Part A

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 7.

Express your answer in units of ℏ to three significant figures.

Part B

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 26.

Express your answer in units of ℏ to three significant figures.

Part C

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 191.

Express your answer in units of ℏ to three significant figures.

Answer:

a. [tex]F_2=31.76N[/tex]

b. [tex]F_2=185.86N[/tex]

Explanation:

Given:

[tex]F_1=27800N[/tex]

[tex]r_1=5.07x10^{-3}m[/tex]

[tex]r_2=0.150 m[/tex]

[tex]p=8.81x10^2 kg/m^3[/tex]

Using the equation to find the force so replacing

a.

[tex]F_1*A_2=F_2*A_1[/tex]

[tex]A=\pi*r^2[/tex]

[tex]F_2=F_1*\frac{A_2}{A_1}=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}[/tex]

[tex]F_2=31.76N[/tex]

b.

[tex]F_2=F_1+F_p[/tex]

[tex]F_2=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^{-3})m^2)[/tex]

[tex]F_2=185.86N[/tex]

Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles

Explanation:

Answer:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

Calculate the EAC of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, consider this equation 1

Calculate the EAC of CFL

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, consider this equation 2

Equate 1 and 2 to find the amount of C

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

Answer:11.18 m/s

Explanation:

Given

mass of particle m=0.2 kg

Potential Energy U(x) is given by

[tex]U(x)=8x^2+2x^4[/tex]

at x=1 m

[tex]U(1)=8+2=10 J[/tex]

kinetic energy at x=1 m

[tex]K.E.=\frac{1}{2}mv^2=\frac{1}{2}\times 0.2\times 5^2[/tex]

[tex]K.E.=2.5 J[/tex]

Total Energy =U+K.E.

[tex]Total=10+2.5=12.5 J[/tex]

at x=0, U(0)=0

as total Energy is conserved therefore K.E. at x=0 is equal to Total Energy

[tex]\frac{1}{2}\times 0.2\times v^2=12.5[/tex]

[tex]v^2=125[/tex]

[tex]v=\sqrt{125}[/tex]

[tex]v=11.18 m/s[/tex]

By conserving energy, we find that the speed of the particle at the origin is 25 m/s.

The question is asking for the speed of the particle at the origin given its potential energy function and speed at x = 1.0 m. This can be solved using the principle of energy conservation which states that the total energy (kinetic energy + potential energy) of the particle is conserved unless acted upon by an outside force.

In this case, we first find the total energy of the particle at x = 1.0 m. The kinetic energy is (1/2)mv² = (1/2)*0.2 kg*(5 m/s)² = 2.5 J. The potential energy at x = 1.0 m, according to the given function, is U(1) = 8(1)² + 2(1)⁴ = 10 J. So, the total energy at x = 1.0 m is 2.5 J + 10 J = 12.5 J.

At the origin (x = 0), the potential energy is U(0) = 0. So, the kinetic energy at the origin is equal to the total energy, which is 12.5 J. From the kinetic energy, we can find the speed using the equation KE = (1/2)mv², which gives v = sqrt((2*KE)/m) = sqrt((2*12.5 J)/0.20 kg) = 25 m/s.

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The value of all options are mathematically given as

a) a=1.5 rad/s^2

b)theta=248 rad

c) theta '=248 radians

d) angles in degree=14207.511

Question Parameter(s):

Generally, the equation for the final angular speed  is mathematically given as

w2=w+at

Therefore

37=25+a 8

a=1.5 rad/s^2

b)

Average angular speed

[tex]\theta =\omega _1t+\frac{1}{2}\alpha t^2[/tex]

[tex]\theta =25\times 8+\frac{1}{2}\times 1.5\times 8^2[/tex]

theta=248 rad

Hence

theta =200+48

theta=248 rad

In conclusion,

c) theta '=248 radians

d) angles in degree=14207.511

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Final answer:

The angular acceleration is 1.5 rad/s², the average angular speed is 31 rad/s, the angle through which the wheel rotates is 320 rad, and the angle in degrees is 18333 degrees.

Explanation:

(a) To find the angular acceleration, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time interval. Plugging in the given values, we get: angular acceleration = (37 rad/s - 25 rad/s) / 8 s = 1.5 rad/s².

