Choose the correct answer/s

Choose The Correct Answer/s

Answers

Answer 1
I’m pretty sure it’s 1 and 3 only that are true, hope this helps
Answer 2

refer to the attachment for the explanation

Choose The Correct Answer/s

Related Questions

formula of minimmum pressure​

Answers

Answer:

pressure=force/area

please help me to solve the problem
?cm=1m​

Answers


1m is equivalent to 100cm

1m = 100cm

As light from a star spreads out and weakens, do gaps form between the photons?​

Answers

Answer:

There are no gaps in space between the photons as they travel. If you look at light as a wave, then there no gaps unless specifically placed there on purpose. ... The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.

Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many cycles are completed in T/2 sec? (c) What is the maximum emf produced when the armature completes 180 0 rotation?

Answers

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

A carpenter measured the lengeth of a small piece of timber as 24.6cm .Calculate the relative error in the measurement if the true length is 24.5cm​

Answers

ANSWER:

0.4081%

Explanation:

Difference=24.6-24.5=0.1

Relative error = 0.1/24.5*100=0.4081%

Relative error is equal to the = difference between both the values/The true value *100

In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball

Answers

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

The fan on a personal computer draws 0.3 ft3/s ofair at 14.7 psia and 708F through the box containing the CPU and other components. Air leaves at 14.7 psia and 838F.Calculate the electrical power, in kW, dissipatedby the PCcomponents

Answers

Answer:

0.12 kW

Explanation:

Given that

The flow rate of air (V)=0.3 ft³/s

V=0.008 m³/s

Pressure, P=14.7 psia

P=1.013529 atm=101.325  kPa

Inlet temperature = 70° F=294.261 K

Exit temperature = 83° F=301.483 K

We know that , specific heat capacity of the air

Cp=1.005 kJ/kg.K

The mass flow rate of air is given as

[tex]\dot{m}=\dfrac{P\times V}{R\times T}\\\dot{m}=\dfrac{101.325\times 0.008}{0.287\times 294.261}\\\dot{m}= 0.0095\ kg/s[/tex]

By using energy conservation

[tex]Electric\ power =\dot{m}\times C_p\times (T_2-T_1)\\Electric\ power =0.0095\times 1.005\times (83-70)=0.12\ kW[/tex]

Therefore electric power dissipate by components will be 0.12 kW.

The total mass of eight identical
building blocks is 31.52 kg. Find the
mass of 1 block.​

Answers

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

Part E Now convert the time from seconds to years. This value is the approximate age of the universe. Write the age in scientific notation. Use the conversion, 1 year = 3.154 × 107 seconds.

Answers

Explanation:

The age of the universe is 13.8 billion years.

We know that,

1 billion years = 10⁹ years

So,

[tex]13.8\ \text{billion years}=13.8\times 10^9\ \text{years}[/tex]

We need to convert the age of the universe to the scientific notation.

Since, [tex]1\ \text{year}=3.154\times 10^7\ s[/tex]

So,

[tex]13.8\times 10^9\ \text{years}=13.8\times 10^9\times 3.154 \times 10^7\\\\=4.35\times 10^{17}\ s[/tex]

So, the age of the universe is [tex]4.35\times 10^{17}\ s[/tex].

Answer:

So the universe is approximately 1.34 x 10^10 years old

Explanation:

Edmentum

Unpolarized light, with an intensity of I0, is incident on an ideal polarizer. A second ideal polarizer is immediately behind the first and its axis of polarization is oriented at an angle of 60° relative to the first polarizer’s. How much of the light will be transmitted through the system?

Answers

Answer:

The light transmitted through the system will be 0.125*I₀.  

