Question 12 (1 point)
[04.01)
What must happen for liquid water to become water vapor? (1 point)
a
Ob
The water molecules must lose energy and stop moving.
The water molecules must exert less pressure on their container.
The water molecules must gain kinetic energy
The water molecules must lose heat and gain volume.
С
Od

Answer:

Frequency and tfcvvbyfgffhhi

Answer:

I think frequency

Explanation:

A house is designed to have passive solar energy features, Brickwork is to be incorporated into the interior of the house to act as a heat absorber. Each brick weighs approximately 1.8 kg. The specific heat of the brick is 0.85 j/g-K. How many bricks must be incorporated into the interior of the hose to provide the same total heat capacity as 1.0 103 gal of water.

The heat capacity of the water can be calculated using the specific heat capacity formula as follows:Q = m * C * ΔTWhere Q is the amount of heat energy, m is the mass of water, C is the specific heat of water, and ΔT is the change in temperature.To find the mass of water, we need to convert the volume from gallons to kilograms.1 gal of water = 3.78541 kg of waterTherefore, 1.0 x 103 gal of water = 3.78541 x 1.0 x 103 = 3.78541 x 103 kg of waterNow, let us calculate the heat capacity of water.Qwater = m * C * ΔTQwater = 3.78541 x 103 * 4.18 * (20 - 10)Qwater = 1.58 x 105 J/KNow, let us calculate the heat capacity of one brick.Qbrick = m * C * ΔTWe are looking for the number of bricks required to have the same heat capacity as 1.0 x 103 gal of water.

Thus, we can equate the two heat capacities as follows:Qbrick = QwaterTherefore,mbrick * Cbrick * ΔT = mwater * Cwater * ΔTmbrick * Cbrick = mwater * Cwatermbrick = (mwater * Cwater) / CbrickSubstituting the values,mbrick = (3.78541 x 103 x 4.18) / (0.85 x 1000)Note that we have converted the specific heat of the brick from J/g-K to J/kg-K. This is because the mass of the brick is given in kilograms.Thus,mbrick = 19.03Therefore, 19 bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.0 x 103 gal of water.Explanation:The mass of water is calculated as:1 gal of water = 3.78541 kg of waterTherefore, 1.0 x 103 gal of water = 3.78541 x 1.0 x 103 = 3.78541 x 103 kg of waterHeat capacity of water is calculated as:Qwater = m * C * ΔTQwater = 3.78541 x 103 * 4.18 * (20 - 10)Qwater = 1.58 x 105 J/KThe heat capacity of one brick is calculated as:Qbrick = m * C * ΔTWe equate the two heat capacities to find the mass of the bricks.mbrick * Cbrick * ΔT = mwater * Cwater * ΔTmbrick * Cbrick = mwater * Cwatermbrick = (mwater * Cwater) / CbrickThe specific heat of the brick is given as 0.85 j/g-K, so we need to convert this to J/kg-K because the mass of the brick is given in kg. Thus,mbrick = (3.78541 x 103 x 4.18) / (0.85 x 1000)Note that mbrick is rounded to two decimal places.Therefore, 19 bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.0 x 103 gal of water.

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A DC motor is powered with Ea = 400V, the armature resistance is Ra = 20, the torque constant of the motor is K+ = 1 Nm/A.

The load torque intercepts the motor torque in two places, only one will be positive, drop you will again need to solve = 0. d the given details are Ea = 400VRa = 20K+ = 1 Nm/A

To find; Gear ratio G required to drive this load at its maximum operating speed.

Formula used: θ=K+ * Ia The motor torque, Tm = K+ * Ia The load torque, TL = J * dw / G The armature current, Ia = (Ea - V) / Ra By equating the motor torque and load torque, we have; K+ * Ia = TL The load torque is positive, so we have;TL = J * dw / G Since the load is operating at maximum speed, d w = 0So, TL = 0 = J * 0 / G

Thus, G is not found in the equation.

So, we can conclude that G does not affect the maximum speed of the motor. The maximum speed is determined by the back EMF, Ea, and the motor constants, K+ and Ra.

So, the gear ratio G is irrelevant in determining the maximum speed of the motor.

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The substituent should be added at different locations on the compounds in electrophilic aromatic substitution are: Phenol: ortho or para position to the -OH group, Benzaldehyde: ortho or para position to the -CHO group.