(b) The average angular speed can be found by taking the average of the initial and final angular velocities. So, the average angular speed = (25 rad/s + 37 rad/s) / 2 = 31 rad/s.

(c) The angle through which the wheel rotates can be found using the formula: angle = initial angular velocity * time + (1/2) * angular acceleration * time². Plugging in the given values, we get: angle = 25 rad/s * 8 s + (1/2) * 1.5 rad/s² * (8 s)² = 320 rad.

(d) To convert radians to degrees, we use the conversion factor: 1 radian = 180 degrees / pi. So, the angle in degrees = 320 rad * 180 degrees / pi = 18333 degrees (rounded to the nearest whole number).

Answer:

Explanation:

The energy of the particle in a square power well is given by the Schrödinger equation, with

       E₁ = RA (h²/ 8m L²)   n²

With h the Planck constant the mass of the particle, L the length of the box and n an integer starting 1

We find the energy for each situation presented

1) For the fundamental state n = 1

       E₁ = RA (h/ 8m L²)

2) the first excited state corresponds to n = 2

       E₂ = RA (h/ 8m L²) 2²

       E₂ = E₁ 4

3)    E₁ = RA (h/ 8m L²)

4) The length of the box is L = 2L

      E’= RA (h’ / 8m (2L)²)

      E’= RA (h’ / 8m L²) Ra ¼

      E’= E₁ / 2

5) L = 2l first excited state

      n = 2

      E ’’ = RA (h/ 8m (2L)²2) 2²

      E ’’ = RA (h ’/ 8m (L)²) 4/2

      E ’’ = E₁ 2

All energies are in relation to the fundamental state (E1), milking from least to greatest

       E1 / 2 <E1 = E1 <2E1 <4E1

       4 <3 = 1 <5 <2

Final answer:

The correct order from lowest to highest energy for a particle in a square well, considering both quantum state and box length, is: (5) < (4) < (3) < (2) < (1). This order is based on the quantum energy level formula E_n = n^2 * pi^2 * h-bar^2 / (2 * m * L^2), where energy increases with the square of the principal quantum number and decreases with the square of the box length.

Explanation:

To determine the proper order from lowest energy to highest energy for the particle in a square well, we must consider the principles of quantum mechanics, particularly the energy levels of a particle in a one-dimensional box. The energy levels are quantized and are given by the expression En = n2π2ħ2/(2mL2), where ħ is the reduced Planck constant, m is the mass of the particle, L is the length of the box, and n is the principal quantum number (n = 1 for the ground state, n = 2 for the first excited state, and so on). Given this, here's the reasoning for the order:

Therefore, the order from lowest energy to highest energy is as follows:

(5) < (4) < (3) < (2) < (1)

Answer:

20.13841 rad/s²

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]

[tex]\omega_f[/tex] = Final angular velocity = 0

t = Time taken = 2.6 s

[tex]\alpha[/tex] = Angular acceleration

Equation of rotational motion

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]

The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

Answer:

hi sandra!!

Explanation:

the number 3 is the correct!!

The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth’s atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.

Auroras that occur in the northern hemisphere are called ‘Aurora Borealis’ or ‘northern lights’ and auroras that occur in the southern hemisphere are called ‘Aurora Australis’ or ‘southern lights’.

Auroral displays can appear in many differents colours, but green is the most common. Colours such as red, yellow, green, blue and violet are also seen occasionally. The auroras can appear in many forms, from small patches of light that appear out of nowhere to streamers, arcs, rippling curtains or shooting rays that light up the sky with an incredible glow.

Answer:

Option-(3): The aurora borealis are the formation of the colored patterns on the pole of the planet, but other celestial bodies also experience such effects due to the heat flares or the sun flares when exposed to them.

Explanation:

Aurora borealis:

The natural phenomenon which is caused in most of the celestial bodies, but when our planet earth is exposed to such electromagnetic rays or flares released from the Sun, when the energy level on the surface increases and the flare is cut off from the environment of the sun. As, starts motion towards the different celestial bodies present in the solar system.

