Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Answers

Answer 1

Answer:

Final answer:

The total work done in moving an alpha particle from the center to the side of a square with electrons at its corners involves finding the potential energy change, which can be calculated using the charges, distances, and Coulomb's constant.

Explanation:

The question deals with the fundamental concepts of electrostatics and the energy associated with moving charges in an electric field. Given the aforementioned question, we are required to find the work done moving an alpha particle (a helium nucleus, having a charge of +2e) from the center of a square to one of its sides, with electrons (each having a charge of -e) being situated at its corners.

To determine the work done, we must consider the potential energy changes resulting from moving the alpha particle. The potential energy associated with two point charges is given by the formula: U = k*q1*q2/r, where q1 and q2 are charges, r is the distance between them, and k is Coulomb's constant.

First, we calculate the potential energy at the center due to all four electrons then find the potential energy at the midpoint of the side. The work done is the difference between these two potential energies. As the electrons are all at an equal distance from the alpha particle (in the center and on the side), the calculations would involve plugging in the values for the charge of an electron, the charge of an alpha particle, the given distance values, and Coulomb's constant into the aforementioned formula.

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Answer 2

Answer:

Final answer:

The work required to move the alpha particle from the midpoint to the midpoint of one of the side of the square with four electrons at its corners would be zero as the net electric field at the midpoint due to the electrons is zero.

Explanation:

The subject of this question pertains to the concept of electrostatics and potential energy in physics. In this scenario, the alpha particle is initially at the midpoint of a square with four electrons at its corners. As per Coulomb's Law, the electrostatic force between two charges is inversely proportional to the square of the distance between them.

Since the alpha particle placed in the center of the square and four electrons at the corners form a symmetrical system, the net force and hence the net electric field at the midpoint due to the electrons is zero. Thus, no work would be required to move the alpha particle to the midpoint of one of the sides of the square as work done is calculated by the formula W = F x d x cos(θ), where F is force, d is the displacement, and θ is the angle between the force and displacement. Since F is equal to zero, the work done will also be zero.

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Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration

∑ F = m a

in this case there are two forces on the x axis

F_applied - fr = 0

since they indicate that the velocity is constant, consequently

F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0m/s , and the distance between them is 52.0m . After t1 = 5.00s , the motorcycle starts to accelerate at a rate of 5.00m/s^2. a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words. b. Find t2−t1

Answers

Answer:

$t\approx4.561\ s$

Explanation:

Given:

initial speed of car and motorcycle, $v=19\ m.s^(-1)$

initial distance between the car and motorcycle, $s=52\ m$

time after which the motorcycle starts to accelerate, $t_1=5\ s$

rate of acceleration of motorcycle, $a=5\ m.s^(-2)$

The initially the relative velocity of the motorcycle is zero with respect to car.

Now using the equation of motion in the relative quantities:

$s=u.t+(1)/(2) .a.t^2$

here:

$s =$ relative distance of motorcycle with respect to the car

$u=$ initial relative velocity of the motorcycle with respect to the car

$t=$ time taken to cover the distance gap from the car.

$52=0+0.5* 5* t^2$

$t\approx4.561\ s$

If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2

Answers

Answer:

the answers the correct one is a 4

Explanation:

The centripetal acceleration is by

a = v² / R

angular and linear velocities are related

v = w R

let's substitute

a = w² R

for initial condition

a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

a = (2w₀)² R

a = 4 (w₀² R)

a = 4 a₀

when reviewing the answers the correct one is a

While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?

Answers

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

W = mg

let's use the concept of density

ρ_body = m / V

m = ρ_body V

W = ρ_body V g

The thrust of the body is given by Archimedes' law

B = ρ_liquid g V_liquid

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

ρ_body V g = ρ_liquid g V_liquid

ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

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