Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposing each other) in a circuit. If you wanted to add in a capacitor to charge it from the batteries, would you be able to get more charge onto the capacitor or less charge, than if there was only one battery. (hint: start this problem by aligning the batteries positive to negative, and think of it from conservation of energy perspective).

Answers

Answer 1

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.


Related Questions

Which of the following helps prevent and cope with heat-related conditions?

Drinking water


Wear proper clothing


Rest frequently


all of the above

Answers

drinking water is ur answer.

An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.

What is heat?

Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.

This ultimately implies that, heat is most likely to cause dehydration and high body temperature.

In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.

Read more on heat here: brainly.com/question/12072129

A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding

Answers

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

please answer correctly

Answers

Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

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) 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

What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?

Answers

Answer:

speed = 10.1215 m/s

Explanation:

speed = distance / time

speed = 100 / 9.88 = 10.1215 m/s

It's time to get a little more specific. Based on the velocity (Vx) graph for the car and the velocity data in the table, divide the total
motion of the car into rough time periods that tell a different "chapter" of the story for this car trip. In each of these time
periods, the car's velocity will be notably different from the previous period. Enter a brief description of the car's motion in each
period. The first one is done for you. Use it as an example to identify and describe the remaining time periods. Note: You can
define as many periods as you think appropriate.
s
B
1
U X
X х.
Font Sizes
А • А
E
E 를 들
E 3
Numbered list
Time Period
Motion Description
0.2 - 4.6 seconds increasing speed in positive direction

Answers

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

A car travels 150 kilometers west in 3 hours. What is its average velocity?
Your answer:
150 km/hr

50 km/hr

50 km/hr west

150 km/hr west

Answers

Answer:

C= 50km/hr west

Explanation:

150/3= 50

Because it asks for velocity, make sure to include the direction as well.

Q4. (a) An acre-foot is the volume of water that would cover 1 acre of flat land to a depth of 1
foot. How many gallons are in 1 acre-foot?

Answers

Answer:

326,000

Explanation:

One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.

The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the
Equator bring cooler waters to some regions, while currents flowing away from the Equator bring warmer waters to other regions.
North
British
Isles
Askan
North Atlantic
Azor
U.S.A
California
Gulf Stream
Loop
n
Canbean
North Equatorial
North Equatorial CC
North Fuatorial
Equator
South Equatorial
Not
South Equatorial
Image courtesy of NOAA
Judging from the map, which region probably has cooler summers than it would without the effect of a nearby ocean current?
A the Central U.S.
B. the British Isles
C. the U.S. East Coast
D. the US West Coast

Answers

Answer:

d

Explanation:

the US West Coast region probably has cooler summers than it would without the effect of a nearby ocean current.

what is ocean current ?

ocean current can be defined as the horizontal movement of seawater which is produced by gravity, wind, and water density, it play an major role in the determination of climates of coastal regions.

The movement of ocean water is continuous which can be up three types such as Waves, Tides, Currents

The streams of water which flow continuously on the ocean surface in specific directions are called ocean currents, it affect the temperature of ocean water as Warm ocean currents increase the temperature whereas Cold ocean currents decrease the temperature.

The magnitude of the ocean currents is about  few centimeters per second to as much as 4 metres per second and the intensity of the ocean currents generally decreases with increasing depth.

There are two types of ocean currents such as Warm Ocean Currents

and Cold Ocean Currents

For more details ocean current,  visit

https://brainly.com/question/21654036

#SPJ2

The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight

Answers

Answer:

D. Weight

Explanation:

Hope that helps:)

D. Weight is the correct answer.
HOPE IT HELPS U!!!!!!

an object of 5kg is attached to a rope of length 4m is Rotating horizontally at 8m/s horizontally 20m above the ground if the rope is suddenly cut what is the horizontal distance travelled by the object? Please guys help

Answers

Answer:

16 meters

Explanation:

When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.

First we need to find how much time de object take to reach at the ground.

