A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer 1

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3


Related Questions

You spaceship has a snazzy lounge called Ten-Forward that needs a new spacecouch. Sadly the couch you bought is too long and won't fit into your car when it is stationary. The couch has a length Lc = 14.0 m and mass m = 49 kg, and your shuttlecraft has a length of L = 9.0 m How fast would you need to run, in terms of the speed of light c, in order to get the couch to fit inside the length of the shuttlecraft?

Answers

Answer:

v = speed

c = speed of light

using the equation

L = Lc Sqrt( 1 - (v/c)2)

9 = 14 Sqrt( 1 - (v/c)2)

v/c = 0.765

v = 0.765 c

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a distance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?

Answers

Answer:

[tex]D_l=d[/tex]

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

[tex]V_{e1}=V_{e2}[/tex]

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

[tex]D=\frac{at^2}{2}[/tex]

Where

Acceleration is given as

[tex]a=\frac{V_o}{2d}[/tex]

And

Time

[tex]T=\frac{d}{v_0}[/tex]

Therefore horizontal displacement towards the left is

[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]

[tex]D_l=d[/tex]

A bus moving on a straight road increases its speed uniformly from rest to 20m's over a time period of 1 min. The distance travelled during the time is (a) 150 m (b) 300 m (c) 600 m (d) 900 m​

Answers

Explanation:

Given that,

Initial velocity (u) = 0 m/sFinal velocity (v) = 20 m/sTime taken (t) = 1 minute = 60 seconds

In order to find the distance travelled, firstly we need calculate the acceleration.

v = u + at

→ 20 = 0 + 60a

→ 20 = 60a

→ 20 ÷ 60 = a

→ ⅓ m/s² = a

Now, by using the 2nd equation of motion :

s = ut + ½at²

→ s = 0(60) + ½ × ⅓ × (60)²

→ s = ⅙ × 3600

→ s = 1 × 600

→ s = 600 m

Hence, the distance travelled is 600 m.

A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor

Answers

Explanation:

Given:

[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]

[tex]V = 12 V[/tex]

[tex]I = 1.2 A[/tex]

Recall that power P is given by

[tex]P = VI[/tex]

so the amount of energy dissipated [tex]\Delta E[/tex] is given by

[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]

[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]

A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J

Answers

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

Vector a has a magnitude of 8 and makes an angle of 45 with positive x axis vector B has also the same magnitude of 8 units and direction along the

Answers

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

The answer is:

A + B = 6,123 units at angle 112,5 degrees.
A - B = 14,782 units at angle 22,5 degrees.

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long after the asteroid hit the Moon, which is 3.84 x 10^5 km away, would the light first arrive on Earth?

Answers

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

A solid metal sphere of radius 3 m carries a total charge of -5.5 uc. What is the magnitude of the
electric field at a distance from the sphere's center of (a) 2.9 m and (b) 8 m? How would the answers
differ if the sphere was (c) a thin shell.
IN​

Answers

Answer:

2.9::: 5.87*10*3 N/C

8: 7.73 × 10 ^2  N/C

Explanation: https://study.com/academy/answer/a-solid-metal-sphere-of-radius-3-00-m-carries-a-total-charge-of-5-50-muc-what-is-the-magnitude-of-the-electric-field-at-each-of-the-following-distances-from-the-sphere-s-center-a-3-10-m-b-8-00-m.html

HELP ME ASAP PLSSS!!​

Answers

I hope this helped !

Can someone log into my acc FOR ME I will pay you to complete my physics assignments for money or points?!!