Explanation:

The light transmitted through the system can be found using Malus Law:

[tex] I = I_{0}cos^{2}(\theta) [/tex]    (1)

Where:

I: is the intensity of the light transmitted

I₀: is the initial intensity

θ is the angle relative to the first polarizer’s = 60°

Because the light transmitted by the first polarizer is dropped by half, the equation (1) results as:  

[tex] I = \frac{I_{0}}{2}cos^{2}(\theta) [/tex]

[tex] I = \frac{I_{0}}{2}cos^{2}(60) [/tex]

[tex] I = 0.125I_{0} = \frac{1}{8}I_{0} [/tex]

Therefore, the light transmitted through the system will be 0.125*I₀.  

I hope it helps you!

HELP me pleaseeee somebody

an object is placed 30cm from a mirror of focal length 15 cm the object is 7.5cm tall. where is the image located? how tall is the image??

Answers

Explanation:

It is given that,

Object distance from the mirror, u = -30 cm

Focal length of the mirror, f = +15 cm

Size of the object, h = 7.5 cm

We need to find the image distance and the size of the image.

Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]

Let h' is the size of the image. So,

[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]

So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.

On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.

Answers

Answer:

(a) By small angle approximation, we have;

F = -2×T×Δy/l

(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]

Explanation:

(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy

Therefore, the component of the tension T that form part of the restoring force is given as follows;

Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅

Then we have;

String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅  = 2×T×sin∅

Whereby the angle is small, we have;

sin∅ ≈ tan∅ = Δy/l

Which gives;

[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)

Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l

F = -2×T×Δy/l

(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;

By Hooke's law, F = -k×x

Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;

Which gives;

F = M×a = -k×Δy

a =  -k×Δy/M

d²(Δy)/dt² =  -k×Δy/M

When we put angular frequency as follows;

ω² = k/M

We get;

d²(Δy)/dt² =  -ω²×Δy

Which gives;

Δy(t) = A×cos(ωt + Ф)

The angular frequency is thus, ω = √(k/M)

Period of oscillation = 2·π/ω = 2·π/√(k/M)

The frequency of oscillation, f = 1/T = √(k/M)/(2·π)

Where:

k = 2·T/l, we have;

f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)

The frequency of oscillation is given as follows;

[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]


In your own words, discuss how energy conservation applies to a pendulum.
.
Where is the potential energy the most?
Where is the potential energy the least?
• Where is kinetic energy the most?
• Where is kinetic energy the least?

Answers

Answer:

when x=0 kinetic energy is maximum and potential energy is high

when it is in mean position the potential energy is maximum and vise versa

Explanation:

And x=0 when the pendulum is at highest point when moving like velocity become zero when ball reached highest point

mean position is the point when pendulum comes back to original position  

I hope u understand

The large-scale distribution of galaxies in the universe reveals Group of answer choices a smooth, continuous, and homogenous arrangement of clusters large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of universe a central void with walls of galaxies at the edge of the universe

Answers

Question

 

The large-scale distribution of galaxies in the universe reveals

A) a smooth, continuous, and homogenous arrangement of clusters

B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe

C) a central void with walls of galaxies at the edge of the universe

Group of answer choices

Answer:

The correct answer is B)

Explanation:

The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.

The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.

Other phenomena associated with the nature fo the universe is the existence of void spaces.

Cheers!

A 140-Hz sound travels through pure carbon dioxide. The wavelength of the sound is measured to be 1.92 m. What is the speed of sound in carbon dioxide?

Answers

Answer:

V = 268.8 m/s

Explanation:

The speed of a wave in general is given by the following formula:

V = fλ

where,

V = Speed of that wave

f = Frequency of the wave

λ = wavelength of the wave

In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:

V = Speed of Sound in Carbon dioxide = ?

f = frequency of sound wave = 140 Hz

λ = wavelength of sound wave = 1.92 m

Therefore,

V = (140 Hz)(1.92 m)

V = 268.8 m/s

The transfer of charge from clouds to the earth or cloud to cloud is called

Answers

That's called "lightning".

Answer:

The lightning itself is the transfer of charge from one region of a cloud to another or between the cloud and Earth. The narrow channel within which the flash of lightning occurs is heated suddenly to ~ 30,000 K, with essentially no time to expand.

