Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electric fields; tests show that they can detect a change in field as small as 0.77 N/C. Honeybees seem to use this sense to determine the charges on flowers in order to detect whether or not a flower has been recently visited, so they can plan their foraging accordingly. As a check on this idea, let's do a quick calculation using typical numbers for charges on flowers.If a bee is at a distance of 24 cm, can it detect the difference between flowers that have a +30 pC charge and a +40 pC charge?

Answers

Answer 1

Answer:

difference between the field  = 1.56 N/C

as; 1.56 N/C is greater than 0.77 N/C;

Hence, Honeybees can detected the difference

Explanation:

Given the data in the question;

distance r = 24 cm = 0.24 m

charge 1 Q1 = +30 pC = 30×10⁻¹² C

charge 2 Q2 = +40 pC = 40×10⁻¹² C

Now, electric field due to +30 pC charge

E1 = kQ1/r²

where coulomb constant k is 9 × 10⁹ N.m²/C²

so we substitute

E1 = [( 9 × 10⁹ ) × (30×10⁻¹²)] / (0.24)²

E1 = 0.27 / 0.0576

E1 = 4.69 N/C

electric field due to +40 pC charge

E2 = kQ1/r²

E2 = [( 9 × 10⁹ ) × (40×10⁻¹²)] / (0.24)²

E2 = [( 9 × 10⁹ ) × (40×10⁻¹²)] / (0.24)²

E2 = 0.36 / 0.0576

E2 = 6.25 N/C

Now,

E2 = E1 =  6.25 N/C - 4.69 N/C = 1.56 N/C

difference between the field  = 1.56 N/C

as; 1.56 N/C is greater than 0.77 N/C;

Hence, Honeybees can detected the difference


Related Questions

Which of the following is a part of both geocentric model and heliocentric model

Answers

Answer:

These planets rotate around the sun in a circular path. Likewise in a heliocentric model it is believed that the sun is at the center of the universe and the planet earth along with all other planet move around it. Thus in both geocentric model and heliocentric model bodies in space move in circular orbits.

Answer:

The bodies in space move in circular orbits

Explanation:

I got it right on my test

A flat screen tv uses 120 watts. How much energy is used up if it is left on for 15 min?
A.) 4j
B.) 15j
C.) 0.67j
D.) 108,000j

Answers

Answer:

d

Explanation:

Greatest to least order

Answers

Answer:

Explanation:

FBEDAC

Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits

Answers

Answer:

a)  A' =  0.345  m,  b)  f = 2,800 Hz

Explanation:

b) The angular velocity of a simple harmonic motion is

        w =[tex]\sqrt{\frac{k}{m} }[/tex]

angular velocity and frequency are related

        w = 2π f

we substitute

        f = 1 /2π   √k/m

indicates that the initial frequency value f = 3.96 Hz

in this case the mass is reduced by half

       m ’= m / 2

we substitute

       f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]

       f = √1/2    (2π √k/m)

       f = 1 /√2  3.96

       f = 2,800 Hz

a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.

When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment

initial instant. Right before the division

        p₀ = (m₁ + m₁) v

final instant. Right after the split

        p_f = m₁ v '

        p₀ = p_f

        (2 m₁) v = m₁ v ’

        v ’= 2v

At this point we can use conservation of energy for the system with only half the block.

Starting point. Where the block divides

         Em₀o = K = ½ m v'²

Final point. Point of maximum elongation

          Em_f = Ke = ½ k A²

how energy is conserved

         Em₀ = Em_f

         ½ m’ v’² = ½ k A’²

we substitute the previous expressions

         ½ m/2 (2v)² = ½ k A’²

         A’² = 2  m v² / k                       (1)

Let's use the conservation of energy with the initial conditions, before dividing the block

          ½ m v2 = ½ k A2

          A² = mv² / k = 5.95 10⁻² m²

we substitute in 1

         A'² = 2 A²

           

          A ’²=  2 5.95 10⁻²

          A ’²= 11.9 10⁻² m

          A' =  0.345  m

Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.

Answers

Answer:

a. of liquid at the time bubbles first emerge slowly from the liquid.

Explanation:

Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.

Which statement best compares coal and ores?

Both are burned for energy.
Both take millions of years to form.
Both require oxygen to form.
Both are used to make coins.

