If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mix? Explain your answer.

Answers

Answer 1

Answer:

Mix

Explanation:

A battery has two electrodes at both of its end terminals namely the anode which is the negatively charged electrode and the anode which is the positively charged electrode.

Now, Electrons usually travel through the battery circuit from the anode to the cathode, and this motion of travel is the propelling force that makes electricity flow through the circuit.

Now, the anode will need to have a low electron affinity because it needs to easily release electrons during discharge while the cathode needs to have a high electron affinity because it normally readily accept electrons during discharge.

Thus, for a battery to be strong, it is a combination of high electron affinity and low electron affinity.

Answer 2

Answer:

Answer: mix

Explanation:

you cant pair 2 of the same electron affinities

Related Questions

The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.a. What is the electric field strength and direction between the plates? b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity] c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have? d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
How does activity on the Sun affect natural phenomena on Earth?
An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)
The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?
Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

A suspended platform of negligible mass is connected to the floor below by a long vertical spring of force constant 1200 N/m. A circus performer of mass 70 kg falls from rest onto the platform from a height of 5.8 m above it. Find the maximum spring compression

Answers

Answer:

The maximum spring compression = 3.21 m

Explanation:

The height of the circus performer above the platform connected to string material = 5.8 m

Let the maximum compression of the spring from the impact of the circus performer be x.

According to the law of conservation of energy, the difference in potential energy of the circus performer between the initial height and the level at which spring is compressed to is equal to the work done on the spring to compress it by x

Workdone on the spring by the circus performer = (1/2)kx²

where k = spring constant = 1200 N/m

Workdone on the spring by the circus performer = (1/2)(1200)x² = 600x²

The change in potential energy of the circus performer = mg (5.8 + x)

m = mass of the circus performer = 70 kg

g = acceleration due to gravity = 9.8 m/s²

The change in potential energy of the circus performer = (70)(9.8)(5.8 + x) = (3978.8 + 686x)

600x² = 3978.8 + 686x

600x² - 686x - 3978.8 = 0

Solving this quadratic equation

x = 3.21 m or - 2.07 m

Since the negative answer doesn't satisfy the laws of physics, our correct answer is 3.21 m

Hope this Helps!!!

When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum

Answers

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

$E = hf = h(c)/(\lambda) \n\nwhere;\n\nh \ is \ Planck's \ constant = 6.626 * 10^(-34) \ Js\n\nc \ is \ speed \ of \ light = 3 * 10^(8) \ m/s\n\nE = ((6.626* 10^(-34))* (3* 10^8))/(248* 10^(-9)) \n\nE = 8.02 * 10^(-19) \ J$

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)

Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

$\phi = hf = (hc)/(\lambda_(max)) \n\n\lambda_(max) = (hc)/(\phi) \n\n\lambda_(max) = ((6.626* 10^(-34)) * (3 * 10^8) )/(6.546 * 10^(-19)) \n\n\lambda_(max) = 3.037 * 10^(-7) m\n\n\lambda_(max) = 303.7 \ nm$

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answers

Answer:

The electric force on the proton is 8.2x10^-10 N

Explanation:

We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N

A student and his lab partner create a single slit by carefully aligning two razor blades to a separation of 0.530 mm. When a helium–neon laser at 543 nm illuminates the slit, a diffraction pattern is observed on a screen 1.55 m beyond the slit. Calculate the angle θdark to the first minimum in the diffraction pattern and the width of the central maximum.

Answers

Answer:

angle = 0.058699 degree

width of central maximum is 3.170566 × $10^(-3) ) mExplanation:Given data separation d = 0.530 mm = 0.530×[tex]10^(-3)$ m

distance D = 1.55 m

wavelength w = 543 nm = 543× $10^(-9)$ m

to find out

angle θ and width of the central maximum

solution

we know according to first condition first dark that mean

wavelength = dsinθ

so put value and find θ

543× $10^(-9)$ = 0.530× $10^(-3)$ ×sinθ

sinθ = 543× $10^(-9)$ / 0.530× $10^(-3)$

sinθ = 1.02452 × [tex]10^{-3}

θ = 0.058699 degree

and

we can say

tanθ = y/D

here y is width of central maximum Y = 2y

put all value we get y

so y = D tanθ

y = 1.55 (tan0.0586)

y = 1.58528 × [tex]10^{-3} m =

so Y = 2 ( 1.58528 × [tex]10^{-3} )

so width of central maximum is 3.170566 × [tex]10^{-3} ) m

In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant

Answers

Answer:

i believe that it is d

Explanation:

Final answer:

In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.

Explanation:

In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.

Learn more about Super heater here:

brainly.com/question/32665042

#SPJ2

Other Questions

Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.