Explanation:

Hope it helps

You stand at the top of a tall building with a stopwatch. You drop a rock off the side of the building, and it takes the rock 3.2 seconds to hit the ground. (Assume no air resistance.) solved for: [please help check!] the building is 50.176 m tall the impact speed is 31.36 m/s If you were standing at the bottom of this building, at what speed would you have to throw the rock for it to reach the top of the building?

Answers

Answer:

50.176m ; 31.36m/s ; 31.36m/s

Explanation:

Given the following :

Time (t) = 3.2s

Height(s) of the building

The initial velocity (u) will be 0

Using the equation:

S = ut + 0.5at^2

Acceleration due to gravity (a) = g = +9.8m/s^2 (downward)

S = 0*t + 0.5(9.8)(3.2^2)

S = 0 + 50.176

S = 50.176m

B.) speed impact when it touches the ground:

v^2 = u^2 + 2aS

where v is the final Velocity

v^2 = 0^2 + 2(9.8)(50.176)

v^2 = 0 + 983.4496

v = √983.4496

v = 31.36m/s

C) Speed of throw from the ground to reach the top of the building

Here we need the initial velocity

Height (s) = 50.176m

Acceleration due to gravity g = a = - 9.8m/s ( upward)

Using the third equation of motion:

S = ut + 0.5at^2

50.176 = u*3.2 + 0.5(-9.8)(3.2^2)

50.176 = 3.2u - 50.176

50.176 + 50.176 = 3.2u

100.352 = 3.2u

u = 100.352 / 3.2

u = 31.36m/s

A skateboarder rides down the street. When his feet push down on the
skateboard, what is the reaction force?

Answers

Answer:

the skateboard pushes up

Explanation:

newton's 3rd law says for every action, there's an equal and opposite reaction, so when his feet push on the skateboard, the skateboard pushes back up.

Answer:

the skateboard pushes up

Explanation:

explain why energy is important to us? ​

Answers

Answer:

we need it to work and without it we dont have strength to do anything

Answer:

energy is important to all living organisms. energy for producers comes from the sun, and energy for consumers comes from other living organisms. the abundance of energy available for organisms impacts the population.

The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is the kinetic energy of the protons when they are ejected from the cyclotron

Answers

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = [tex]\frac{v^2}{r}[/tex]             [v = linear velocity, r = radius of circular path]

=> F = m[tex]\frac{v^2}{r}[/tex]           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m[tex]\frac{v^2}{r}[/tex] = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m[tex]\frac{v^2}{r}[/tex] = q v B sin 90°

=> m[tex]\frac{v^2}{r}[/tex] = q v B

Divide both side by v;

=> m[tex]\frac{v}{r}[/tex] = qB

Make v subject of the formula

v = [tex]\frac{qBr}{m}[/tex]

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = [tex]\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}[/tex]

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = [tex]\frac{1}{2}[/tex]mv²

m = mass of proton

v = velocity of the proton as calculated above

K = [tex]\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )[/tex]

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 7 in. long and increasing at the rate of 0.3 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time

Answers

Answer:

my big long peen for ur mom

Explanation:

(a) In electron-volts, how much work does an ideal battery with a 17.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.88 × 1018 electrons pass through each second, what is the power of the battery?

Answers

Answer:

(a) 17.0eV

(b) 10.55W

Explanation:

(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;

W = q x V                 --------(i)

Where;

q = charge on the electron = 1e

From the question;

V = 17.0 V

Substitute the values of q and V into equation (i) as follows;

W = 1e x 17.0

W = 17.0eV

Therefore, the work done in electron volts is 17.0

(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;

P = (n q V) / t            --------------(ii)

Where;

n = number of electrons = 3.88 x 10¹⁸

t = 1s

q = 1.6 x 10⁻¹⁹C

V = 17.0V

Substitute these values into equation (ii) as follows;

P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1

P = 10.55W

Therefore the power of the battery is 10.55W

In a normal use bicycle, the angular speed of the minor gear has been found to be 25 rad / s. Find the angular speed of the larger gear, knowing that the small gear has a diameter of 4 R and the large gear 10 R.

