four examples of compounds which are classed as carbohydrate​

Answers

Answer 1

Answer:

Carbohydrates are divided into four types: monosaccharides, disaccharides, oligosaccharides, and polysaccharides. Monosaccharides consist of a simple sugar; that is, they have the chemical formula C 6 H 12 O 6. Disaccharides are two simple sugars. Oligosaccharides are three to six monosaccharide units, and polysaccharides are more than six.


Related Questions

how is a trench and a tsunami related? 6-8 sentences

Answers

Answer: A tsunami is a very long-wavelength wave of water that is generated by sudden displacement of the seafloor or disruption of any body of standing water.  Tsunami are sometimes called "seismic sea waves", although they can be generated by mechanisms other than earthquakes.  Tsunami have also been called "tidal waves", but this term should not be used because they are not in any way related to the tides of the Earth.  Because tsunami occur suddenly, often without warning, they are extremely dangerous to coastal communities. Ocean trenches are steep depressions in the deepest parts of the ocean [where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.  

Explanation:

identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.​

Answers

Answer:

Explanatation

1 is just walking

2 spinging

3 roatating

4 rolling

5 stomping there feet

Those should be right but if im wrong then just someone eles the brainly

A ball is thrown so that its speed increases by 20 m/s in 10 seconds. What is the ball’s acceleration?

Answers

Answer: a= 2 m/s²

Explanation: acceleration = change of speed/ time = 20 m/s / 10 s

How can a mutation result in a new trait

Answers

They are called beneficial mutations. They lead to new versions of proteins that help organisms adapt to changes in their environment. Beneficial mutations are essential for evolution to occur. They increase an organism's changes of surviving or reproducing, so they are likely to become more common over time.

Mutation is when there is a particular change in the bases of the DNA, this changes how enzymes and proteins work, and there may be a chance that this change will be helpful.

In the figure, a 32 cm length of conducting
wire that is free to move is held in place
between two thin conducting wires. All of the
wires are in a magnetic field. When a 6.0 A
current is in the wire, as shown in the figure,
the wire segment moves upward at a constant
velocity.
The acceleration of gravity is 9.81 m/s?.
a) Assuming the wire slides without friction
on the two vertical conductors and has a mass
of 0.13 kg, find the magnitude of the minimum
magnetic field that is required to move the
wire.
Answer in units of T. b) What is the direction?

Answers

Answer:

.66354 T

Explanation:

Use F=ILB

B =  [tex]\frac{F}{IL}[/tex]

B = Magnetic field

F= force due to magnetic

I= current

L= length in meters

F = mg

Final formula:

B=[tex]\frac{mg}{IL}[/tex]

B=[tex]\frac{(.13)(9.8)}{(6)(.32)}[/tex]

B= .66354

ayo btw ion know how to find direction, my b G

The minimum magnetic field required to move the wire is 66354 T.

The direction of magnetic field is normal to the page outwards.

What is magnetic field?

The region surrounding a magnet that experiences the effects of magnetism is known as the magnetic field. When describing the distribution of the magnetic force within and around a magnetic object in nature, the magnetic field is a useful tool.

Given parameter:

Current passing through the wire, I = 6.0 A.

Length of the wire ,L = 32 cm = 0.32 m.

Mass of the wire, m = 0.13 Kg.

Acceleration due to the gravity, g = 9.8 m/s².

We know that, force acting on a current caring wire due to magnetic field is,  F=ILB

Where,

B = Required magnetic field.

To find the minimum magnetic field that is required to move the

wire, force acting on a current caring wire due to magnetic field is equal to weight the wire, that is, mg.

Hence, we can write,

mg = ILB

⇒ B = mg/IL

= (0.13 * 9.8)/(6.0 * 0.32)

=0.66354 Tesla

Hence, the minimum magnetic field is 0.66354 Tesla.

b) By using Maxwell's right hand thumb Rule along current flow, the direction of magnetic field is  determined as normal to the page pointing outwards.

Learn more about magnetic field here:

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Please help !!!!!!!!!!!!