And when it reaches earth, the earth surface reflects the high in energy atoms coming towards its atmosphere. As the magnetic field present on the earth surface interacts with the Sun's flare and then it cut off's into two moving towards the two poles of the planet.

Answer:

138.18 minutes

Explanation:

[tex]L_v[/tex] = Latent heat of water at 0°C = 80 cal/g

m = Mass of water = 570 g

Heat removed for freezing

[tex]Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal[/tex]

Let N be the number of cycles and each cycle removes 56 cal from the freezer.

So,

[tex]55\times N=45600\\\Rightarrow N=\frac{45600}{55}[/tex]

Each cycle takes 10 seconds so the total time would be

[tex]\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes[/tex]

The total time taken to freeze 138.18 minutes

Answer:

b = 1.2 m

L = 4.8 m

d= 1.2 m

Explanation:

Lets take

Width of the tank = b

Depth = d

length = L

Given that L = 4 b

v= 2 cm/s

Flow rate Q= 10,000 m³/d

We know that 1 d = 24 hr = 24 x 3600 s

1 d= 86400 s

Q= 0.115  m³/s

Flow rate Q

Q= Area x velocity

Q= L . b .v

0.115 = 4 b . b . 0.02

4 b² = 5.78  

b = 1.2 m

L = 4.8 m

d= 1.2 m

This is the dimension of the tank.

To design the settling tank with the given water flow and settling velocity, we use conservation of mass and settling speed to calculate the dimensions. We convert the flow rate to m3/s, relate it to the tank's cross-sectional area, and given settling characteristics to find the tank's width, length, and depth.

To design a rectangular settling tank for settling sand particles with a settling velocity of 2.0 cm/s and a water flow of 10,000 m3/d, we will apply the principles of conservation of mass and the given settling velocity to determine the dimensions of the tank. The problem states that the tank length (L) should be four times the width (W), and the width should be equal to the depth (D), therefore L = 4W and W = D.

To find the dimensions, we first convert the water flow rate into m3/s: flow rate = 10,000 m3/d × (1 day/86400 s) ≈ 0.1157 m3/s. Since the sand particles settle at 2 cm/s, for a particle to settle before reaching the end of the tank, the residence time should be at least the depth divided by the settling velocity. From here, we use the flow rate (Q), which is also equal to the cross-sectional area (A) times the water velocity (V), and we know that A = W² for this tank because W = D.

Therefore, Q = W²V. Using V = L / (residence time) and substituting for L = 4W, we find that Q = W² × (4W / (W/0.02 m/s)) which simplifies to Q = 0.08 m/s × W³. Solving for W gives us W = ∛(Q / 0.08 m/s) and subsequently we find L = 4W and D = W. Finally, we can plug in the flow rate and solve for the dimensions in meters.

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Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2[/tex]

v = r ω

[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}[/tex]

[tex]I = \dfrac{m(2gh - v^2)r^2}{v^2}[/tex]

[tex]I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}[/tex]

[tex]I =mr^2(0.287)[/tex]

I = 0.287 MR²

Answer:

D.The potential energy per unit charge

Explanation:

Electric potential of a charged particle:

It is scalar quantity because it has magnitude but it does not have direction.

It is the amount of work done required to move a unit positive charge from reference point to specific point in the electric field without producing any acceleration.

Mathematical representation:

[tex]V=\frac{W}{Q_0}[/tex]

Where W= Work done

[tex]Q_0[/tex]= Unit positive charge

Other formula to calculate electric field:

[tex]V=\frac{KQ}{r}[/tex]

Where K=[tex]\frac{1}{4\pi \epsilon_0}[/tex]

It can be defined as potential energy per unit charge.

Hence, option D is true.

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m[/tex]

The image distance is 0.051 m

When u = 50 cm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m[/tex]

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

To focus on the closer object, the lens must move away from the object. By applying the lens formula, we calculate that the lens must move exactly 7mm away from the initial position.

This problem comes down to the concept of the lens formula in optical physics. The lens formula is 1/f = 1/v - 1/u where f is the focal length, v is the image distance, and u is the object distance.

Part A: To focus on the second object which is closer, the lens must move away from the object.

Part B: To calculate how far the lens must move, we need to apply the lens formula. Let's calculate the image distances for the two object distances.