VERTICALLY EQUATION:[tex]h(t)=h-v*t-\frac{g}{2} t^{2} \\[/tex]

g=acceleration of gravity=10m/s²

v= vertical velocity =0m/s

h=vertical altitude =20m

We will find t such that h(t)=0

[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]

HORIZONTALLY EQUATION:*horizontally we do not have acceleration[tex]D(t)=v*t[/tex]

v=horizontal velocity

D(t=2) is the horizontal distancetravelled by the object:

[tex]D(2)=8*2\\\\D(2)=16m[/tex]

A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is?​

Answers

Answer:

h over 2 dg

Explanation:

brainliest!!!!!!!

Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each other. What law of motion is this. (Newton's laws)

Answers

Answer:

forcing in act

Explanation:

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change

Answers

Complete question is;

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change?

Answer:

182.84 %

Explanation:

Formula for rate of heat transfer of an infinite log fin is given as;

q_f1 = (π/2) × (hk)^(½)) × D^(3/2)) × (T_b - T_∞)

Where D is diameter.

Now, if the diameter of the rod is doubled, it means Diameter is now 2D.

Thus;

q_f2 = (π/2) × (hk)^(½)) × (2D^(3/2)) × (T_b - T_∞)

To find how much the rate of heat removal will change, we will calculate as follows;

((q_f2/q_f1) - 1) × 100

Plugging in the relevant expressions, we have;

([[(π/2) × (hk)^(½)) × ((2D)^(3/2)) × (T_b - T_∞)]/[(π/2) × (hk)^(½)) × (D^(3/2)) × (T_b - T_∞)]] - 1) × 100

Upon simplifying, we have;

(((2D)^(3/2))/(D^(3/2)) - 1) × 100

((2^(3/2)) - 1) × 100

This gives;

182.84 %

how does the strength of the forces that hold the basic particles of a substance together relate to the temperature at which the substance changes state​

Answers

The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.

Select the correct answer.
The oceanic Nazca plate is being subducted beneath the continental South American plate. Which type of plate boundary is this?
OA continental-oceanic convergent
ОВ. oceanic-oceanic convergent
OC divergent
OD. strike-slip
ОЕ.
transform
Reset
Next

Answers

Answer:

A. continental-oceanic convergent

Explanation:

I knew it couldn't be B because it's oceanic and continental, not oceanic and oceanic.

Next, I noticed the word convergent, which implies "coming together" to me.

I looked it up and noticed the term convergent referred to a plate boundary where a plate slips under (subducted) another, so I knew it was A.

Hopefully, this helps you understand the question better. Have a great day!

Explain the working and principle of perisocope.​

Answers

Answer:

a periscope use total internal reflection to allow us to see things

the reflection happens at 45°

Explanation:

help me pls it’s a usa test prep pretty easy

Answers

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

something that orbiys other things in space​

Answers

Answer: well we all orbit the sun all the planets do so the

SuN

Explanation: two words common sense

Jack weighs 170 lbs and is 72 inches tall. He is pulling horizontally on a door handle situated at his shoulder height. Actually, it is his body weight and lean that creates this pulling action (a hint). His center of mass while standing erect is 61 percent of his body height, measured from the floor upwards. The door handle is 60 inches above the ground, and again he is pulling purely horizontally on this handle.

If Jack's lean angle is 20 degrees and he is leaning back - pivoting about his heels, how much force does he apply to the door handle?

Include units in your answer, lbs.

Express your answer to the nearest 0.1 lbs.

Answers

Answer:

He is pulling horizontally on a door handle situated at his shoulder height. ... His Center Of Mass While Standing Erect Is 61 Percent Of His Body Height, Measured ... Actually, it is his body weight and lean that creates this pulling action (a hint).

72ibs

Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.
1) What is Cole's acceleration?
2) What is his velocity when he reaches school?

Answers

Explanation:

this is the answer for your question. if you have any doubt.

you can send your doubt to:6369514784(what's app)

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.