Answers

Answer:

152 Volts

Explanation:

First, the Resistors are in series. So, the net resistance, R = 3 + 4 +3 + 4 +5

= 19 ohms

Using, V = IR;

V (the needed P.D) = 8 x 19 = 152 Volts

will be needed to successfully transport 8 amps of current round the circuit

Answer:

4q

Explanation:

wait nvm i dont know

Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 5.0 A .
How close can a credit card be held to this wire without damaging its magnetic strip?

Answers

Answer:

his distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Explanation:

For this exercise let's use the ampere law.

Let's use a cylinder as the circulating surface

          ∫ B. ds = μ₀ I

in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1

          B 2π r =  μ₀ I

          r =  [tex]\frac{\mu_o I}{2\pi B}[/tex]

The field needed to demagnetize the card is B = 1000 gauss = 0.1 T

          r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]

           r = 2 10⁻⁷  5.0/0.1

          r = 1  10⁻⁵ m

this distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Roger drives his car at a constant speed of 80 km/hr. How far can he travel in 2 hrs. and 30 minutes?

Answers

Answer:

200 km/hr

Explanation:

Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.

80 x 2.5 = 200 km/hr.

A marshmallow is fired from ground level with an initial speed of 41.5 m/s at an
angle of 33.5° above the horizontal. (a) Determine the maximum height reached
by the marshmallow. (b) Determine the horizontal range that the marshmallow
travels during its flight.

Answers

Look at picture first I couldn’t fit it all in but after you get the time get all the x values that you have and sub into s=ut+1/at^2

S=? t=2.34 a=0 u=41.5cos(33.5)
s= (41.5cos(33.5))(2.34)+1/2(0)
s=80.98 meters
this is the horizontal range

Need help ASAP on number 4 pls

Answers

The Arrows on both sides that has equal length is wrong and the rest of the boxes are the correct

Find the projection of vector vector A=2i-8j+k in the direction of vector B=3i-4j-12k

Answers

Use the dot product.
(a•b)=(2x3)+(-8x-4)+(1x-12)
=26

...........................

Answers

Answer: kooi kooi

how r u and thanks for the free points :)

Answer:............
lol thanks for the free points as well

A farmer plants the same crop in a field year after year. Every year there are months during which the field is left without any plants. The farmer notices a decline in soil
quality during these months. What causes this decrease in soil quality?
Erosion
Drought
Desertification
Consumption of nutrients by the crops

Answers

Answer:

Drought

Explanation:

recall that a centripetal force is not a new type of force: it's a role that is taken by one of the four types of force that we already use. if we see an object moving in a curved path, we know that at least one of those forces is pointing inward, acting centripetally. describe at least one real-life example of an object moving in a curved path where its centripetal force is provided at least partially by each of the following types of forces: gravity, normal force, tension,

Answers

Answer:

Answer in explanation.

Explanation:

GRAVITY AS CENTRIPETAL FORCE:

When a satellite orbits around a planet, it is in a circular motion. In this scenario, the centripetal force is provided by the gravity force.

NORMAL FORCE AS CENTRIPETAL FORCE:

When a spacecraft is sent into space, it rotates about its own axis to create artificial gravity. During this rotating motion, the centripetal force is provided by the normal reaction (force) of the walls of the satellite.

TENSION AS CENTRIPETAL FORCE:

When a mass attached to a rope is swirled in a circular motion, the tension in rope acts as the centripetal force.  

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. At each end of the swing the ship hangs motionless for a moment before the ship swings down under the influence of gravity. Assume that this motionless point occurs when the bar connecting the pivot point and the ship is horizontal.

Required:
a. Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.
b. What is the centripetal acceleration at the bottom of the arc?
c. Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
d. Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
e. Discuss whether the answer seems reasonable.

Answers

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

let's calculate

        v = [tex]\sqrt{2 \ 9.8 \ 14.0}[/tex]

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

Which of the following would fill in the table where "A" is?
Symbol
Element Name
Atomic Number Mass Number
#nº
#p
#e
Net Charge
Si
Silicon
A (#13)
27
B (#14 14 C (#15)
0
D (#16)
Potassium
18
39
19
18 E (#17)
A. 3
B. 13
C. 14
D. 28

Answers

Answer:

.lyrjhg.