For an object at 2.5 m, 1/v = 1/f + 1/u = 1/0.05 + 1/2.5 = 20 + 0.4 = 20.4. So, v = 0.049 m or 49 mm.

For an object at 0.5 m, 1/v = 1/f + 1/u = 1/0.05 - 1/0.5 = 20 - 2 = 18. So, v = 0.056 m or 56 mm.

Hence, the lens must move 56mm - 49mm = 7mm away from the initial position.

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Answer

given,

length of wire = 10 cm = 0.1 m

resistance of the wire =  0.330 ohms

speed of pulling = 4 m/s

Power = 4.40 W

a) Force of pull = ?

 [tex]P = F_{pull} v[/tex]

 [tex]F_{pull} =\dfrac{P}{v}[/tex]

 [tex]F_{pull} =\dfrac{4.40}{4}[/tex]

 [tex]F_{pull} =1.1\ N[/tex]

b) using formula

  [tex]P = \dfrac{B^2l^2v^2}{R}[/tex]

  where B is the magnetic field

   v is the pulling velocity

   R is the resistance of the wire

  [tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]

  [tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]

  [tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]

  [tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]

  [tex]B =\sqrt{9.075}[/tex]

      B = 3.01 T

The pulling force in this magnetic field is equal to 1.1 Newton.

Given the following data:

Mathematically, the pulling force in a magnetic field is given by this formula:

[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]

F = 1.1 Newton.

To determine the strength of the magnetic field, we would apply this formula:

[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]

B = 3.012 T.

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The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

[tex]PE=\frac{GMm}{r}[/tex]

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

[tex]m = \rho V[/tex]

[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]

[tex]m = 7.210*10^{11}Kg[/tex]

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]

Then the potential energy at this point would be,

[tex]PE_1 = \frac{GMm}{r_1}[/tex]

[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]

[tex]PE_1 = 9.05*10^{15}J[/tex]

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]

The we calculate the Potential gravitational energy,

[tex]PE_2 = \frac{GMm}{r_2}[/tex]

[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]

[tex]PE_2 = 9.361*10^{15}J[/tex]

The final temperature of the mixture is approximately 29.91°C.

The heat exchange between the tea and the ice. The temperature of the final mixture will be somewhere between 0°C and 31°C, and we need to determine that final temperature. The heat transfer can be calculated using the principle of conservation of energy:

Qin = Qout

The heat gained by the ice as it melts is given by:
Qice = m2 ⋅ Lf

The heat gained by the tea as it cools down is given by:
Qtea = m1 ⋅ c ⋅ (T1 - Tfinal)

The negative sign is used because the tea is losing heat. Setting these equal to each other and solving for Tfinal, we get:
m1 ⋅ c ⋅ (T1 - Tfinal) = m2 ⋅ Lf

Now, let's plug in the given values:

0.73 kg ⋅ 4186 J/(kg ⋅ K) ⋅ (31°C - Tfinal) = 0.095 kg ⋅ 33.5 × 10^4 J/kg

Now, solve for Tfinal:

0.73 ⋅ 4186 ⋅ (31 - Tfinal) = 0.095 ⋅ (33.5 × 10^4)

3050.78 ⋅ (31 - Tfinal) = 3182.5

94602.78 - 3050.78 ⋅ Tfinal = 3182.5

-3050.78 ⋅ Tfinal = -91420.28

Tfinal = 91420.28/3050.78

Tfinal ≈ 29.91°C

So, the final temperature of the mixture is approximately 29.91°C.

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

[tex]f_o[/tex] = Focal length of objective = 0.9 cm

[tex]f_e[/tex] = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

[tex]m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52[/tex]

The angular magnification of the compound microscope is -252.52

Answer:

68.6 m/s

Explanation:

v = Speed of sound in air = 343 m/s

u = Speed of train

[tex]f_1[/tex] = Actual frequency

From the Doppler effect we have the observed frequency as

When the train is approaching

[tex]f=f_1\frac{v+u}{v}[/tex]

When the train is receeding

[tex]\frac{2f}{3}=f_1\frac{v-u}{v}[/tex]

Dividing the above equations we have

[tex]\frac{f}{\frac{2f}{3}}=\frac{f_1\frac{v+u}{v}}{f_1\frac{v-u}{v}}\\\Rightarrow \frac{3}{2}=\frac{v+u}{v-u}\\\Rightarrow 3v-3u=2v+2u\\\Rightarrow v=5u\\\Rightarrow u=\frac{v}{5}\\\Rightarrow u=\frac{343}{5}\\\Rightarrow u=68.6\ m/s[/tex]

The speed of the train is 68.6 m/s

Final answer:

To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.