Answers

Answer:

Option 2 both take millions of years to form

Explanation:

Both  coal and ores take millions of years to form.

What are ores?

Ore is a naturally occurring rock or silt that has precious minerals in it that may be extracted, processed, and sold for a profit. These minerals are usually metals. Mining is the process of removing ore from the soil. The valuable metals or minerals are then removed by treating or refining the ore, frequently through smelting.

The concentration of the desired ingredient in an ore is referred to as its grade. To decide if a rock has a high enough grade to be worth mining and is therefore regarded as an ore, the value of the metals or minerals it contains must be evaluated against the expense of extraction.

Typically, oxides, sulphides, silicates, or native metals like copper or gold are the minerals of interest. To separate the valuable components from the waste rock, ore must be treated. Numerous geological processes collectively known as ore genesis are responsible for the formation of ore deposits.

Learn more about ore here:

https://brainly.com/question/4712893

#SPJ2

PLEASE HELP QUICK which statement describes a primary difference between an electromagnetic wave and mechanical wave?​

A. electromagnetic waves can travel through empty space

B. electromagnetic waves can be transverse longitudinal or surface waves

C. electromagnetic waves can only travel through solids liquids or gases

D. electromagnetic waves need a medium to transfer energy

Answers

Answer:

A.

Explanation:

An electromagnetic wave is produced by the interaction between a variable electric field, and a magnetic electric field, which propagates in space, even in vaccuum, at a fixed speed, whilst the mechanical waves require a medium in order to transfer energy.

Answer: A

Explanation:

Why are carbon atoms able to form many organic compounds?

A. Carbon atoms have strong attraction to other elements.

B. Carbon atoms attract electrons from other atoms.

C. Carbon atoms can form many types of bonds with other carbon.

D. All of the above​

Answers

Answer:

yo imma so I dunno find out yourself

Explanation:

dhdhdhdnndsisijjsksskskekekekkekssisisieieieiiwiwiwieiwieidjdjddi?

Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.

Answers

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = [tex]\frac{kq}{r}[/tex]

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m

So that,

V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]

   = [tex]\frac{2.7*10^{4} }{0.5}[/tex]

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.

2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.

Answers

Answer:

3.277 m

Explanation:

Given :

Maximum Height (Hmax) = (u²sin²θ) / 2g

Xv = Xh + Uv * t + 0.5gt²

Xv and Xh are vertical and horizontal distances

-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2

-1 = 0 + 0.903Uv - 11.025

-1 + 11.025 = 0.903Uv

10.025 = 0.903Uv

Uv = 10.025 / 0.903

Uv = 11.10 m/s

Hmax = 1 + (u²sin²θ) / 2g

= (11.10^2 * (sin37)^2) / 2*9.8

= 44.624360 / 19.6

= 2.277

Hmax = 1 + 2.277

Hmax = 3.277 m

Consider the air over a city to be a box that measures 100 km per side that reaches up to an altitude of 1.0 km. Wind (clean air) is blowing into the box along one of its sides with a speed of 4 m/s. An air pollutant is emitted into the box at a rate of 10.0 kg/s; the pollutant degrades with a rate constant k = 0.20/hr. a. Find the steady state concentration of the pollutant (µg/m3 ) in the box if the air is assumed to be completely mixed. b. If the wind speed suddenly drops to 1 m/s, estimate the concentration of the pollutant (µg/m3 ) two hours later.

Answers

Answer:

a)  ρ = 6.25 10⁵ μg / m³, b) ρ  = 1 10⁷ μg / m³

Explanation:

Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.

Let's start by finding the volume of air that goes into the box

               V = Lh x

Let's find the distance of air that enters per unit of time, as it goes at constant speed

               x = v₀ t

we substitute

               V₀ = Lh v₀ t

At this same time, a quantity of pollutant is distributed

              Q₀ = r t  

the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is

               Q = Qo t

we substitute

               Q = r t²

the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state

              [tex]Q_{net}[/tex]= Q - k t

 

the pollutant concentration is

              ρ = Q_net / V

              V = L L h

              ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]

              ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]

               ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]

let's reduce the magnitudes to the SI system

           r = 10 kg / s

           L = 100 km = 100 10³ m

           h = 1 km = 1 10³ m

           k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵  1/s

           v₀ = 4 m / s

let's calculate

The volume of the box

             V = (100 100 1) 109

             V = 1 10¹³ m³

            ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]

            ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]

            ρ = 6.25 10⁻⁴ kg / m³

       

let's reduce to μg / m³

               ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)

               ρ = 6.25 10⁵ μg / m³

 

b) in case the air speed decreases to v₀ = 1 m / s

             

             ρ= \frac{10}{ 1^2 \  1\  10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3  1 \ 10^3}

             ρ = 1 10⁻² - 5.5556 10⁻¹³

             ρ =  1 10⁻² kg / m³

             ρ  = 1 10⁷ μg / m³

Mischievous Joey likes to play with his family's lazy susan (this drives Mom crazy because it is an antique). He puts the salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 79 grams. She also measures the radius of the turntable to be 0.23 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.7222 m/s.

Required:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

Answers

Answer:

0.179 N

Explanation:

What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.

So, F = mv²/r where m  = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m

So, substituting the values of the variables into the equation, we have

F = mv²/r

F = 0.079 kg (0.7222 m/s)²/0.23 m

F = 0.0412 kgm/s² ÷ 0.23 m

F = 0.179 kgm/s²

F = 0.179 N

Please help! Will mark brainliest.

Answers

Answer:

1122.8

Explanation:

12.73 kg x 9.8 m/s^2 x 9m

=1122.786

Rounded=1122.8

types of aerobic activities?​

Answers

Answer:

swimming, cycling, jump rope, brisk walking, gardening, jogging

An iron block of 12 kg undergoes a process during which there is a heat gain from the block at 2 kJ/kg, an elevation increase of 32 m, and a decrease in velocity from 40 m/s to 7 m/s. During the process, which also involves work transfer, the internal energy of the block increases by 70 kJ. Suppose the total energy of the system remains constant. Determine the work transfer during the process in kJ and indicate whether the work is done on/by the system.

Answers

Answer:

Explanation:

Total heat gain by the block ΔQ = 2 x 12 kJ = 24 kJ .

Gain of potential energy = mgh = 12 x 9.8 x 32 = 3.763 kJ

Decrease in kinetic energy KE = 1/2 x 12 ( 40² - 7² )

= 9.306 kJ

increase in internal energy ΔE = 70 kJ

ΔQ =  ΔE + PE - KE + W , W is work done by the gas

Putting the values

24 = 70 + 3.763 - 9.306 + W

W = - 40.457 kJ .

Since W is negative that means work is done on the system .

A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil can be poured into the glass to keep it still sinking? The density of the oil is 900 kg / m​

Answers

Answer:

Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...

In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position

Answers

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the train platform hears beats with a frequency of 2.00 beats/s when the whistles sound at the same time. What are the two possible speeds and directions that the moving train can have?

Actual answers :3.85 m/s away from the station and 3.77 m/s towards the station from the book. I just need to know how to get to the answers.

Answers

Answer:

-3.77 m/s

3.85 m/s

Explanation:

given that

Frequency at stationary = 180 Hz

Beat frequency = 2 Hz

Using Doppler effect, we know that

f' = f[(v ± v0) / (v ± vs)], where

v = speed of sound, 343 m/s

v0 = speed of the observer, 0

vs = speed of light, ?

f = stationary frequency, 180 Hz

f' = stationary ± beat frequency, 180 ± 2

Applying the formula, we have

f' = f[(v ± v0) / (v ± vs)]

182 = 180 [(343 + 0) / (343 + vs)]

182/180 = 343 / 343 + vs

343 + vs = 343 * 180/182

343 + vs = 339.23

vs = 339.23 - 343

vs = -3.77 m/s

Again, using

f' = f[(v ± v0) / (v ± vs)]

178 = 180 [(343 + 0) / (343 + vs)]

178/180 = 343 / 343 + vs

343 + vs = 343 * 180/178

343+ vs = 346.85

vs = 346.85 - 343

vs = 3.85 m/s

A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2

Answers

Answer:

i think 692m/s2 is the correct answer

Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?

An electrically charged object
An uncharged object
A positively charged object
A negatively charged object

Answers

Answer:

its An uncharged object.

if its not charged the electrically wont go on it

Answer:

uncharged object

Explanation:

A block of wood 3 cm on each
side has a mass of 27 g. What is the
density of the block? (Hint, don't
forget to find the volume of the
wood first using lx W h.)