Explanation:

qqlqfgtjnfh

The certain forest moon travels in an approximately circular orbit of radius
14,441,566 m with a period of 6 days 10 hr, around its gas giant exoplanet host. Calculate the mass of the exoplanet from this
information. (Units: kilograms)

Answers

Answer:

Mass of Exoplanet =  0.58 kg

Explanation:

First, we will calculate the speed of the forest moon:

[tex]speed = v = \frac{Circumference}{time}\\[/tex]

circumference = 2πr = 2π(14441566 m) = 90739035.3 m

time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)

time = 554400 s

Therefore,

[tex]v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s[/tex]

We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:

[tex]Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}[/tex]

Mass of Exoplanet =  0.58 kg

What happens during heat
transfer?
A. Heat always flows from cool to warm.
B. Heat always flows from warm to cool.
C. Heat always flows from warm to hot.

Answers

Answer:

B

Explanation:

Help me please I have other ones like this too on my page please help!

Answers

Balanced equation is:
2Mn+4CuCl——->4Cu+2MnCl2
a=2
b=4
c=4
d=2

SCIENCE
The growth of algae in ocean water is limited by their need for
a.
warm ocean currents.
b.
carbon dioxide and sunlight.
c.
dissolved oxygen.
d.
low salinity.

Answers

Answer:

c is trueeeeeeeeeeeeeee

The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.

What is algae?

The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.

Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.

Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.

To learn more about algae, click here:

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Part one: Multiple choices
1) A person sitting in the compartment of moving train is:
a) in the state of rest with respect to surroundings of the compartment,
b) in the state of motion with respect to surroundings of the compartment.
C) in the state of rest with respect to surroundings outside of the compartment
d)all of them

2) The motion of tuning fork prongs on vibration is:
a) Linear motion
b) periodic motion
c) circular motion
d) projectile motion
3) All the following are periodic motion except
a) moving car in straight line
b) Earth's rotation
c) pendulum
d) Swing
4) The rate of change of displacement is:
a) Acceleration
b) force
c) distance
d) velocity
5) When an object moves at negative acceleration in a straight line its:
a) displacement equals zero
b) velocity decrease
C) velocity increase
d) none of them
6) When the object speeds up its acceleration:
a) decreases
b) increases c) it has no acceleration d) All of them
7) The rate of change of velocity is:
a) force
b) variable velocity
c) instantaneous velocity
d) acceleration
8) If a train is moving in a straight line to cover a distance of 600 m in a minute its
velocity is:
a) 600 m/s
b) 60 m/s c) 100 m/s d) 480 m/s
9) The division between total displacement and total time is the:
a) variable velocity b) average velocity c) speed d) Instantaneous velocity
10) The rate of change of displacement at a given instant is called the
a) average velocity
b) instantaneous velocity
C) average velocity
d) instantaneous acceleration
11) A body completes one circular revolution in a roundabout whose diameter
140 m.
Find its displacement,
a) 439.6 m
b) 440 m
c) zero
d) 879.2 m​

Answers

Answer:

1)

a) in the state of rest with respect to surroundings of the compartment,

2)

b) periodic motion

3)

a) moving car in a straight line

4)

d) velocity

5)

b) velocity decrease

6)

b) increases

7)

d) acceleration

8)

10 m/s

9)

b) average velocity

10)

b) instantaneous velocity

11)

a) 439.6 m

Explanation:

1)

With respect to the inside surrounding the person will be at rest. Because the person is not moving inside the compartment.

2)

The vibration motion follows periodic motion.

3)

The car moving in a straight line is an example of rectilinear motion and its wheels are in rotational motion. They are not in periodic motion.

4)

Definition of velocity.

5)

Acceleration is the rate of change of velocity. So negative acceleration means a decrease in velocity.

6)

Acceleration is the rate of change of velocity. So an increase in velocity means an increase in acceleration.

7)

Definition of acceleration.

8)

[tex]velocity = \frac{Distance}{Time}\\\\velocty = \frac{600\ m}{1\ min}\frac{1\ min}{60\ s}\\\\velocity = 10\ m/s[/tex]

Hence, none of the options is correct. The correct answer is 10 m/s.