For each of the compounds that undergo electrophilic aromatic substitution, the substituent should be added to the ring. The location of the substituent, however, differs depending on the specific compound. The location is dependent on the presence of any functional groups already present in the ring.
Phenol: The substituent is added to the ortho or para position to the -OH group.
Benzaldehyde: The substituent is added to the ortho or para position to the -CHO group.
Benzoic acid: The substituent is added to the ortho or para position to the -COOH group.
Bromobenzene: The substituent is added to the ortho or para position to the bromine atom.
Nitrobenzene: The substituent is added to the meta position to the nitro group.
Toluene: The substituent is added to the ortho or para position to the methyl group.

Therefore, when each compound undergoes electrophilic aromatic substitution, the substituent should be added to the ring. The specific location is determined by the presence of any functional groups already present in the ring.

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The acceleration of the car is approximately 0.2625 m/s².

To calculate the acceleration of the car, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

Given:

Initial Velocity (u) = 6.4 m/s (forward)

Final Velocity (v) = 10.6 m/s (forward)

Time (t) = 16 s

Plugging in the values into the formula:

Acceleration = (10.6 m/s - 6.4 m/s) / 16 s

Acceleration = 4.2 m/s / 16 s

Acceleration ≈ 0.2625 m/s²

Therefore, the acceleration of the car is approximately 0.2625 m/s².

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Based on the given information, the most appropriate classification for the state of the volcano would be "Active-Dormant."

The volcano's history of generating multiple lava flows over the past 10,000 years indicates that it has been active in the relatively recent past. The presence of lava flows that have spread over large areas suggests ongoing volcanic activity, although not necessarily in the immediate present.

Additionally, the mention of very little tephra (volcanic ash and fragments) indicates that the volcano is not currently experiencing explosive eruptions. Instead, it has been primarily characterized by effusive eruptions that produce lava flows.

The fact that people in the nearby town remember the last eruption, which caused some forest fires, further supports the classification of the volcano as active-dormant. While the volcano may not be actively erupting at the moment, the memory of the recent eruption and the potential for future volcanic activity suggests that it is still in a state of activity, albeit in a relatively dormant phase.

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The given statement is: True.

When a boulder at rest on top of a hill suddenly rolls down the hill, there is a conversion of potential energy to kinetic energy. Potential energy is the energy possessed by an object due to its position or state, and in this case, the boulder has gravitational potential energy while it is at rest on top of the hill.

As the boulder rolls down, this potential energy is converted into kinetic energy, which is the energy of motion. The boulder gains speed and its kinetic energy increases as it moves downhill, while its potential energy decreases.

This conversion occurs because of the force of gravity acting on the boulder, causing it to accelerate and gain kinetic energy.

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Hi! Scientists use models for a lot of reasons. Primarily to study objects in detail that they wouldn't be able to normally. For example, to study something small like a cell, making a model could be easier. Also, things that are too dangerous to study in real life can be turned into models as well.

To put it shortly, scientists use models to observe things in better detail.

I can include a link to a website if that will help :)

https://study.com/academy/lesson/why-scientists-use-models-simulations.html

Good luck!

Answer:

ndndnskiaidjdjjdndndkkskdldo

If a person gets into the rowboat, the water level at the edge of the pool will rise slightly. This is because the weight of the person adds to the weight of the boat and displaces more water, causing the water level to rise.

The amount of the rise in water level will depend on the weight of the person and the size of the boat. Similarly, if objects are added to the boat, the water level at the edge of the pool will also rise. However, if the boat is empty and just floating in the pool, it will displace a certain amount of water and the water level at the edge of the pool will be at the marked level.

Overall, the water level in the pool is determined by the amount of water displaced by the rowboat and any additional weight that is added to it.

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Answer:

heart, lungs, blood vessels

Explanation:

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

The thickness of asbestos required is 0.122 ft.

The heat flux will be -70 Btu/h ft²

And the temperature of the interface is 633.33 °F.  

(i) the rate of heat loss :

dQ/dt = kAΔT/d

where k = thermal conductivity, A = area, ΔT = temperature gradient, and

d = thickness of insulation.

[dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft²,

k = 0.10 Btu/h ft ℉,

ΔT = 500 °F - 1600 °F = -1100 °F

We have to find the thickness of asbestos that is d.

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft is the thickness required.

(ii) a 3-in thick Kaolin is added to the outside of the asbestos

outside temperature of the asbestos is 250℉,

the heat loss due to the Kaolin is:

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

(iii) temperature at the interface

the total heat flux :

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d'

where  [dQ/dt]/A = -900 Btu/h ft²,

k = 0.10 Btu/h ft ℉  (for asbestos),

k' = 0.07 Btu/h ft ℉  (for Kaolin),

T₁ = 1600 °F and T₃ = 250℉.