Explanation:

To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:

Mass flow rate = density x velocity x area

Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:

Area = mass flow rate / (density x velocity)

However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:

Pressure x Volume = n x R x Temperature

Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.

Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.

The exit area of the nozzle is approximately [tex]\( 0.00140 \text{ m}^2 \) or \( 1.40 \text{ mm}^2 \)[/tex].

The continuity equation for a steady-state flow is given by:

[tex]\[ \dot{m} = \rho \cdot A \cdot v \][/tex]

To find the density [tex]\( \rho \)[/tex], we need to use the ideal gas law, which is a good approximation for steam under these conditions:

[tex]\[ P = \rho \cdot R \cdot T \][/tex]

where:

- P is the absolute pressure at the nozzle exit (2 MPa or 2000 kPa),

- R is the specific gas constant for steam (0.4615 kJ/kg·K),

- T is the absolute temperature at the nozzle exit (400°C + 273.15 = 673.15 K).

Rearranging the ideal gas law to solve for [tex]\( \rho \)[/tex]:

[tex]\[ \rho = \frac{P}{R \cdot T} \][/tex]

Now, we can substitute the density [tex]\( \rho \)[/tex] back into the continuity equation to solve for the exit area A:

[tex]\[ A = \frac{\dot{m}}{\rho \cdot v} \][/tex]

Substituting the values we have:

[tex]\[ \rho = \frac{2000 \text{ kPa}}{0.4615 \text{ kJ/kg·K} \cdot 673.15 \text{ K}} \] \[ \rho = \frac{2000}{310.56} \text{ kg/m}^3 \] \[ \rho \approx 6.44 \text{ kg/m}^3 \][/tex]

Now, we can find the exit area A:

[tex]\[ A = \frac{2.5 \text{ kg/s}}{6.44 \text{ kg/m}^3 \cdot 280 \text{ m/s}} \] \[ A = \frac{2.5}{1787.2} \text{ m}^2 \] \[ A \approx 0.00140 \text{ m}^2 \][/tex]

Answer:L=109.16 m

Explanation:

Given

initial temperature [tex]=20^{\circ}C[/tex]

Final Temperature [tex]=80^{\circ}C[/tex]

mass flow rate of cold fluid [tex]\dot{m_c}=1.2 kg/s[/tex]

Initial Geothermal water temperature [tex]T_h_i=160^{\circ}C[/tex]

Let final Temperature be T

mass flow rate of geothermal water [tex]\dot{m_h}=2 kg/s[/tex]

diameter of inner wall [tex]d_i=1.5 cm[/tex]

[tex]U_{overall}=640 W/m^2K[/tex]

specific heat of water [tex]c=4.18 kJ/kg-K[/tex]

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

[tex]\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)[/tex]

[tex]2\times (160-T)=1.2\times (80-20)[/tex]

[tex]160-T=36[/tex]

[tex]T=124^{\circ}C[/tex]

As heat exchanger is counter flow therefore

[tex]\Delta T_1=160-80=80^{\circ}C[/tex]

[tex]\Delta T_2=124-20=104^{\circ}C[/tex]

[tex]LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}[/tex]

[tex]LMTD=\frac{80-104}{\ln \frac{80}{104}}[/tex]

[tex]LMTD=91.49^{\circ}C[/tex]

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

[tex]\dot{m_c}c(80-20)=U\cdot A\cdot[/tex] (LMTD)

[tex]A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2[/tex]

[tex]A=\pi DL=5.144[/tex]

[tex]L=\frac{5.144}{\pi \times 0.015}[/tex]

[tex]L=109.16 m[/tex]

The length of a counterflow double-pipe heat exchanger required can be calculated by first determining the heat needed to raise the water temperature, then calculating the necessary surface area, and finally dividing by the inner circumference of the pipe to find the length.

