The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer 1

Answer:

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery.Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.

I hope it helps you!

Related Questions

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =
Why do societies stratify?
Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.
A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.
An ordinary drinking glass is filled to the brim with water (268.4 mL) at 2.0 ° C and placed on the sunny pool deck for a swimmer to enjoy. If the temperature of the water rises to 32.0 ° C before the swimmer reaches for the glass, how much water will have spilled over the top of the glass? Assume the glass does not expand.

Which one defines force?

Answers

Answer:

Explanation:

a push or a pull that occurs when an object interacts with another object or field.

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The normal is a line perpendicular to the reflecting surface at the point of incidence.

Answers

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3neutrons. The mass of the uranium-235 is greater than the total mass of the
products. Which statement explains this difference in mass?
A. Some of the mass was transformed into neutrons during the
process.
O B. Mass was destroyed and disappeared during the process.
C. Some of the mass was transformed into gases during the
process.
D. Mass was transformed into energy during the process.

Answers

Answer:

D. Mass was transformed into energy during the process.

Answer:

Explanation:

Some of the mass

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force, F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = $(F)/(x)$

= $(150)/(0.5)$

= 300 N/m

The potential energy, EP of the spring is

EP = $(1)/(2) kx^(2)$

the kinetic energy, EK of the spring

EK = $(1)/(2) mv^(2)$

According to conservative energy,

EP = EK

$(1)/(2) kx^(2)$ = $(1)/(2) mv^(2)$

$kx^(2)$ = $mv^(2)$

$v^(2)$ = $(kx^(2) )/(m)$

v = $x\sqrt{(k)/(m) }$

= $0.5\sqrt{(300)/(0.05) }$

= 38.73 m/s

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

Learn more about speed of the arrow here:

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A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is at rest just after the separation. How much work was done by the internal forces that caused the separation

Answers

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_(f)$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_(f)$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_(f)$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

Final answer:

The principle of conservation of momentum implies that no work is performed by the internal forces during the separation of the space vehicle. This is granted that external forces are ignored and the total momentum and kinetic energy of the closed system remain constant.

Explanation:

The subject you're asking about centers around the principle of conservation of momentum. In the case of this space vehicle, before separation, the momentum of the whole system is given by the product of the mass and velocity, mv. After separation, one piece is at rest, leaving the other piece with momentum mv. As there is no external force, the total momentum does not change, so no work is performed by the internal forces causing the separation.

In more detail, the principle of conservation of momentum states that the total linear momentum of a closed system remains constant, regardless of any interactions happening within the system. The system is 'closed' meaning that no external forces are acting upon it. In this case, the space vehicle and the two smaller pieces it separates into form a closed system. This is consistent with your question's stipulation to ignore external forces, such as gravitational forces.

This can also be understood from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. If we consider the vehicle before and after the separation, the kinetic energy of the system remains the same: initially all the energy is concentrated in the moving vehicle, and after the separation, all the kinetic energy is transferred to the moving piece while the at-rest piece has none. Therefore, the work done by the internal forces - which would change the kinetic energy - must be zero.