Answers

Answer:

1g/cm3

Explanation:

volume of block is 3 cubed which is 27 cm3

we know density is m/v so d= 27g/27cm3

which is 1g/cm3

if my answer helps please mark as brainliest

HELPPPPP
What can you infer about the strength and direction of forces experienced by the pod and space station when they collided? What evidence from today’s activities supports your inference?

Answers

Answer:

In the collision, the strength of the force exerted on the pod is greater than the strength of the force exerted on the space station, but those forces are exerted in opposite directions.

Explanation:

. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential

Answers

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.

Solution:

The relation between electric field and the electric potential is:

E = [tex]\frac{dV}{dr}[/tex]

So, making dV the subject, we have:

dV = E x dr

Integrating the above equation, we get.

V = [tex]\int\limits^_ {} \,[/tex]E x dr      Equation 1

Now, in 2-D

E is inversely proportional to the radius r.

E ∝ 1/r

So, we can write: replacing E ∝ 1/r in the equation 1

V ∝  [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr

Which implies that,

V ∝  log (r)

Hence, distance dependence expected for the electric potential =  ln (r)

Which two statements help explain why digital storage of data is so reliable?

A. Memory chips are sturdy.

U B. Digital data usually deteriorate over time.

C. It is usually possible to recover data from a memory chip even

when the device containing it is broken.

D. Digital data are easier to copy than analog data are, making them

more accessible to thieves.

Answers

Answer:

A. Memory chips are sturdy.

C. It is usually possible to recover data from a memory chip even when the device containing it is broken.

Explanation:

Digital storage of data refers to the process which typically involves saving computer files or documents on magnetic storage devices usually having flash memory. Some examples of digital storage devices are hard drives, memory stick or cards, optical discs, cloud storage, etc.

A reliable storage ensures that computer files or documents are easily accessible and could be retrieved in the event of a loss.

The two statements which help explain why digital storage of data is so reliable are;

A. Memory chips are sturdy: they are designed in such a way that they are compact and firm.

C. It is usually possible to recover data from a memory chip even when the device containing it is broken.

Answer:

A and C

Explanation:

got it right on a p e x

How many planets on the solar system?

Answers

Answer:

8

Explanation:

tsijtsiztuztuistizrizturzurz

Answer:

8

Explanation:

Mercury, Venus, earth , Mars, jupiter, saturn , Uranus,Neptune

3.
What part of your eye is responsible for regulating the amount of light that enters your eye?

Answers

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

Answer:

I hope this helps.

Explanation:

the ____ is a particle with one unit of positive change

a. proton
b. positron
c. electron
d. nucleus

Answers

Answer:

a proton because it has a positive charge

Answer:

The answer is

B)  

A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine at the rate of 3 gpm, containing 2 lbs of dissolved salt per gallon. Assuming that the mixture is kept uniform by stirring, a drain pipe draws out of the tank the mixture at 2 gpm. Find the amount of salt in the tank at the end of 30 minutes.
A. 171.24 lbs
B. 124.11 lbs
C. 143.25 lbs
D. 105.12 lbs

Answers

Answer:

A. 171.24 Ibs

Explanation:

To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

  = e ^ 2㏑ (100 + t)

     = e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

 Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

  When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

 C = -1.5 * 10 ^6

therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

 Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6

    Q = 171.24 Ibs

The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.

The given parameters:

Initial volume of the tank, i = 100 gallonsRate of gain of salt = 3 gpmRate of loss of salt = 2 gpm

The linear differential equation of the salt solution is calculated as follows;

[tex]\frac{dx}{dt} = Gain - loss[/tex]

where;

x is the salt concentration

The salt concentration at time t, is calculated as follows;

[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]

Apply the general solution of linear differential equation as follows;

[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]

[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]

When t = 0 and X = 50

[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]

When t = 30 min, the concentration is calculated as;

[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]

Learn more about solution of Linear differential equation here: https://brainly.com/question/5508539

What is the correct definition of amplitude

Answers

Answer:

In my textbook's words-

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.

Explanation:

The correct definition of amplitude is that it is a maximum displacement

that occurs on a vibrating body from one point to the other.

The initial point of the wave is regarded as its equilibrium position which is

equal to one-half the length of the vibration path.

Amplitude helps to calculate the peak value of different types of waves

such as water waves and in electrical appliances so as to know the peak

current suitable for it.

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these

Answers

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

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