May you please help?​

The electron's distance from the nucleus in the 1s state of a hydrogen atom is A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * (∫ e^(-2r/a) dr)^2] from 0 to ∞ = [(r * (a^3 / 4√π) * (a/2)) - (a^3 / 4√π) * (a/2)^2].

To find the expectation value of the electron's distance from the nucleus in the 1s state of a hydrogen atom, we need to calculate the quantity Ar, where A is given by:

A = ⟨r⟩ = ∫ r * ψ(1s) * r^2 dr from 0 to ∞

First, let's express ψ(1s) in terms of the radial wavefunction R(r):

ψ(1s) = R(r) * Y(0,0)  [Since the 1s state has azimuthal quantum number l = 0 and magnetic quantum number m = 0]

The radial wavefunction R(r) for the 1s state is given by:

R(r) = (1 / √π * a^(3/2)) * e^(-r/a)

Now, let's calculate A:

A = ∫ r * ψ(1s) * r^2 dr from 0 to ∞

 = ∫ r * (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr from 0 to ∞

We can use integration by parts to simplify this integral. Let's define u = r and dv = (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr.

Taking the differentials, we have du = dr and v = ∫ (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr.

Integrating v, we can use the given integrals:

v = ∫ (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr

 = (a^3 / 4√π) * ∫ e^(-2r/a) dr  [Using the given integral]

Now, applying integration by parts:

∫ u dv = uv - ∫ v du

∫ r * (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr

= (r * v) - ∫ v du

= (r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - ∫ [(a^3 / 4√π) * ∫ e^(-2r/a) dr] dr

Simplifying the expression:

A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * ∫ ∫ e^(-2r/a) dr dr] from 0 to ∞

The integral ∫ ∫ e^(-2r/a) dr dr can be expressed in terms of the given integrals:

∫ ∫ e^(-2r/a) dr dr = (∫ e^(-2r/a) dr)^2

                     = (a/2)^2   [Using the given integral]

Now, substituting the values and evaluating the limits:

A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * (∫ e^(-2r/a) dr)^2] from 0 to ∞

 = [(r * (a^3 / 4√π) * (a/2)) - (a^3 / 4√π) * (a/2)^2]

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Answer:

Earth's atmosphere is a layer of gases surrounding the planet Earth and retained by the Earth's gravity. It contains roughly 78% nitrogen and 21% oxygen 0.97% argon and carbon dioxide 0.04% trace amounts of other gases, and water vapor. This mixture of gases is commonly known as air.  Explanation:

Answer:

the earth's atmosphere

Explanation:

hope that helps

Answer:

In absorption, the frequency of the incoming light wave is at or near the energy levels of the electrons in the matter. The electrons will absorb the energy of the light wave and change their energy state.

Explanation:

Answer:

Why does it take sometimes to be brighter again after eclipse? The moon looks dimmer before lunar eclipse because the moon enters into shadow of penumbra region of earth. As a result, the brightness of moon decreases and looks dimmer.

Explanation:

Answer: The moon looks dimmer before lunar eclipse because the moon enters into shadow of penumbra region of earth. As a result, the brightness of moon decreases and looks dimmer.

Explanation:

Answer:

B. There are no forces acting on the ball.

Explanation:

There are no forces acting on the ball.

Answer:

No forces acting upon the ball

Explanation:

newtons law, an object at rest will remain at rest until an force is acted upon the object

Answer:

You can breathe in too much carbon monoxide, which will eliminate the flow of oxygen to your bloodstream and can kill you.

Explanation:

No, because some scientist are bad and they must have a scientific attitude like honesty, curiosity, open-mindedness, optimistic and more. And they should follow others sometimes because for their own good.

Answer:

No

Explanation:

Science operates in the context of national and international laws, agreements and conventions. It requires scientists to conduct and communicate scientific work for the benefit of society, with excellence, integrity, respect, fairness, trustworthiness, clarity, and transparency.