Answers

Answer: no sorry../

Explanation:

a youthful person run at 7km/h in a north- west direction across the derk of a ship in which is streaming due east at 40 km/h .ifind the velocity of the boy relative to the Sea
ii,And the velocity of the sea relative to the boy​

Answers

Answer:

velocity of ship wrt sea= 7i∧

velocity of women on deck= 40j∧

velocity of women relative to sea will be resultant of above two velocoties,

7i∧ + 40j∧ magnitude is

square root (7 x 7 + 40x40)

=√49+1600

=√1612

=40.14 m/s

An oscillating particle has the equation x = 10cos(8πt +π ) (x in cm, t in s). The number of times the object passes through the equilibrium position in one second is

Answers

Answer:

x = A sin (wt + theta)        where w = angular frequency - basic SHM equation

w = 8 pi = 2 pi f

f = 4         basic frequency

N = 8     number of times thru origin

Each cycle the particle will pass thru the origin +x and -x    twice

A river is 87. meters wide and its current flows northward at 6 meters per second. A boat is launched with a velocity of 1.0 meters per second eastward from the west bank of the river. Determine the magnitude and direction of the boat’s resultant velocity as it crosses the river.

Answers

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

 

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran

Answers

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor​

Answers

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

pls mark me as a brainlist


. A patient on the infectious disease floor takes 10 mL of levofloxacin syrup bid. If the product is only
available as a 5 mL unit-dose oral syringe, how many syringes will the technician prepare for a 24-
hour supply?

Answers

Answer:

The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.

Explanation:

brainliest plz . . .

A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage of 0.0320 V is applied across the ends of the wire. From this information calculate the resistance of the wire.

Answers

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

One of the asteroids, Ida, looks like an elongated potato. Surprisingly it has a tiny (compared to Ida) spherical moon! This moon called Dactyl has a mass of 4.20 × 10^16 kg, and a radius of 1.57 × 10^4 meters, according to Wikipedia. Ida has a radius of 3.14 x 10^4 meters.
Find the acceleration of gravity on the surface of this little moon.

Answers

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

[tex]g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}[/tex]

g = 0.0114 m/s²

What are the multiple of meter ?
What are the sub multiple of meter?

Answers

Answer:

To cover a larger distance we use km(kilometre) , hm (hectometre) , and dac(decametre). These are called the multiples of Metre. For the distance is smaller , we use Dm (decimetre , cm(centimetre) and mm (millimetre) . These are called the submultiples of

1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia. ​

Answers

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.

Answers

Answer:

always runs slower than normal.

Explanation:

The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.

According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.

In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.

A person runs up the stairs elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.
Determine the work done by the person climbing the stair case.

Answers

Answer:

Work done = 2289.084 Joules

Explanation:

Given the following data;

Mass = 102 Kg

Height = 2.29

Time = 1.32 seconds

We know that acceleration due to gravity, g = 9.8 m/s²

a. To find the work done by the person;

Here, work would be done in the form of gravitational potential energy.

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

Work done = 102 * 2.29 * 9.8

Work done = 2289.084 Joules

Describe sound and record​

Answers

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

What do scientists use to determine the temperature of a star?

Answers

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?

a.) a loudspeaker

b.) a headphone

c.) a light bulb

d.) a microphone

I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure

Answers

Answer:

I THINK C

Explanation:

BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?

Answers

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

                       

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant speed. If the driver of the car hears the horn at a frequency of 246 Hz, then what is the speed of the car? Use 340 m/s for the speed of the sound

Answers

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

0
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ZOOLS
6) The mass of a motorcycle is 250 kg. What is?
A) Its weight on Earth in Newtons?
B) Its weight on the moon (in Newtons)?
ges
C) The mass of your motorcycle on the moon?

Answers

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

The speed of a sound wave

A. Depends on wavelength.
B. Depends on the medium.
C. Depends on amplitude.
D. None of the above.

Answers

Answer:

B) the medium

Explanation:

B. Depends on the medium.

A cowgirl ties one end of a 10.0 m long rope to a fence post (to the left of the cowgirl) and pulls on the other end so the rope is stretched horizontally with a tension of 140 N. The mass of the rope is

Answers

Answer:

Mass is zero if the whole rope is horizontal. Imaginary rope.

Explanation:

With any mass at all, only a small section of the rope will be truly horizontal. The rope curve will be a a catenary.