To determine the length of the heat exchanger required to heat water from 20°C to 80°C using geothermal water at 160°C, we can apply the basic principles from thermodynamics and heat transfer. The equation that relates the heat transfer Q with the overall heat transfer coefficient U, the surface area A, and the temperature difference ΔT is Q = U * A * ΔT. We can determine the heat needed to raise the temperature of water by using the specific heat of water and the mass flow rate of water.

First, the amount of heat needed to raise the temperature of water (Q) can be calculated using the specific heat capacity of water (c_p, which is approximately 4.18 kJ/kgK), the mass flow rate of water (m_dot), and the temperature rise (ΔT = T_f - T_i), using the formula Q = m_dot * c_p * ΔT.

Next, we can rearrange the heat transfer equation to solve for the surface area A required for the heat exchanger: A = Q / (U * ΔTLMTD), where ΔTLMTD is the logarithmic mean temperature difference, which can be calculated for a counterflow heat exchanger knowing the inlet and outlet temperatures of both the hot and cold fluids.

Finally, we can find the length of the heat exchanger (L) by dividing the surface area A by the inner circumference of the pipe (π * d), with d being the inner pipe diameter.

By calculating the heat necessary for the temperature rise of the water and applying the given overall heat transfer coefficient, we can determine the required length of the exchanger.

Explanation:

It is given that,

Force exerted by a boy, F = 11.8 N

Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]

Mass of the sled, m = 6.15 kg

Distance moved, d = 2.75 m

Initial speed, u = 0.37 m/s

Let W is the work done by the boy. Using the expression for the work done to find it as :

[tex]W=Fd\ cos\theta[/tex]

[tex]W=11.8\times 2.75\ cos(28)[/tex]

W = 28.65 joules

Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]

[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]

v = 3.07 m/s

So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.

The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.

To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.

First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:

[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]

[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]

Now, we can calculate the work done:

[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]

[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]

\[ W \approx 28.64 \, \text{J} \]

Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):

[tex]\[ W = \Delta KE \][/tex]

[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]

Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:

[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]

[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]

[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]

Answer:

A,

Explanation:

To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which

[tex]E = \frac{hc}{\lambda}[/tex]

Where,

h = Plank's constant [tex](6.626*10^{-34}J.s)[/tex]

c = Speed of light [tex](3*10^8m/s)[/tex]

[tex]\lambda[/tex] =Wavelength

If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.

The larger the amplitude of the wave, the smaller the energy.

On the other hand we have the frequency value defined as

[tex]f = \frac{c}{\lambda}[/tex]

In this case the frequency is also inversely proportional to the wavelength.

In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.

In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.

Answer:

mass percent of chlorine is 56.5%

Explanation:

Given data:

Volume V = 581 mL = 0.581 L

Pressure P = 752 mm of Hg = 752/760 = 0.9894 atm

Temperature T = 298 K

Molar gas constant R = 0.08206 atm.L/mol.K

from Ideal gas equation we have following relation

PV = nRT solving for n

[tex]n = \frac{PV}{RT}[/tex]

[tex]n = \frac{0.9894 \times 0.581}{0.08206 \times 298}[/tex]

n = 0.0235 mole

mas of [tex]Cl_2[/tex]  = moles × molar mass of [tex]Cl_2[/tex]

                        [tex]= 0.0235 \times 70.906 = 1.66 g[/tex]

Mass percent of chlorine  [tex]=  \frac{mass\ of\ Cl}{mass\ of\ sample}[/tex]

[tex]=\frac{1.66}{2.95} = 0.565 = 56.5 %[/tex]

Answer:

The force the bat applies to the ball is 340 N

Explanation:

given information:

mass of the ball, m = 400 g = 0.4 kg

initial velocity, [tex]v_{i}[/tex] = 30 m/s

final velocity, [tex]v_{f}[/tex]  = 38 m/s

time, t = 80 ms = 0.08 s

the correlation between the change in momentum and the force is shown by the following equation

ΔP = FΔt

F = ΔP/Δt

where,

Δp = mΔv

     = m([tex]v_{f} - v_{i}[/tex])