9)

Definition of average velocity.

10)

Definition of instantaneous velocity.

11)

[tex]Displacement = Circumference = \pi d\\Displacement = \pi(140\ m)\\Displacement = 439.6\ m[/tex]

What is the equation for Hookes law ?

Answers

Answer:

Fs = -kx is the formula

Explanation:

Fs = spring force

k = spring constant

x = spring compression

Answer:

Equation of Hooks law

F=kx

Where,

F=Force applied

k= spring constant

x=extension of spring

In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into castles. You are asked to study the motion of such a projectile for a group of local enthusiast planning a medieval war reenactment. Unfortunately an actual trebuchet had not been built yet, so you decide to first look at the motion of a thrown ball as a model of rocks thrown by a trebuchet. Specifically, you are interested in how the horizontal and the vertical components of the velocity for a thrown object change with time. 1. Make a large rough sketch of the trajectory of the ball after it has been thrown. Draw the ball in at least five different positions; two when the ball is going up, two when it is going down, and one at its maximum height. Label the horizontal and vertical axes of your coordinate system.
2. On the sketch, draw and label the expected acceleration vectors of the ball (relative sizes and directions) for the five different positions. Decompose each acceleration vector into its vertical and horizontal components.
3. On the sketch, draw and label the velocity vectors of the object at the same positions you chose to draw your acceleration vectors. Decomposes each velocity vector into its vertical and horizontal components. Check to see that the changes in the velocity vector are consistent with the acceleration vectors.
4. Looking at the sketch, how does someone expect the ball's horizontal acceleration to change with time? Could you give a possible equation giving the ball's horizontal acceleration as a function of time? Graph this equation. If there are constants in your equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
5. Looking at the sketch, how does someone expect the ball's horizontal velocity to change with time? Is it consistent with the statements about the ball's acceleration from the previous question? Could you give a possible equation for the ball's horizontal velocity as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
6. Could you give a possible equation for the ball's horizontal position as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph? Are any of these constants related to the equations for horizontal velocity or acceleration?
7. Repeat questions 4-6 for the vertical component of the acceleration, velocity, and position. How are the constants for the acceleration, velocity and position equations related?

Answers

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

What is the difference between the isotopes Hydrogen-2 and Hydrogen-3?

Answers

Answer:

They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. The isotopes of hydrogen have, respectively, mass numbers of one, two, and three.

Explanation:

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.40 cm. If the potential difference across the plates was 23.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

Answers

Answer:

E =  1.64 x 10⁶ V/m

Explanation:

The electric field in the region between the plates can be given by the following formula:

[tex]E = \frac{\Delta V}{d}[/tex]

where,

E = Electric Field = ?

ΔV = Poetential Difference across the plates = 23 KV = 23000 V

d = distance between plates = 1.4 cm = 0.014 m

Therefore, using these values in the equation, we get:

[tex]E = \frac{23000\ V}{0.014\ m}[/tex]

E =  1.64 x 10⁶ V/m

Joe is carrying a load of building supplies and likely injuring his back by supporting 882 in of weight. Calculate the mass of what he is carrying

Answers

Answer:

Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy

Explanation:

Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy

1. If airbags reduce the impact force from an accident why has there been questions over their safety?

2. Are airbags the safest option to prevent serious injury or death from a car accident?

Answers

Answer:Air bags can leave you in even more injury, From the impact they give

You could end up with a broken nose,arm

concussion

Suppose a rocket in space is accelerating at 1.5 m/s2. If, at a later time, the rocket quadruples its thrust (i.e., net propelling force), what is the new acceleration?

Answers

Nshsjdjdjdjdjjdjehrbdbdbdbdbdvdydiejeje he’s he’s right

A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closest point. (a) Calculate the semimajor axis of the orbit. Incorrect: Your answer is incorrect. m (b) Calculate the eccentricity of the orbit. Incorrect: Your answer is incorrect. Did you find the semimajor axis a from the greatest and smallest radii

Answers

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the
things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.

Answers

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]

            E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex]   [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]

             E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

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