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The magnificent creatures appeared in storybooks I read in the library, paintings I saw in museums, movies I watched in the theater, and the dreams I had in my sleep. By the time I was thirteen, one question consumed me. I wanted to know if dragons ever existed, so I set out on a quest for facts.2As I started my research, I discovered many skeptics. Scientists presented evidence to show why dragons could notand did notexist. They explained that it would be impossible for dragons to fly because they would be too big. They laughed at the idea of dragons breathing fire. They pointed out that no other animal has ever done this. They said that if dragons had lived, someone would have found remains somewhere in the world. No bones about it, there were plenty of logical explanations. It would have been easy for me to accept that the only place dragons ever existed was in the imaginations of those who believed.3I could have given up, but I thought about my grandmother. She always told me that "people who believe that science is the answer to everything are missing out on everything else." With her words in mind, I searched some more. There were many facts that hinted that dragons may not be fictional. I noticed that cultures across the world all described dragons in similar ways. This was odd because they had no way to communicate with each other. I found dragons mentioned in more than just stories. They appeared in old legal papers, in the travel logs of Marco Polo, and in the Bible. I saw that the Chinese calendar uses a different animal each year. Dragons are included along with eleven real animals. I began to believe it was a real possibility that all of these people were talking about a creature that actually existed.4With renewed hope that there was some truth to the legends, I looked for new research. I found that some experts disagreed with popular arguments against dragons. They suggested that a dragon could have four stomachs like a cow. If it created stomach gases like birds, it might create enough to lift itself off the ground. This would give it the ability to fly. If it forced out air when diving toward the earth, it might release gases which could ignite into flame. When the animal died, the stomachs would release strong acids that would dissolve its dead body over time. Biologists backed up these ideas with sketches and models based on known animals. Not everyone agreed with these ideas, but many of the things we accept about dinosaurs and other extinct species started the same way.5I doubt we will ever truly know whether dragons existed. There may always be two sides to the fiery debate. Some will say the stories come from active imaginations. Some will believe with all their hearts that the legendary creatures roamed our ancient world. I don't know for certain which side to believe, but the sound and fury of a night like this makes me smile. It rekindles my childhood dreams and keeps the exciting possibility alive.In paragraph 5, the author discusses two sides to the dragon debate. Which statement best summarizes the ideas in the paragraph? The author implicitly states that she believes that dragons once existed. The author explicitly states that she believes that dragons once existed. The author implicitly states that she believes that dragons did not exist. The author explicitly states that she believes that dragons did not exist. Cc yu t no cn c so snh khi tin hnh nh gi cng vic Describe the test of divisibility for:a) 6 b) 4 I cant figure out which formula goes with which sequence. Help. Please???1st sequence:1,3,9,27,. What is the formula?2nd sequence:400,200,100,50,What is the formula?3rd sequence:4,8,16,32,What is the formula?4th sequence:400,100,25,6.25,What is the formula?5th sequence:1,5,25,125,What is the formula?6th sequence:1000,500,250,125,What is the formula?7th sequence:2,10,50,250What is the formula?I need to know the correct geometric sequence formula for each of these sequences. Thanks!!! Read this excerpt from a passage."Her purchase got me thinking about the early days of online shopping, when people were anxious about orderingmerchandise from the Internet. What if the items never arrived? Was putting all my personal information online safe? Should Ireally disclose my credit card information to some unknown entity? Is this online company legitimate? And last but not least,who would buy something without seeing it first? But my grandmother's transaction made it clear: online shopping hasbecome as commonplace as brushing your teeth."As millions of consumers flock to online shopping, I wondered how and why people make the decision to shop online orin person. To get a better understanding of how e-commerce has affected our buying decisions, I decided to gain some insightfrom friends and family. The reviews were varied but insightful."Which of the following subjects will be compared in the rest of the passage?o delivery to a home and delivery to a storeO shopping online and shopping in personO online shopping and brushing your teethO prices of items online and prices of items in storesP Urgent how was kristallnacht carried out ?