Answers

Answer:

a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c)  F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]

         F = k Q1 λ ([tex]-\frac{1}{x}[/tex])  

we evaluate the integral

        F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]

        F = k Q₁ λ  [tex]( \frac{L}{d \ (d+L)})[/tex]

we change the linear density by its value

      λ = Q2 / L

       F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]

       F = -1.09 N

the sign indicates that the force is attractive

Answer:

a)Toward the rod

b)|dF| = k|Q1|Q2(dx/L)/x^2

c)|F| = k|Q1|Q2/(d(d+L))

d)Plug in for answer c and solve

Explanation:

A)

Q1 is negative and Q2 is positive so it is an attractive force to  where the rod is located.

B)

The formula for Force due to electric charges is F=kQ1Q2/r^2

In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.

The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.

The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.

The final formula is |dF| = k|Q1|Q2(dx/L)/x^2

C)

Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:

F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2

factor out constants

F = kQ1Q2/L * integral d to d+L(1/x^2)dx

F = kQ1Q2/L * (-1/x)| from d to d+L

F = kQ1Q2/L * (-1/d+L - -1/d)

F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))

F = kQ1Q2/L * (L)/(d(d+L))

F = kQ1Q2/(d(d+L))

D)

Plug in the given values into c and you have your answer.

what is permittivity​

Answers

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:

Answers

Answer:

in oil film        λ = 303.57 10⁻⁹ m

in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

             c = λ₀ f

we divide the two expression

            c / v = λ₀ / λ

the refractive index is

             

              n = c / v

              n = λ₀ /λ

              λ = λ₀ / n

let's calculate

in oil film

            λ = 425 10⁻⁹ / 1.40

            λ = 303.57 10⁻⁹ m

in the water film

            λ = 425 10⁻⁹ / 1.33

            λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

two small identical conducting spheres have charges of 2.0x10-9C and - 0.5x109 C respectively when they are placed 4cm apart, what is the force between them? If they are brought into contact and then separated by 4cm, what is the force between them? ​

Answers

Answer:

6

Explanation:

nothingnsbejejjdbsbzbawkje


How can you tell whether an object is neutral
or charged? What would you have to do to test
that object?

Answers

Answer:

The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral

Explanation:

The ancient Egyptians used the stars to _____.

predict annual flooding

make their calendar

harvest their crops

choose their pharaoh

Answers

Answer:

predict annual flooding

Answer:

I'm pretty sure it is b make their calander

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C. (drag on jet / drag on transport)

Answers

Answer:

[tex]2.267[/tex]

Explanation:

Drag force is given by

[tex]F=\dfrac{1}{2}\rho Av^2C[/tex]

C = Drag coefficient is constant

A = Area is constant

[tex]v_1[/tex] = Velocity of the passenger jet = 1200 km/h = [tex]\dfrac{1200}{3.6}\ \text{m/s}[/tex]

[tex]v_2[/tex] = Velocity of the prop plane = [tex]\dfrac{1}{4}v_1[/tex]

[tex]\rho_1[/tex] = Density of the air where the jet was flying = [tex]0.38\ \text{kg/m}^3[/tex]

[tex]\rho_2[/tex] = Density of the air where the prop plane was flying = [tex]0.67\ \text{kg/m}^3[/tex]

[tex]F\propto \rho v^2[/tex]

[tex]\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267[/tex]

The ratio of the drag forces is [tex]2.267[/tex].

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not equally sensitive to all frequencies of sound. In other words, a particular noise may sound louder or softer depending on the frequency of the sound wave being transmitted. Because of this variation, scientists have defined a unit of loudness, called a phon, to represent the intensity of sound waves with a frequency of 1000 Hz. A 60-phon sound is one that is perceived by the human ear to have the same loudness as a sound wave with an intensity of 60 dB and a frequency of 1000 Hz.

Required:
a. At approximately what frequency do most people perceive the least intense sounds? Answer numerically in hertz to two significant figures.
b. Normal conversation has a sound level of about 60 dB. How many times more intense must a 100-Hz sound be compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness?

Answers

Answer:

20 Hz

15.8 times

Explanation:

A

Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.

B

A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness

An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is​

Answers

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Other Questions
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