Answer: The half-life is the amount of time it takes for a given isotope to lose half of its radioactivity. If a radioisotope has a half-life of 14 days, half of its atoms will have decayed within 14 days. In 14 more days, half of that remaining half will decay, and so on.

Explanation:

mass=force*acceleration

mass=3000*10

mass=30,000

The mass of the bodies in the elevator is 400 kg.

The acceleration of the elevator is 2.5 m/s².

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Given parameters:

Mass of the elevator: M = 250 kg.

Mass of two persons: m₁ = 50 kg and m₂ = 100 kg.

Force exerted by the motor: F = 3000N.

g = 10 m/s².

Let, the acceleration of the  elevator = a.

the mass of the bodies in the elevator :m= 250 kg. + 50 kg +100 kg. = 400 kg.

Now, F = mg - ma

⇒ 3000 = 400×10 - 400a

⇒ a = 1000/400 = 2.5 m/s²

Hence, the acceleration of the elevator is 2.5 m/s².

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A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,

• the net horizontal force on the car is

∑ F = N cos(45°) = m a = m v ² / R

• the net vertical force on the car is

∑ F = N sin(45°) - m g = 0

where

• N = magnitude of the normal force

• m = mass of the car

• a = v ² / R = centripetal acceleration of the car

• v = tangential speed of the car

• R = 100 m = radius of curvature

• g = 9.8 m/s² = acceleration due to gravity

From the net vertical force equation, we get

N = m g / sin(45°)

and substituting this into the net horizontal force equation and solving for v gives

(m g / sin(45°)) cos(45°) = m v ² / R

v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s

We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

From the question we are told

a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:

Generally the equation for the Velocity is mathematically given as

[tex]V=\sqrt{rgtan\theta}[/tex]

Therefore

[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]

Therefore

A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

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Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

Answer:

Answer:

A. it's got everything set.... correct connection

The setup that will light the bulb in the given different circuit diagram is circuit A, because it is the only circuit is that is closed.

A complete circuit or closed circuit is the circuit that will allow the flow of electric current through it.

When a circuit is closed or complete, the connection of the different components of the circuit is complete.

From the image given, only option A has complete connection of different components of the circuit.

Thus, the setup that will light the bulb in the given different circuit diagram is circuit A, because it is the only circuit is that is closed.

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Answer:

[tex]f=4.70\times 10^{14}\ Hz[/tex]

Explanation:

Given that,

The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]

We need to find the frequency of the light. We know that,

[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]

So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].

Answer:

D

Explanation:

its D

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2

distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m

Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

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Answer:

a) f = 4.76 10¹⁴ Hz, b) d = 2.73 10⁻⁴ m, c) θ = 6.923 10⁻³ rad

Explanation:

a) In this problem the frequency of light is asked, let's use the relationship between the speed of the wave, its wavelength and its frequency

           c = λ f

           f = c /λ

           f = [tex]\frac{3 \ 10^8}{630 \ 10^{-9}}[/tex]

           f = 4.76 10¹⁴ Hz

b) slit separation (d)

the expression for the constructive interference of the double-slit experiment is

          d sin  θ = m λ

let's use trigonometry

          tan θ = y / L

          tan θ = [tex]\frac{sin \theta}{cos \theta}[/tex]

in general the angles are small, so we can approximate

          tan θ = sin θ

          tan θ = y/L

we substitute

          d y / L = m λ

          d = m L λ / y

we calculate

          d = 3  1.3  630 10⁻⁹ /0.90 10⁻²

          d = 2.73 10⁻⁴ m

c) the angle

           tan θ = y / L

           θ = tan⁻¹ y / L

           θ = tan⁻¹ 0.9 10⁻² / 1.3

           θ = tan⁻¹ 6,923 10⁻³

let's find the angle in radians

           θ = 6.923 10⁻³ rad

Answer:

5.03

Explanation:

trust me

Answer:

c) The impulse is the same in both cases.

Explanation:

What drives the car is the amount of movement of the car during its journey. In this case, we can consider that in the car, presented in the question above, the movement is generated through speed and friction on the track, the impact that makes the car stop, is not part of the movement and therefore, not part of the impulse. In the two situations presented, the speed of the car is the same, the same car and the same track were also used, which shows us that the impulse, in both situations, is the same.

Answer:

into the page

Explanation:

Since the uniform electric field is in the negative y direction so its is -E and the electron is travelling in the positive x direction, it experiences an electric force F = -e × -E = + eE, so the electric force is in the positive y direction. Now since the net force on the electron is zero in the region of the magnetic field, it follows that the direction of the magnetic force is opposite to that of the electric force. Since the electric force is in the positive y direction, the magnetic force is in the negative y direction.

By the right hand rule, since the magnetic force is in the negative y direction and the electron moves in the positive x direction, it follows that the magnetic field is in the positive z direction, into the page.

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

Answer:

1.78×10×10^10 electron

Explanation:

Electric field outside the sphere can be calculated using below expression

E= kq/ r^2..........eqn(1)

Where k= 9 × 10^9 NM^2/C^2

q= charge

E= 1440 N/C

Diameter= 30.0 cm= 0.3 m

r= radius= 0.3/2= 0.15m

If we make q subject of formula from eqn(1) we have

q= Er^2/k............eqn(2)

q= 1440 × (0.15)^2 /(9 × 10^9 )

= 2.85×10^-9C

Total charge is an integer of electron charge , then we can calculate the number of the electron using the expression below

q= Ne

Where

N = number of electron

Making N subject of the formula we have

N= q/e

Where e= electron value= 1.6× 10^-19

N=2.85×10^-9 /1.6× 10^-19

= 1.78×10×10^10 electron

If a Catatlyst was added the values affected would be 2, 3, 4

A catalyst is a substance that speeds up the rate of a chemical reaction without undergoing any permanent change itself. It facilitates the reaction by providing an alternative pathway with lower activation energy, which allows the reaction to occur more readily.

Catalysts work by interacting with the reactant molecules and weakening the existing bonds or creating new ones, making it easier for the reaction to proceed.

Catalysts are typically specific to certain reactions and can be used in small amounts relative to the reactants. They play a crucial role in many industrial processes and are essential for efficient chemical transformations.

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Answer:

the time taken for the motor to lift the object is 21.17 s.

Explanation:

Given;

mass of the object, m = 45 kg

height through which the object was lifted, h = 9.0 m

electrical power used by the motor, P = 750 W

Efficiency of the electrical motor, n = 25% = 0.25

The electrical energy used by the motor in lifting the object is calculated as;

E = P x t

where;

t is the time taken for the motor to lift the object

E = 750 x t

E = 750t

The electrical energy converted by the motor to mechanical energy is calculated as;

P.E = 0.25(750t)

P.E = 187.5t

Recall, P.E = mgh

mgh = 187.5t

45 x 9.8 x 9 = 187.5t

3969 = 187.5t

t = 3969/187.5

t = 21.17 s

Therefore, the time taken for the motor to lift the object is 21.17 s.

Answer:

14 μm

Explanation:

The magnetic field due to a long straight wire is B = μ₀i/2πr where  μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current = 7.0 A and r = distance of credit card from magnetic field.

So r = μ₀i/2πB since B = 1000 gauss = 1000 G × 1 T/10000 G  = 0.1 T

r = 4π × 10⁻⁷ H/m × 7.0 A/(2π × 0.1 T)

r = 2 × 10⁻⁷ H/m × 7.0 A/0.1 T

r = 14 × 10⁻⁷ H/m × A/0.1 T

r = 140 × 10⁻⁷ m

r = 1.4 × 10⁻⁵ m

r = 14 × 10⁻⁶ m

r = 14 μm

Answer: C. Stored energy

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west