-900 = 0.10(T₂ - 1600 °F)/0.122 + 0.07(250℉ - T₂)/0.25

-900 = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

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The final speed of the UFO after experiencing a constant acceleration for 63 seconds is 351.3 m/s.

Acceleration is the rate at which an object increases its speed, velocity or rate of change. It is the rate of change of velocity over time, or the rate of change of the position of an object. Acceleration can be either positive, when the speed of an object is increasing, or negative, when the speed of an object is decreasing.

The final speed of the UFO can be determined by using the equation, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. So, in this case, v = 64 + (21,892.5/63) = 351.3 m/s. Therefore, the final speed of the UFO after experiencing a constant acceleration for 63 seconds is 351.3 m/s.

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From the given hypothesis, Jessica assumes that the two trains have equal mass and initial speeds, and therefore the total initial momentum of the system will be zero.

She also assumes that the net momentum of the system will be conserved, and the final momentum will also be zero, making the two trains come to rest independent of the degree.

However, Josh concludes that Jessica's hypothesis is incorrect. In the case of an inelastic collision, the two trains will be at rest, but not for any degree of inelasticity. In an elastic collision, kinetic energy conservation will not hold if the two trains are at rest after the collision. In this case, Jessica's hypothesis would be incorrect.

Therefore, Jessica's hypothesis is incorrect. The two trains are at rest after the collision only for the completely inelastic case. In the inelastic case, kinetic energy is lost, meaning that at least one of the trains will be moving after the collision. Both trains must have zero velocity after the collision in order to make the final net momentum be zero.

So, Jessica's assumption that the degree of elasticity of the collision does not matter for the final velocity of the trains is not correct.

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(a) The lens is a converging lens.

(b) Its power is +6.06 diopters.

(a) When the sunlight focuses at a point behind the lens, it indicates that the lens is a converging lens. Converging lenses have a thicker center and are capable of converging parallel rays of light to a focal point.

(b) To determine the power of the lens, we can use the lens formula:

1/f = 1/v - 1/u,

where f is the focal length of the lens, v is the image distance, and u is the object distance. In this case, the image distance is given as v = -16.5 cm (negative sign indicates that the image is formed on the opposite side of the object) and the object distance (u) is usually assumed to be large for most practical purposes.

Since we know that the focal length (f) is related to the power (P) of the lens as P = 1/f (in diopters), we can rearrange the lens formula as:

1/f = 1/v,

Substituting the values, we get:

1/f = 1/(-16.5 cm).

Solving for f, we find:

f = -16.5 cm.

Therefore, the power of the lens is +6.06 diopters (since P = 1/f). The positive sign indicates that the lens is a converging lens.

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Answer:

It is larger than the Sun.

Explanation:

Brainliest pls

:3

The centripetal force required to hold the Earth in orbit around the Sun is approximately 3.52 x \(10^{22}\) Newtons.

To calculate the centripetal force required to hold the Earth in orbit around the Sun, you can use the formula:

F = (m * v²) / r

where F is the centripetal force, m is the mass of the Earth, v is the orbital velocity, and r is the distance between the Earth and the Sun. The mass of Earth (m) is approximately 5.97 x \(10^{24}\)kg. The average distance between the Earth and the Sun (r) is approximately 1.496 x \(10^{11}\) meters (also known as 1 astronomical unit). The Earth's orbital velocity (v) is approximately 29,500 meters per second.

Using these values, you can calculate the centripetal force:

F = (5.97 x \(10^{24}\) kg * (29,500 m/s)²) / (1.496 x \(10^{11}\) m)

F ≈ 3.52 x \(10^{22}\) N

So, the centripetal force required to hold the Earth in orbit around the Sun is approximately 3.52 x \(10^{22}\) Newtons.

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Answer:

C. It can increase the chance of asthma attacks

Explanation:

The volumetric strain rate is zero, the vorticity is -1/2μ (∂p/∂x) y, the rate of angular deformation is 1/2μ (∂p/∂x) y, and the shear stress at the plate surface is -1/2 (∂p/∂x) \(h^2\).

Given:

x-velocity: u = 1/2μ (∂p/∂x) (\(y^2 - h^2\))

y-velocity: υ = 0 (no variation in y-direction)

where,

μ = dynamic viscosity

p = pressure

x, y = coordinates

h = distance between the plates

To determine the volumetric strain rate, we can start by considering the continuity equation for incompressible flow, which states that the product of velocity and cross-sectional area is constant:

u × (h-y) = Q/A

where A is the cross-sectional area and Q is the volumetric flow rate.

