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Nady [450]
2 years ago
7

Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri

c 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.
Part A If a bee is at a distance of 20 cm, can it detect the difference between flowers that have a 24 pC charge and a 32 pC charge
Physics
1 answer:
Vilka [71]2 years ago
8 0

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

You might be interested in
You work for the police. An accident has happened. A farmer was taking his dairy cows for milking across a road. The crossing po
ikadub [295]

Answer:

Since the stopping distance is less than the thinking distance, Mr Hoyland is therefore responsible for the accident

Explanation:

. In normal road conditions and with an alert and sober driver the car

could have stopped before the crash.

2. The reaction time of an alert and sober driver is 0.6s.

3. The reaction time of a drunk driver could be 2s.

4. The initial speed of the car was 30m/s

5. From the time of applying the brakes the total braking distance in

normal conditions is 70m

6. From the time of applying the brakes the total braking distance in bad

conditions is 90m

7. From the time of applying the brakes in normal conditions the car would

reach 15m/s in 52m

You are a forensic expert working for the police. You must analyse the facts and produce a

report stating whether or not Mr Hoyland was responsible for the accident.

1. Calculate the thinking distance for a car travelling at 30m/s

2. Calculate the braking distance for a car travelling in normal conditions at 30m/s

3. Calculate the total stopping distance.

4. REMEMBER TO SHOW YOUR WORKING OUT

Challenge:

1. Is Mr Hoyland responsible for the accident?

2. Explain your answer giving CLEAR reasons for your answer and likely causes.

The Thinking distance = 100m

The breaking distance = 70m

A breaking distance = Stopping distance = 70m

Since the stopping distance is less than the thinking distance, Mr Hoyland is therefore responsible for the accident

8 0
2 years ago
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
docker41 [41]

1) At x = 6.6 cm,  E_x=3.47\cdot 10^6 N/C

2) At x = 6.6 cm, E_y=0

3) At x = 1.45 cm, E_x=-3.76\cdot 10^6N/C

4) At x = 1.45 cm, E_y=0

5) Surface charge density at b = 4 cm: +62.75 \mu C/m^2

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: +65.25 \mu C/m^2

8) None of these regions

Explanation:

1)

The electric field of an infinite sheet of charge is perpendicular to the sheet:

E=\frac{\sigma}{2\epsilon_0}

where

\sigma is the surface charge density

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}

where

\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C

And the negative sign indicates that the direction is to the right.

2)

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), dE_y, is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, -dE_y. Therefore, the net field along the y-direction is always zero.

3)

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

E=E_1-E_2

Substituting the same expressions of part 1), we find

E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C

where the negative sign indicates that the direction is to the left.

4)

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

5)

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

\sigma=+64\mu C/m^2

And this the average of the surface charge density on both sides of the slab, a and b:

\sigma=\frac{\sigma_a+\sigma_b}{2} (1)

Also, the infinite sheet located at x = 0, which has a negative charge \sigma_1=-2.5\mu C/m^2, induces an opposite net charge on the left surface of the slab, so

\sigma_a-\sigma_b = +2.5 \mu C/m^2 (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2

6)

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

7)

We  calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is \sigma_a = +65.25 \mu C/m^2

8)

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0
2 years ago
A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0
2 years ago
where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun
creativ13 [48]

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

8 0
2 years ago
Which of the following statements correctly describes the models put forth by Ptolemy and Copernicus to explain the universe? A.
FrozenT [24]
The correct option is D. 
The model developed by Ptolemy has a lot of inconsistency and during the middle age additional explanation was offered for the claims made by the model. The model was very complicated because it was based on erroneous assumptions.
Copernicus model was simpler and some of his claims were correct.<span />
8 0
2 years ago
Read 2 more answers
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