Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm

Answers

Answer 1

This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V[tex]_D[/tex] = ω[tex]_c[/tex] × r[tex]_c[/tex]  ----- let this be equ 1

where V[tex]_D[/tex] is velocity of cylinder,  ω[tex]_c[/tex] is the angular velocity of drum C and r[tex]_c[/tex] is the radius of drum C

Now, Angular velocity of gear B is;

ω[tex]_B[/tex] = ω[tex]_C[/tex]

ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]  -------- let this equ 2

so;

V[tex]_D[/tex] / 0.1 m = 10V[tex]_D[/tex]

Next, we determine the angular velocity of gear A;

from the diagram;

ω[tex]_A[/tex]( 0.15 m ) = ω[tex]_B[/tex]( 0.2 m )

from equation 2; ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]

so

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / r[tex]_c[/tex] ) 0.2 m

substitutive in value of radius r[tex]_c[/tex] (0.1 m)

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / 0.1 m ) 0.2 m

ω[tex]_A[/tex]( 0.15 ) = 0.2V[tex]_D[/tex] / 0.1

ω[tex]_A[/tex] =  2V[tex]_D[/tex]  / 0.15

ω[tex]_A[/tex] = 13.333V[tex]_D[/tex]   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑[tex]U_{1-2[/tex] = T₂

0 + m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}I_A[/tex]ω²[tex]_A[/tex] + [tex]\frac{1}{2}I_B[/tex]ω²[tex]_B[/tex]

so

m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}[/tex](m[tex]_A[/tex]k[tex]_A[/tex]²)(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex](m[tex]_B[/tex]k[tex]_B[/tex]²)(10V[tex]_D[/tex])²

we given that; m[tex]_D[/tex] = 50 kg, h = 2 m, m[tex]_A[/tex] = 10 kg, k[tex]_A[/tex] 125 mm = 0.125 m, m[tex]_B[/tex] = 30 kg, k[tex]_B[/tex] = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( [tex]\frac{1}{2}[/tex] × 50 × V[tex]_D[/tex]²) + [tex]\frac{1}{2}[/tex]( 10 × (0.125)² )(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex]( 30 × (0.15)²)(10V[tex]_D[/tex])²

981 = 25V[tex]_D[/tex]² + 13.888V[tex]_D[/tex]² + 33.75V[tex]_D[/tex]²

981 = 72.638V[tex]_D[/tex]²

V[tex]_D[/tex]² = 981 / 72.638

V[tex]_D[/tex]² = 13.5053

V[tex]_D[/tex] = √13.5053

V[tex]_D[/tex] = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Determine The Speed Of The 50-kg Cylinder After It Has Descended A Distance Of 2 M, Starting From Rest.

Related Questions

Someone please help me with this question
How are mass and weight different?

Answers

Answer:

The mass is the amount of matter in something. Weight is mass times gravitational field strength and so also includes gravity. Mass is measured in grams (g) and weight is measured in Newtons (N)

Yellow light (600 nm) passes through a
diffraction grating with d = 2.35 x 10-6 m.
What is the angular separation between
the second (m = 2) and third (m = 3)
maximum?

Answers

Answer:

19.284

Explanation:

The answer is right on Accellus.

A ray of light has a wavelength of
284 nm in glass (n = 1.51). What
is its wavelength in a vacuum?

Answers

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, [tex]\lambda=284\ nm[/tex]

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

[tex]n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm[/tex]

So, the wavelength in vacuum is equal to 428.8 nm.

If 12 coulombs of electric charge pass a point in 4.0
seconds, the current is
A) 8.0 amperes
B) 16 amperes
C) 3.0 amperes
D) 48 amperes

Answers

Answer:

AS

Explanation:

Which statement can be made about the amplitude of any transverse wave?

It is the length from the midpoint to the crest.
It is half the length from the midpoint to the trough.
It is the length of the wavelength.
It is half the length of the wavelength.

Answers

Answer:

A) It is the length from the midpoint to the crest.

Explanation:

The correct statement about the amplitude of any transverse wave is: It is the length from the midpoint to the crest. The correct option is A.

What is a transverse wave?

A transverse wave is a type of wave in which the displacement of the medium is perpendicular to the direction of wave propagation.

Examples of transverse waves include light waves and water waves in which the surface of the water oscillates up and down while the wave moves horizontally.

The properties of a transverse wave include:

Amplitude: The maximum displacement of the medium from its rest position. This represents the intensity or strength of the wave.

Wavelength: The distance between two adjacent points in the wave that are in phase with each other, for example, between two consecutive crests or troughs. It is typically represented by the symbol λ.

Frequency: The number of complete wave cycles that pass a point in a given amount of time, usually measured in hertz (Hz). The frequency is inversely proportional to the wavelength and is represented by the symbol f.

Period: The time taken for one complete wave cycle to pass a given point, usually represented by the symbol T. The period is directly proportional to the wavelength and inversely proportional to the frequency.

Speed: The speed at which the wave propagates through the medium, usually represented by the symbol v. The speed is directly proportional to the frequency and wavelength.

These properties are related to each other by the wave equation:

v = fλ,

Where v = the speed of the wave,

f = the frequency,

and λ = the wavelength.

Therefore, The correct statement about the amplitude of any transverse wave is A. It is the length from the midpoint to the crest.

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Which kind of kinetic energy is a rolling ball?

Answers

rotational, hope this helps!

Please help help help !!!!! ( science) is

Answers

Answer:

I believe it's applied force (push)

It’s applied force (push) hope this help you have a good time

If a ball is travelling in a circle of diameter 10m with velocity 20m/s, find the angular velocity of the ball.​

Answers

Answer:

4 rad/s

Explanation:

The angular velocity is velocity over time:

[tex]\displaystyle{\omega = \dfrac{v}{r}}[/tex]

We know that the velocity is 20 m/s and the radius is half of diameter which is 10m/2 = 5 m. Hence, substitute in the formula:

[tex]\displaystyle{\omega = \dfrac{20 \: \text{m/s}}{5 \: \text{m}}}\\\\\displaystyle{\omega = 4 \: \text{rad/s}}[/tex]

Therefore, the angular velocity is 4 rad/s

What two pigments would be subtracted to reflect a red object?

a. yellow and magenta

b. magenta and cyan

c. green and blue

d. cyan and red

Answers

(C)Green and blue

Just took a test in science with this!

A mass on a
string is swung in a circle of radius 1.0m
at a speed of 5.0mls. What is the period of
revolution?​

Answers

Answer:

S = 2 * pi * 1 m = 6.28 m = distance traveled

V = S / T   or   T = S / V = 6.28 m / 5 m/s = 1.26 sec

This will be the time for 1 revolution or the period of the motion.    

Explained Kepler's laws, satellites motion and weightlessness

Answers

Johannes Kepler was a German mathematician and astronomer who formulated three laws of planetary motion in the early 17th century. These laws describe the motion of planets around the Sun and are based on observations made by Tycho Brahe.

1- The first law, also known as the law of elliptical orbits, states that the orbit of a planet around the Sun is an ellipse, with the Sun at one of the two foci. This means that the distance between the planet and the Sun varies as the planet moves in its orbit.

2- The second law, also known as the law of equal areas, states that the line connecting the planet to the Sun sweeps out equal areas in equal amounts of time. This means that the planet moves faster when it is closer to the Sun and slower when it is farther away.

3- The third law, also known as the law of harmonic motion, states that the square of the period of a planet's orbit is proportional to the cube of its average distance from the Sun. This means that the farther a planet is from the Sun, the longer it takes to complete one orbit.

A satellite is an object that orbits around a planet or other celestial body. The motion of a satellite is determined by the gravitational force of the planet or body it is orbiting. The satellite is constantly falling towards the planet due to gravity, but it is also moving horizontally, which keeps it from crashing into the surface. This combination of falling and horizontal motion is what creates the orbit.

Weightlessness, also known as zero gravity, is the state of being weightless or having very little weight. It is experienced by objects in orbit around a celestial body, such as a planet or moon. In this environment, objects do not experience the normal force of gravity that they would on the surface of the body. This can create the sensation of weightlessness for objects and humans inside a spacecraft or on a space station. Weightlessness can also occur in other situations, such as during freefall or in an aircraft following a parabolic trajectory.


Hi, can anyone help me with this question I have?


Assume a tornado is in the vicinity of your neighborhood. Which would
be the most useful information to determine your safety - speed or
velocity? Explain your answer.

Answers

Answer:

speed

Explanation:

you need to move fast so the tornado can't reach you

While skiing in Jackson, Wyoming, your friend Ben (of mass 63.2 kg) started his de- scent down the bunny run. 11.5 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben's kinetic energy at the bottom of the bunny run? Use g = 9.8 m/s Answer in units of J.​

Answers

Answer:

Approximately [tex]7.1 \times 10^{3}\; {\rm J}[/tex] (given: [tex]g = 9.8\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

To find the change in the gravitational potential energy ([tex]\text{GPE}[/tex]), use the formula:

[tex]\begin{aligned}& (\text{change in GPE}) \\ &= (\text{mass})\, (g)\, (\text{change in height})\end{aligned}[/tex].

Assume that gravitational field strength [tex]g[/tex] is constant (e.g., [tex]g = 9.8\; {\rm m\cdot s^{-2}}[/tex].) For an object of mass [tex]m[/tex], if the altitude of the object changes by [tex]\Delta h[/tex], the [tex]\text{GPE}[/tex] of that object would change by [tex]m\, g\, \Delta h[/tex].

In this question, the mass of Ben is [tex]m = 63.2\; {\rm kg}[/tex]. It is given that [tex]g = 9.8\; {\rm m\cdot s^{-2}} = 9.8\; {\rm N\cdot kg^{-1}}[/tex] and is constant. Since change in the altitude of Ben is [tex]\Delta h = 11.5\; {\rm m}[/tex], the change in the ([tex]\text{GPE}[/tex]) of Ben would be:

[tex]\begin{aligned} m\, g\, \Delta h &= (63.2\; {\rm kg}) \, (9.8\; {\rm N\cdot kg^{-1}})\, (11.5\; {\rm m}) \\ &\approx 7.1\times 10^{3}\; {\rm N\cdot m} = 7.1\times 10^{3}\; {\rm J} \end{aligned}[/tex].

Which are dangerous places to put a flammable substance? Check all that apply.

near a flame
on a scale
near a ruler
in a sink
near a hot plate

Answers

Answer:

near a flame and a hot plate

Explanation:

An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon and the wavelength (in nm) of each.

Answers

Answer:

Explanation:

An electron accelerates through a 12.5 V potential difference, starting from rest, so it will acquire kinetic energy of 12.5 eV .

In hydrogen atom energy of n th orbit in terms of eV is given as follows

En = -13.6 / n² eV

Total energy of 1 st orbit  E₁ = - 13.6 eV

Total energy of 2 st orbit E₂ = - 13.6 eV / 2² = - 3.4 eV

Total energy of 3 st orbit E₃ = - 13.6 eV / 3² = - 1.5  eV

Total energy of 4 st orbit E₄ = - 13.6 eV / 4² = - 0.85 eV

E₄ - E₁ = 13.6 - 0.85 = 12.75 eV

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV .

The electron has energy of 12,5 eV so it can excite electron from E₁ to E₃ . .

Jump possible = E₃ to E₂ , E₂ to E₁ and E₃ to E₁

Energy of E₃ to E₂ = 3.4 - 1.5 eV = 1.9 eV

wavelength = 1237 / 1.9 nm = 651 nm

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

wavelength  = 1237 / 12.10  nm = 102.23 nm

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV

wavelength  = 1237 / 10.2  nm = 121.27  nm

wavelength of photon possible are 651 nm , 121.27  nm , 102.23 nm .

Q12. How big is a Moon? How big is a Mars? What is therefore the weight of the person from Q11 on the Moon? What is the person's weight on Mars?

Answers

Answer:

The moon is 1,079.4 mi.

Mars is 2,106.1 mi

Multiply your weight by the moon's gravity relative to earth's, which is 0.165. Solve the equation. In the example, you would obtain the product 22.28 lbs. So a person weighing 135 pounds on Earth would weigh just over 22 pounds on the moon

Being that Mars has a gravitational force of 3.711m/s2, we multiply the object's mass by this quanitity to calculate an object's weight on mars. So an object or person on Mars would weigh 37.83% its weight on earth.

Explanation:

~Hope this helps

PLS THIS IS DUE IN 2 MINUTES
Which has more momentum, a 0.5kg toy car moving a 5 m/s or a 1000kg real car that is
parked?

Answers

Answer:

The toy car. An object that isn't moving has no momentum

Explanation:

A lawnmower is pushed with a force of 72 Newtons at an angle of 40 degrees from the horizontal. How much work is done to the mower if it moves 740 meters?

Answers

The work done to the mower is the product of force and the distance covered by the object which is equal to 40,811 Joules.

What is Work done?

Work done can be defined as the product of external force and the distance over which the force is being applied. Work is done on an object when a force is applied to an object and the object is moved through a particular distance.

From this we can see that:

X-Component:

cos(θ) = x/F

Then,

X= F cos(θ).…..(1)

Now putting the values of F and θ in equation (1) we get,

X = 72 N × cos(40)

X = 72 N × 0.766

X= 55.15 N this is the x-component force

The work done to the mower is:

W = F × d

W = 55.15 × 740

W = 40,811 Joules

The work done to the mower is 40811 Joules.

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The reliability or reproducibility of a measurement is its _____

Answers

The degree of data stability when the measurement is replicated under identical circumstances is known as reproducibility or reliability.

What exactly are repeatability and reproducibility?

Reproducibility determines how an entire study an experiment can be replicated, whereas repeatability assesses the variation in data made by a single equipment or human under similar circumstances.

What makes repeatability crucial?

Science needs reproducibility because it enables more in-depth investigation, and replication validates our findings. There are several investigations and experiments, which result in a wide range of variables, unforeseen, and things that are either outside your influence or you cannot guarantee.

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the work a force does is measured in?​

Answers

Answer:

newtons

Explanation:

In the metric system of units, where force is measured in newtons (abbreviated N), work is measured in newton-meters (N-m). For reference, a newton is roughly equal to the force exerted on your hand by a baseball.

Answer:

newtons

In the metric system of units, where force is measured in newtons (abbreviated N), work is measured in newton-meters (N-m). For reference, a newton is roughly equal to the force exerted on your hand by a baseball.

PLS THIS IS DUE IN 2 MINUTES
Which has more momentum, a 0.5kg toy car moving a 5 m/s or a 1000kg real car that is
parked?

Answers

Answer:

The toy car

Explanation:

the real car is parked so yeah but maybe in some way technically the real car has more "momentum"

Tritium, the 3H atom, consists of a nucleus of one proton and two neutrons with a single electron. It is unstable and decays via beta emission to singly ionized helium, 3 He , consisting of a nucleus of two protons and one neutron with a single electron. This decay takes place instantaneously and thus there occurs a sudden doubling of the Coulomb interaction between the atomic electron and the nucleus. If the tritium atom is in its ground state when it decays, determine the probability that the 3He atom is in its ground state immediately after the decay.

Required:
Determine the probability that the 3H^+ atom is in its ground state immediately after decay.

Answers

Answer: hello your question is poorly written attached below is the complete question

P(  ³₂He₊ )  at ground state = 1

Explanation:

Determine the probability that the  3He atom is in its ground state after decay

From the attached solution the coulomb interaction of ( ³₂He)  is double that of H³ . given that coulomb interaction is attractive ( -ve ) this will make  the product to become more stable hence the product ( ³₂He₊ ) will be gotten at the ground state

i.e. P(  ³₂He₊ )  at ground state = 1

attached below is a detailed solution

n experimental vehicle starts from rest (0 = 0) at = 0 and accelerates at a rate given by
= (7m/s
3
). What is (a) its velocity and (2) its displacement 2 later?

Answers

The velocity of the experimental vehicle after 2 seconds is 14 m/s.

The displacement of the experimental vehicle after 2 seconds is 14 m.

What is the velocity and displacement of the experimental vehicle?

The velocity of the experimental vehicle is the speed of the in a given direction starting from rest.

The velocity, v, of the experimental vehicle after 2 seconds is calculated using the formula below:

v = u + at

where;

u is the initial velocity = 0 m/s

a is the acceleration = 7 m/s²

t is time = 2 seconds

v = 0 + 7 * 2

v = 14 m/s

The displacement, s, of the vehicle is calculated below as follows:

v² = u² + 2as

14² = 0² + 2 * 7 * s

s = 196 / 14

s = 14 m

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A 62.3 kg base runner begins his slide into second base while moving at a speed of 4.49 m/s. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s². What is the magnitude of the mechanical energy lost due to friction acting on the run- ner?​

Answers

Answer:

628 J

Explanation:

ME = PE + KE

PE = mgh = 0,  because h = 0

KEi (initial KE) = 1/2mv² = 1/2(62.3 kg)(4.49 m/s)² = 628 J

KEf (final KE) = 0, because v-final = 0

All the initial ME, which is all KE, converts to thermal E due to friction

Ruby was training for a race at the high school track. She took her dog walked 2250 meters in 15 minutes. Calculate the average speed

Answers

Answer:

Average speed [tex]= 2.5[/tex] meter per second

Explanation:

Given

The total distance walked [tex]= 2250[/tex] meters

The total time taken to walk [tex]2250[/tex] meters is [tex]15[/tex] minutes

In one minute there are [tex]60[/tex] seconds

As we know Average speed is equal to total distance travelled divided by total time taken

Hence,

Average speed

[tex]\frac{2250}{15*60} \\2.5[/tex]meter per second

While skiing in Jackson, Wyoming, your friend Ben (of mass 63.2 kg) started his de- scent down the bunny run, 11.5 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben's kinetic energy at the bottom of the bunny run? Use g = 9.8 m/s² Answer in units of J.- answer :7100J part 2 of 2 What is his final velocity? Answer in units of m/s.​

Answers

Answer:

12.3 m/s

Explanation:

To calculate the final velocity of Ben, we can use the equation KE = 1/2mv^2, where KE is the kinetic energy and m is the mass of Ben. Inserting the given values, we get:

7100 = 1/2 (63.2 kg) v^2

Therefore, v^2 = 7100/31.6

v = √(7100/31.6)

v = 12.3 m/s

Please help with this

Answers

A unit vector must have length equal to 1. By calculating the length of (1, 1), we discover that (12 + 12) = 2 = 1.414, which is not the same as 1, is the length.

How do you write unit vector notation?I,J notation is a way to describe a vector. In Cartesian coordinates, a unit vector is a vector with length 1. The unit vectors along the axis are denoted by the letters I and j, respectively. The form ai+bj an I + b j can be used to represent any two-dimensional vector.Numerous operations, including addition, subtraction, and vector multiplication, can be carried out with the aid of the vector form of representation. A=xi+yj+zk is the vector form of the cartesian representation of the three points (x, y, and z). A is equal to x I y j, and z k.Vy = Vyj, Vz = Vz k, and Vx = Vxî. V = Vx + Vy + Vz = Vxî+ Vyj+ Vz k is the result of applying the triangle rule to vector addition twice. The unit vector notation is used in this situation.

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Why is Dimension important in Physics?​

Answers

Explanation:

Dimensions help understand how physical amounts are related to their dependence on basic or fundamental volumes, i.e. how the body's dimensions rely on mass, time, length and temperature.

Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions.

I need help i don’t want to go to summer school

Answers

I think it’s meditation
You have no choice you will go

The quasar that appears the brightest in our sky, 3C 273, is located at a distance of 2.4 billion lightyears. The Sun would have to be viewed from a distance of 1300 light-years to have the same apparent magnitude as 3C 273. Using the inverse square law for light, estimate the luminosity of 3C 273 in solar units.

Answers

Answer:

Explanation:

Let the luminosity of the star be I and luminosity of the sun be Isun.

2.4 billion light years = 2.4 x 10⁹ light years .

brightness = luminosity / (distance)²

Given   Sun would have to be viewed from a distance of 1300 light-years to have the same apparent magnitude as 3C 273 so

For the sun

brightness = Isun / (1300 light years )²

For star

brightness = I / (2.4 x 10⁹ light years  )²

Both these brightness are same

Isun / (1300 light years )²  =  I / (2.4 x 10⁹ light years  )²

I = Isun x (2.4 x 10⁹ light years  )² / (1300 light years )²

= Isun x 3.4 x 10¹² .

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