Taking the derivative of both sides with respect to time and simplifying, we get:

dQ/dt = -u × dA/dy

Since the y-velocity is zero, we have dA/dy = 0, so:

dQ/dt = 0

This means that the volumetric flow rate is constant and there is no change in volume with time. Therefore, the volumetric strain rate is zero.

The velocity vector's curl is used to define the vorticity.

ω = ∇ x v

where ∇ is the del operator. For two-dimensional flow, the vorticity is a scalar and can be expressed as:

ω = (∂υ/∂x) - (∂u/∂y)

Substituting the given values for u and υ, we get

ω = 0 - (∂/∂y)[1/2μ (∂p/∂x) (\(y^2 - h^2\))]

Simplifying and integrating with respect to y, we get:

ω = -1/2μ (∂p/∂x) y

The formula for angular deformation rate is:

D = (∂u/∂y + ∂υ/∂x) = ∂u/∂y

Substituting the given value for u, we get:

D = (∂/∂y)[1/2μ (∂p/∂x) \((y^2 - h^2)\)]

Simplifying and integrating with respect to y, we get:

D = 1/2μ (∂p/∂x) y

To find the shear stress at the plate surface, we can use the following relation:

τ = μ (∂u/∂y)|y=h

Substituting the given value for u, we get:

τ = μ (∂/∂y)[1/2μ (∂p/∂x) (\(y^2 - h^2\))]|y=h

Simplifying, we get:

τ = -1/2 (∂p/∂x) \(h^2\)

Therefore, the volumetric strain rate is zero, the vorticity is -1/2μ (∂p/∂x) y, the rate of angular deformation is 1/2μ (∂p/∂x) y, and the shear stress at the plate surface is -1/2 (∂p/∂x) \(h^2\).

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Women:

They were excluded due to social expectations, they were expected to take a female role that entailed learning domestic skills, getting married, having children.

Minorities:

They were excluded because of limited academic opportunities.

Women were excluded from psychology due to social expectations.

Minorities were excluded from psychology because of limited academic opportunities.

Psychology is the scientific study related to mind and behavior. Psychology involves the study of conscious and unconscious mind, including feelings and emotions.

Women were not allowed or accepted to refer to the Psychologist because they were considered to be always happy with the family and being busy with the house chores and take responsibility of all the family members.

Women were expected to take a female role that include learning domestic skills, getting married and having children.

For minority, they are not allowed to take hold of post or admission in academic institutions. They have very limited amount of opportunities in educational background.

Thus, the reasons are explained for the exemption of women and minority from psychology.

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Answer:

the answer to the first one is B and the second one is c I hope I got them right

The electric field E for the Schottky diode is approximately 3.81 x 10^5 V/m.

To calculate the electric field E, we can use the formula:

E = V / d,

where V is the applied voltage and d is the distance from the interface.

Given:

V = -5 V (negative sign indicates reverse bias)

d = 1.2 μm = 1.2 x 10^-6 m

Substituting these values into the formula, we get:

E = (-5 V) / (1.2 x 10^-6 m)

≈ -4.17 x 10^6 V/m

Since the electric field is a vector quantity and its magnitude is always positive, we take the absolute value of the result:

|E| ≈ 4.17 x 10^6 V/m

≈ 3.81 x 10^5 V/m (rounded to two significant figures)

The electric field for the Schottky diode Au-n-Si at V = -5 V and a distance of 1.2 μm from the interface is approximately 3.81 x 10^5 V/m.

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Answer:

-( 4 *e^4 /u )/u^2

Explanation:

d/du(e^4/u)= -( 4 *e^4 /u )/u^2

After differentiating the function we get,

\(\large f'(u)=\frac{-4}{u^2}*e^\frac{4}{u}\).

What is differentiation?

Apart from integration, differentiation is one of the two key ideas of calculus. A technique for determining a function's derivative is differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of change based on one of its variables. The most typical illustration is velocity, which is the rate at which a distance changes in relation to time. Finding an antiderivative is the opposite of differentiation.

The rate of change of x with respect to y is given by dy/dx if x and y are two variables. The universal representation of a function's derivative is given by the equation f'(x) = dy/dx, where y = f(x) is any function.

\(f(u)=e^\frac{4}{u}\)

\(\large f'(u)=e^\frac{4}{u}*\frac{d(4/u)}{du}\)

\(\large f'(u)=\frac{-4}{u^2}*e^\frac{4}{u}\)

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The kinetic energy of the ball is 100j

100j of work would be required to stop the ball.

Kinetic energy is a form of energy that an particle has by reason of it's motion.if we accelarate a object we must apply a force.