     = m([tex]v_{f} + v_{i}[/tex])

[tex]v_{i}[/tex] = - because in opposite direction, thus

F = ΔP/Δt

  = m([tex]v_{f} + v_{i}[/tex])/Δt

   = (0.4)(38+30)/0.08

   = 340 N

Answer:209.98 kJ

Explanation:

mass of water [tex]m=456 gm[/tex]

Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]

Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]

specific heat of ice [tex]c=2090 J/kg-K[/tex]

Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]

specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]

Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]

[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]

Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]

[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]

heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]

[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]

Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]

[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]

Answer:

108507.02596 Pa

42.42 cm

Explanation:

F = Force

m = Mass of piston = 23 kg

[tex]T_1[/tex] = Initial temperature = 307°C

[tex]T_2[/tex] = Final temperature = 13°C

[tex]P_o[/tex] = Outside pressure

[tex]P_i[/tex] = Pressure inside cylinder

A = Area of pistion

h = Height of piston

Change in pressure is given by

[tex]\Delta P=\frac{F}{A}\\\Rightarrow \Delta P=\frac{mg}{\pi r^2}\\\Rightarrow \Delta P=\frac{23\times 9.81}{\pi 0.1^2}\\\Rightarrow \Delta P=7182.02596\ Pa[/tex]

[tex]P_i-P_o=\Delta P\\\Rightarrow P_i=\Delta P+P_0\\\Rightarrow P_i=7182.02596+101325\\\Rightarrow P_i=108507.02596\ Pa[/tex]

Gas pressure inside the cylinder is 108507.02596 Pa

From ideal gas law we have the relation

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow \frac{Ah_1}{T_1}=\frac{Ah_2}{T_2}\\\Rightarrow \frac{h_1}{T_1}=\frac{h_2}{T_2}\\\Rightarrow h_2=\frac{h_1}{T_1}\times T_2\\\Rightarrow h_2=\frac{0.86}{307+273.15}\times (13+273.15)\\\Rightarrow h_2=0.42418\ m=42.42\ cm[/tex]

The height of the piston at 13°C will by 42.42 cm

(a) The gauge pressure inside the cylinder is 108,500 Pa.

(b) The height of the piston when the temperature is lowered to 13°C is 42.4 cm.

The change in the pressure of the gas inside the cylinder is calculated as follows;

ΔP = F/A

ΔP = mg/πr²

ΔP = (23 x 9.8)/(π x0.1²)

ΔP = 7,174.7 Pa

The gauge pressure inside the cylinder is calculated as follows;

Pi = ΔP + Po

Pi = 7,174.7  +  101325

Pi = 108,500 Pa

The height of the piston is calculated as follows;

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\\frac{Ah_1}{T_1} = \frac{Ah_2}{T_2} \\\\h_2 = \frac{T_2h_1}{T_1} \\\\h_2 = \frac{(13 + 273) \times 0.86}{(307 + 273)} \\\\h_2 = 0.424 \ m[/tex]

h₂ = 42.4 cm

Learn more about ideal gas law here: https://brainly.com/question/12873752

Answer:

The resistance must be 6.67[tex]\Omega[/tex]

Solution:

Resistance, [tex]R_{1} = 20\Omega[/tex]

Resistance, [tex]R_{2} = 10\Omega[/tex]

For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:

Thus in parallel:

[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]

[tex]\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}[/tex]

[tex]R_{eq} = 6.67\ \Omega[/tex]

Answer:

It slows down from 0-31.25s

It speeds up from 31.25-62.507s

Explanation:

If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.

[tex]Y = -16t^2+1000t+7[/tex]

[tex]Y' = -32t+1000 = 0[/tex]  Solving for t:

t = 31.25s

The bullet will slow down in the interval 0-31.25s

Let's now find the moment when it hits ground:

[tex]Y = 0 = -16t^2+1000t+7[/tex]   Solving for t:

t1 = -0.0069s     t2 = 62.507s

The bullet will speed up in the interval 31.25-62.507s

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

[tex]m'=\dfrac{m}{2}[/tex]

[tex]m'=\dfrac{2\times 10^{-6}}{2}\ kg/s[/tex]

m' = 1 x 10⁻⁶ kg/s