There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C/m^2 and the bottom plate has a surface charge density of -10C/m^2. Find the total charge on each plate. Find the electric field at the point exactly midway between the plates. Find the electric potential between the two plates. If an electron was in the middle the two plates, find the force on it.

Answers

Answer 1

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

[tex]V_{tot} = V_{q1} + V_{q2}[/tex]

[tex]V_q = \dfrac{k \cdot q}{r}[/tex]

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]

The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0

3) The electric field, 'E', between plates is given as follows;

[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e

∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N


Related Questions

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:

A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)

Answers

Answer:

h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2  \  t_o^2 =0

The correct result is that of a positive height

Explanation:

For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man

             v_s = y / t

             t = [tex]\frac{y}{ v_s}[/tex]

this height is y = h

             t =  \frac{h}{ v_s}

the man has a response time of t = t₀, therefore

time to move is

             t' =  t - t₀

             

the initial height of flower pot is

           y = y₀ + v₀ t' - ½ g  t'²

when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0

            0 = y₀ +0 - ½ g (t -t₀)²

if the initial height is i = h,

             

            h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2

            [tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²

            [tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0

            [tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]

 

To know the height, you must solve the second degree equation, it is much easier with numerical values.

             The correct result is that of a positive height

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:

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Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangment of the four particles is defined as having zero electrical potential energy No work is required to bring one electron from infitinty to a location defined as the origin, while the other three particles remain at infiniuty. This is because no voltage exists near the origin until the first electron arrives. (a) Now, with the first electron remaining fixed at the origin, how much work is required to bring one of the remaining electrons from infinity to the coordinate (0 m, 2.00 m)? The other three particles remain at infinity. If this second electron was subsequently released, how fast would it be traveling once it returned to infinity? (b) Nļw, considering the two electrons fixed 2.00 m apart, how much work is required to bring the third electron from infinity to the coordinate (3.00 m, 0 m)? The other two particles remain at infinity. If this third electron was subsequently released, how fast would it be traveling once it returned to infinity? (c) Now considering the three fixed electrons at the coordinates described above. How much work is required to bring the last electron from infinity to the coordinate (3.00 m, 4.00 m)? If this forth electron was subsequently released, how fast would it be traveling once it returned to infinity? (d) Now considering the three fixed electrons at the coordinates described above. Finally, how much work is required to bring the proton from infinity to a coordinate of (1.00 m, 1.00 m)? If the proton is subsequently released and we assume that minimum separation distance between a proton and an electron is 1.00 pm, then how fast will the proton be traveling once it crashes into an electron?

Answers

Answer:

a)  W = 1.63 10⁻²⁸ J,  b)  W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d)  W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

          W = ΔU = U_∞ -U

the potential energy for point charges is

           U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]

in this case we only have two particles

           U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]

the distance is

           r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]

           r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)

           r₁₂ = √2= 1.4142 m

     

we substitute

           W = k \sum \frac{q_i q_j}{r_{ij} }

         

let's calculate

            W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142

            W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

             

in this case we have two fixed electrons

            U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]

in this case all charges are electrons

             q₁ = q₂ = q₃ = q

             W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]

the distances are

            r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0

            r₁₃ = 3

            r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2

            r₂₃ = √13

            r₂₃ = 3.606 m

let's look for the job

            W = U

let's calculate

            W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]

            W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

             W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]

all charges are equal q₁ = q₂ = q₃ = q₄ = q

             W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]

             

let's look for the distances

             r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]

             r₁₄ = 5 m

             r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]

             r₂₄ = √13 = 3.606 m

             r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]

            r₃₄ = 4 m

we calculate

           W = 9 10⁹ (1.6 10⁻¹⁹)²  [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]

           W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

            W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]

in this case the charges have the same values ​​but charge 5 is positive and the others negative, so the products of the charges give a negative value

            W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]

we look for distances

            r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2

            r₁₅ = √ 2 = 1.4142 m

            r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]

            r₂₅ = √2 = 1.4142 m

            r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]

            r₃₅ = √5 = 2.236 m

            r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]

            r₄₅ = √13 = 3.606 m

we calculate

           W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]

            W = - 4.93 10⁻²⁸ J

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?​

Answers

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

who has brown hair and brown eyes but is a boy

Answers

Answer:

I have strawberry blonde/brown hair blue eyes and a girl lol

Explanation:

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 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]

How many planets on the solar system?

Answers

Answer:

8

Explanation:

tsijtsiztuztuistizrizturzurz

Answer:

8

Explanation:

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

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:

Velocity time graph and how to draw it

Answers

Answer:

Velocity time graph

Explanation:

Draw on graph paper two straight lines originating at the same point and perpendicular to each other. This is the x-y axis. The x-axis is the horizontal line and the y-axis is the vertical line.

Mark appropriate equally-spaced time intervals on the x-axis so that you can easily graph the time values from the table.

Mark appropriate velocity increments on the y-axis so that you can easily graph the velocity values from the table. If you have negative velocity values, extend the y-axis downward.

Find the first time value from the table and locate it on the x-axis. Look at the corresponding velocity value and find it on the y-axis.

Put a dot where a straight line vertically drawn up through the x-axis value and a straight line horizontally drawn through the y-axis value intersect.

Plot in similar fashion for all other velocity-time pairs in your table.

Draw a straight line with a pencil, connecting each dot you have put down on the graph paper, going from left to right

A diet is to contain at least 2400 mg vitamin C, 1800mg Calcium, and 1200 calories every day. Two foods, a dairy-based meal and a vegan option are to fulfill these requirements. Each ounce of the dairy-based meal provides 50 mg vitamin C, 30 mg Calcium, and 10 calories. Each ounce of the vegan option provides 20 mg vitamin C, 20 mg Calcium, and 40 calories. If the dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce, how many ounces of each food should be purchased to minimize costs? What is that minimum cost (per day)?

Answers

Answer:

The answer is below

Explanation:

Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.

Since the diet must contain at least 2400 mg vitamin C, therefore:

50x + 20y ≥ 2400

Since the diet must contain at least 1800 mg Calcium, therefore:

30x + 20y ≥ 1200

Since the diet must contain at least 1200 calories, therefore:

10x + 40y ≥ 1200

Therefore the constraints are:

50x + 20y ≥ 2400

30x + 20y ≥ 1200

10x + 40y ≥ 1200

x > 0, y > 0

The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:

C(30, 45) and D(48, 18)

dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:

Cost = 0.042x + 0.208y

At C(30, 45);  Cost = 0.042(30) + 0.208(45) = $10.62

At C(48, 18);  Cost = 0.042(48) + 0.208(18) = $5.76

The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan

types of aerobic activities?​

Answers

Answer:

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

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 " .

Greatest to least order

Answers

Answer:

Explanation:

FBEDAC

A motorcyclist is making an electric vest that, when connected to the motorcycle's 12 V battery, will warm her on cold rides. She is using 0.25-mm-diameter copper wire, and she wants a current of 4.2 A in the wire. Part A What length wire must she use

Answers

Answer:

L = 8.35 m

Explanation:

The lenght of a wire L can be calculated using the following expression:

L = R A/ρ  (1)

Where:

R: resistance of the wire

A: Cross section area of the wire

ρ: resistivity of the copper wire.

With this expression we realize that we do not have the area of the cross section, and the resistance of the wire either.

To calculate the area we can use the following expression:

A = πr²    (2)

If the diameter is 0.25 mm, then the radius is half, 0.125 mm. Converting this in meter it will have to be:

0.125 /1000 = 0.000125 m

Replacing we have:

A = π(0.000125)²

A = 4.91x10⁻⁸ m²

The reported resistivity of a copper wire is 1.68x10⁻⁸ Ω.m, so we just need to determine the resistance, which can be found using Ohm's law:

R = V/I  (3)

Replacing (3) into (1) we have:

L = (V * A) / (I * ρ) (4)

So finally, the length of the copper wire will be:

L = (12 * 4.91x10⁻⁸) / (4.2 * 1.68x10⁻⁸)

L = 8.35 m

Hope this helps

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

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:

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

A toy car can go 5 mph. How long would it take to go 12 miles?

Answers

60 or 1 hour because 5 times 12 equals 60

he nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons). (a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N (b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2

Answers

Answer:

A) F = 21.134 N

B) a = 3180.76 × 10^(24) m/s²

Explanation:

A) We are given;

Mass of alpha particle; m = 4.0026 u

Now, 1u = 1.66 × 10^(-27) kg

Thus; m = 4.0026 × 1.66 × 10^(-27)

Distance apart; r = 6.60 × 10^(−15) m

Charge on the alpha particle is;

q = 2e = 2 × 1.6 × 10^(-19) C

Formula for the force between the two alpha particles is;

F = kq1.q2/r²

k = 8.99 × 10^(9) N.m²/C²

q1 = q2 = 2 × 1.6 × 10^(-19) C

F = 8.99 × 10^(9) × (2 × 1.6 × 10^(-19))²/(6.60 × 10^(−15))²

F = 21.134 N

B) acceleration is given by;

a = F/m

Thus; a = 21.134/(4.0026 × 1.66 × 10^(-27))

a = 3180.76 × 10^(24) m/s²

Name the state of matter that diffusion happens the fastest in.

Answers

Answer:

Liquids

Explanation:

Diffusion occurs fastest in liquids.

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.

PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.



The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.


Which number label represents the cell membrane?


1

2

4

6

(this is middle school science)

Answers

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

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:

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:

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

The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium withthe mixture. The piston is then moved inward very slowly, that thegas is always in thermal equilibrium with the ice-water mixture,what happens to the following(increase, decrease, same)?

a. volume of gas
b. temperature of gas
c. internal energy of gas,
d. pressure of gas

Answers

Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive position. Assuming that the diver's drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position (Aslower / Afaster).

Answers

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

[tex]v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\[/tex]

[tex]\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55[/tex]

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