Here the mass (m) is 0.5 kg

It's velocity (v) is 20 m/s

The kinectic energy is

KE = 1/2 mv^2 = 1/2(0.5)20.20= 100 J

Again the workdone to stop it is calculated by work energy theorem

Change in KE = workdone

Or 100 J - 0 J or 100 J needed to stop the ball.

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The isolines are always opposite to the electric field vectors in the region around q1q1 and q2q2.

isolines - The lines known as isolines link points with equal values.

To draw Isolines of Electric Potential :

The isolines are always opposite to the electric field vectors in the region around q1q1 and q2q2.

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a) The total resistance of the circuit in which three 45 ohm lightbulbs and three 75 ohm light bulbs connected in series is 360 ohm.

b) The total resistance of the circuit if all six are wired in parallel is 28.57 ohm.

a) In order to find the series total resistance, we need to add all resistances,

Req = R₁ + R₂ + R₃ + R₄ + R₅ + R₆ = 45 + 45 + 45 + 75 + 75 + 75 = 135+ 225 = 360 ohm

b) Now, since the resistances are in parallel, we need to use the expression below,

1/Req = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅ + 1/R₆

Putting in the values,

1/Req = 1/45 + 1/45 + 1/45 + 1/75 + 1/75 + 1/75

1/Req = 1/0.035

Req = 28.57 ohm

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Answer:

86400 seconds is equal to 1 day.

Explanation:

When 1.00 X 10²⁰ electrons flow through a cross-section of a 4.50 mm diameter iron wire in 5.00 s and the electron density of iron is n = 8.5 X 10²⁸. The electron drift speed is approximately 3.26 × 10⁻⁴ m/s.

To find the electron drift speed, we need to use the formula:
Drift speed (v) = Current (I) / (Charge of an electron (e) × Electron density (n) × Cross-sectional area (A))

First, we'll find the current.

Current (I) = Number of electrons / Time
I = (1.00 × 10²⁰ electrons) / (5.00 s)

= 2.00 × 10¹⁹ electrons/s

Next, we'll find the cross-sectional area.

A = π × (Diameter / 2)²
A = π × (4.50 mm / 2)² = π × (2.25 mm)²

= π × 5.0625 mm²
We'll convert the area to m²:

A = π × 5.0625 × 10⁻⁶ m²

Now, we'll use the formula for drift speed:
v = (2.00 × 10¹⁹ electrons/s) / (1.6 × 10⁻¹⁹ C/electron × 8.5 × 10²⁸ electrons/m³ × π × 5.0625 × 10⁻⁶ m²)
v ≈ 2.00 × 10¹⁹ / (1.36 × 10¹⁰ × π × 5.0625 × 1010⁻⁶)
v ≈ 3.26 × 10⁻⁴ m/s

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(A) The pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.
(B) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow

(a) The pressure transients at the mid-point of the pipeline can be calculated using the water hammer equation. Water hammer refers to the sudden changes in pressure and flow rate that occur when there are rapid variations in fluid flow. The equation is given by:

ΔP = (ρ × ΔV × c) / A

Where:

ΔP = Pressure change

ρ = Density of water

ΔV = Change in velocity

c = Wave speed

A = Cross-sectional area of the pipe

First, let's calculate the change in velocity:

ΔV = Q / A

Q = Flow rate = 0.12 m/s

A = π × ((d/2)^2 - ((d-2t)/2)^2)

d = Internal diameter of the pipe = 300 mm = 0.3 m

t = Pipe wall thickness = 5 mm = 0.005 m

Substituting the values:

A = π × ((0.3/2)^2 - ((0.3-2(0.005))/2)^2

A = π × (0.15^2 - 0.1495^2) = 0.0707 m^2

ΔV = 0.12 / 0.0707 = 1.696 m/s

Next, let's calculate the wave speed:

c = √(E / ρ)

E = Young's modulus of steel = 210x10^9 Pa

ρ = Density of water = 1000 kg/m^3

c = √(210x10^9 / 1000) = 4585.9 m/s

Finally, substituting the values into the water hammer equation:

ΔP = (1000 × 1.696 × 4585.9) / 0.0707 = 1,208,277 Pa

Therefore, the pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.

(b) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow. As water moves through the pipe, it encounters frictional forces between the water and the pipe wall. This friction causes a pressure drop along the length of the pipeline.

The presence of friction results in a higher effective wave speed, which affects the calculation of pressure transients. The actual wave speed in the presence of friction can be higher than the wave speed calculated using the Young's modulus of steel alone. This higher effective wave speed leads to a reduced pressure rise during the transient event.

Learn more about friction here:
https://brainly.com/question/13000653

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Answer: D just took the test

Explanation:

Answer:

D

Explanation: