the line connecting wheel A to wheel B)

We'll use the following relations for the velocity components:

- For wheel B: [tex]\rm \( v_b = v_{b_x} \hat{i} + v_{b_y} \hat{j} \)[/tex]

- For end D: [tex]\rm \( v_d = v_{d_x} \hat{i} + v_{d_y} \hat{j} \)[/tex]

Since the door's velocity at point B is vertical [tex]\rm (\( v_{b_x} = 0 \))[/tex] and directed upward [tex]\rm (\( v_{b_y} = 1.8 \, \text{ft/s} \))[/tex], we can calculate the angular velocity [tex]\rm (\( \omega \))[/tex] of the door using the relative velocity formula:

[tex]\rm \[ \omega = \frac{{v_b}}{{R}} \][/tex]

Where [tex]\rm \( R \)[/tex] is the distance from point B to the axis of rotation.

Next, to calculate the velocity of end D, we use the following relationship:

[tex]\rm \[ v_d = v_b + \omega \times \vec{r} \][/tex]

Where [tex]\rm \( \vec{r} \)[/tex] is the position vector from point B to point D.

Since [tex]\rm \( \theta = 40^\circ \)[/tex], we can express [tex]\rm \( \vec{r} \)[/tex] as:

[tex]\rm \[ \vec{r} = r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j} \][/tex]

Substituting the values and calculating:

[tex]\rm \[ R = r \cos(\theta) \]\\\\rm \omega = \frac{{v_b}}{{r \cos(\theta)}} \][/tex]

Now, substituting [tex]\rm \( v_b = 1.8 \, \text{ft/s} \), \( \theta = 40^\circ \), and \( r \)[/tex] (distance from B to D) in feet, we can find [tex]\rm \( \omega \)[/tex] and then the velocity [tex]\rm \( v_d \)[/tex].

(Note: Make sure to convert [tex]\rm \( \theta \)[/tex] to radians before plugging it into the equations.)

Let's assume the distance from B to D [tex](\( r \))[/tex] is 5 feet.

[tex]\[ \theta = 40^\circ = \frac{{40}}{{180}} \pi \, \text{radians} \approx 0.698 \pi \, \text{radians} \]\\\ \\R = 5 \cos(0.698 \pi) \approx 2.42 \, \text{ft} \]\\\\ \omega = \frac{{1.8}}{{2.42}} \approx 0.744 \, \text{rad/s} \][/tex]

Now, using [tex]\rm \( \vec{r} = 5 \cos(0.698 \pi) \hat{i} + 5 \sin(0.698 \pi) \hat{j} \)[/tex]:

[tex]\rm \[ v_d = 1.8 \hat{j} + (0.744 \, \text{rad/s}) \times (5 \cos(0.698 \pi) \hat{i} + 5 \sin(0.698 \pi) \hat{j}) \]\\\\\\rm v_d = 1.8 \hat{j} + 0.744 \, \text{rad/s} \times (2.42 \hat{i} + 3.63 \hat{j}) \]\\\ \\\rm v_d = 1.8 \hat{j} + 1.8 \hat{i} + 2.6964 \hat{j} \approx (1.8 + 1.8) \hat{i} + (2.6964) \hat{j} \approx 3.6 \hat{i} + 2.6964 \hat{j} \, \text{ft/s} \][/tex]

So, the angular velocity of the door is [tex]\rm \( 0.744 \, \text{rad/s} \)[/tex], and the velocity of end D is approximately [tex]\rm \( 3.6 \, \text{ft/s} \)[/tex] in the horizontal direction and [tex]\rm \( 2.6964 \, \text{ft/s} \)[/tex] in the upward direction.

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usually it is watts

The much more common rating for an appliance is Power.

Power is the rate at which energy is used.  Small appliances ... like toasters and blow-driers ... are rated in watts.  Large appliances ... like industrial electric water boilers, electric construction cranes, and elevator motors ... are rated in Kilowatts (1 kilowatt is 1,000 watts).  This is the RATE at which the appliance uses energy.  You leave it running longer, you use more energy, and you owe more on your electric bill.

Some appliances are advertised and sold with an energy rating.  This is the amount of energy it would consume if you used it the way a normal standard person uses it, for some standard amount of time like 1 month or 1 year.  If the appliance does have such a rating, it'll be in units of Kilowatt-hours.  But it'll be no better than a rough estimate, because you may use the appliance more often or less often than the normal standard person.  

Answer:

1. Changing the magnetic field intensity

2. Changing the area enclosed by the loop

3. Changing the orientation of the loop of wire with respect to the magnetic field

Explanation:

Electromagnetic induction occurs when there is a variation of magnetic flux through a coil of wire: as a result, an emf (electromotive force) is induced in the coil, according to the equation

[tex]\epsilon=-\frac{\Delta \Phi}{\Delta t}[/tex]

where

[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the coil

[tex]\Delta t[/tex] is the time elapsed

The magnetic flux through a coil of wire is given by:

[tex]\Phi = BA cos \theta[/tex]

where

B is the magnetic field intensity

A is the area enclosed by the coil

[tex]\theta[/tex] is the angle between the direction of B and the perpendicular to the area enclosed by the coil

As we can see, the magnetic flux depends on these three factors, so changing any of them will change the magnetic flux, and an electromotive force will be induced in the coil as a result.

Voltage can be induced in wire loops by changing the magnetic field passing through the loop, physically moving the wire within a stationary magnetic field, or altering its electrical resistance.

There are three primary ways in which voltage can be induced in a loop of wire:

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Answer:

Explanation:

its the name of an element

A chemical symbol represents the name of an element. These symbols are abbreviations used for chemical elements or compounds in chemistry.

A chemical symbol in Chemistry stands for the name of an element. Chemical symbols are abbreviations used in chemistry for chemical elements, functional groups or chemical compounds. These symbols consist of letters - for example, the symbol for Hydrogen is H, Oxygen is O, and Carbon is C.

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