Answers

Answer:

10 rad/s

Explanation:

The gears have the same linear speed.

ω₁ r₁ = ω₂ r₂

(25 rad/s) (2R) = ω (5R)

ω = 10 rad/s

What is at the very center of our galaxy? This question is difficult to answer because the center is completely obscured to our eyes behind a large amount of dust. On the basis of what you know about dust from previous chapters, what can be done to improve our view of the galaxy’s center?

Answers

Answer:

The answer is "Observe it in the infrared"

Explanation:

The infrared observatory is a device, which helps to track but also fix thermal light from sources out and the front of Earth's atmosphere like nebulae, young stars, and dust and gas in several other galaxies.

The very center of our galaxy is like a star cluster is situated in the central of our universe across the Milky Way supermassive, and the central black hole, around 4 million times our sun's mass.

The main super black hole named the Sagittarius A-star has been located in the nucleus of the galaxy, that's why in this question "Observe it in the infrared" is the correct answer.

Assignment: Ultrasound and Infrasound Research Exploration Ultrasound and infrasound are categories of sound with different frequencies. These categories of sound can be used for a variety of different applications. In this research assignment, you will take a look at the industrial applications of ultrasound and infrasound. Your essay should do the following things: • Define and describe in detail (and in your own words) ultrasound and infrasound • Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples. When conducting your research, remember to gather information from multiple sources. Start your research by defining ultrasound and infrasound. Then find ideas about industrial applications. The essay should be about 350-450 words, which is 1-1.5 typed pages, double-spaced, using 12-pt. font. Check the rubric to review how you will be graded for this assignment. plz no funny answers

Answers

Answer:

Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz.  Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

Explanation:

idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest

Answer:

Ultrasound vs. Infrasound Research Exploration

Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)

-Sav xx

Which of the following object is in dynamic equilibrium?

Answers

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

There is a Full Moon on September 14th. On which date will the New Moon occur? A.September 21st B.September 28th C.13th D.October 31st

Answers

Answer:

the answer B

Explanation:

1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.

Answers

Answer:

1) a    α,  m   I,  W=F.d    W =τ . θ,

2)  a = v²/r

Explanation:

1) The amounts of rotational and translational motion are related

acceleration is

        a = d²x / dt²

    linear displacement is equivalent to angular rotation, therefore angular acceleration is

      α = d²θ / dt²

force in linear motion is equivalent to moment in endowment motion

       F = m a

       τ = I α

the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia

          I = m r²

Work is defined by W = F. d

in rotation it is defined by W = τ . θ

The linear momentum is p = mv

the angular momentum L = I w

momentum the linear motion is I = F dt

in the rotation it is I = τ dt

 

2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).

      [tex]a_{c}[/tex]Ac = v²/r

Show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.

Answers

Explanation:

let us use the explanation below to get the intuition so desired;

According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.

The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the  motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created

Did to gravity, the moon has a much smaller acceleration than earth. How do you think that would affect the period of pendulum?

Answers

I think any pendulum would swing slower on the moon than it would on Earth.

The time it takes a pendulum to go through a complete back and forth swing is:

Time period = 2 π √(length/gravity)

You can see that 'gravity' is in the denominator of the fraction, so the smaller gravity gets, the longer the period gets.

To be a little bit more technical, the period is inversely proportional to the square root of gravity.

So the period for a complete swing on the moon would be  √(9.8/1.6) times as long as the complete swing of the same pendulum on Earth.

That number is roughly 2.47 .

So, for every 1 second that a pendulum takes to swing back and forth once on Earth, the same pendulum would take 2.47 seconds to do it on the moon.

Answer:

based on my opinion....

as we know that gravity in moon are less than gravity in earth.. since the force of gravity is less on the moon, the pendulum would swing slower at the same length, angle

and the frequency would be less.

I hope this helps

Other Questions
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