For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?

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]

A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.

A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.

Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.

Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:

Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:

(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))

(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)

(1/f) = 0.5 * 0.1667cm⁻¹

(1/f) = 0.08335cm⁻¹

Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:

f ≈ 12cm

Answer:

[tex]\Delta P=1060184.8946\ Pa[/tex]

[tex]P_1=124651.2383\ Pa[/tex]

Explanation:

Given:

We know the Bernoulli's equation of in-compressible flow:

[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)

Cross sectional area of pipe at location 2:

[tex]A_2=\pi \frac{d_2^2}{4}[/tex]

[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]

[tex]A_2=0.0246\ m^2[/tex]

Cross sectional area of pipe at location 1:

[tex]A_1=\pi \frac{d_1^2}{4}[/tex]

[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]

[tex]A_1=0.0115\ m^2[/tex]

Using continuity equation:

[tex]A_1.v_1=A_2.v_2[/tex]

[tex]0.0115\times 9.83=0.0246\times v_2[/tex]

[tex]v_2=4.5953\ m.s^{-1}[/tex]

Now apply continuity eq. on both the locations:

[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]

[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]

[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]

[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)

Now the mass flow rate at location 1:

[tex]\dot{m_1}=\rho\times \dot{V}[/tex]

[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]

[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]

Now pressure at location 1:

[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]

[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]

[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)

The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as

dP = 114 kPa

Question Parameter(s):

Generally, the Bernoulli's equation   is mathematically given as

P + ρ*g*y + v² =pipe  constant

Where

A1*v1 = A2*v2

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

v2 = 3.9 m/s

Therefore

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

dP = 1290*9.8*9.01 + 9.91² - 3.9²

dP = 114 kPa

In conclusion, difference between fluid pressure is

dP = 114 kPa

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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

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]

Answer: The silver pellet will release heat

Explanation:

Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.

Answer:

9.15 km/s

Explanation:

rp = 4.5 x 10^8 km

vp = 18.5 km/s

ra = 9.10 x 10^8 km

va = ?

According to the conservation of angular momentum constant.

Let m be the mass of planet

m x rp x vp = m x ra x va

4.5 x 10^8 x 18.5 = 9.10 x 10^8 x va

va = 9.15 km/s

To calculate the speed of the planet at apastron, we can use Kepler's second law and the given values of rp, vp, and ra. Plugging in the values, we find that the planet is moving at a speed of 0.92 km/s at apastron.

To calculate the speed of the planet at apastron, we can use Kepler's second law, which states that the area swept out by a planet in equal time intervals is constant. At periastron, the planet is moving fastest, so we can use the equation:

A1 = A2

where A1 is the area swept out at periastron and A2 is the area swept out at apastron.

Since the areas are equal, we can set up the following equation:

0.5 * rp * vp = 0.5 * ra * va

where rp is the distance at the periastron, vp is the velocity at the periastron, ra is the distance at the apastron, and VA is the velocity at the apastron. We can rearrange this equation to solve for va:

va = (rp * vp) / ra

Plugging in the given values, we get:

va = (4.50 x 10^8 km * 18.5 km/s) / (9.10 x 10^8 km)

va = 0.92 km/s

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Final answer:

The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.

Explanation:

To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).

The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.

Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.

To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

[tex]n_1sin\theta_1 = n_2 sin\theta_2[/tex]

Where,

n_i = Refractive indices of each material

[tex]\theta_1[/tex] = Angle of incidence

[tex]\theta_2[/tex] = Refraction angle

Our values are given as,

[tex]\theta_1 = 38\°[/tex]

[tex]n_1 = 1[/tex]

[tex]n_2 = 1.4[/tex]

Replacing

[tex]1*sin38 = 1.4*sin\theta_2[/tex]

Re-arrange to find [tex]\theta_2[/tex]

[tex]\theta_2 = sin^{-1} \frac{sin38}{1.4}[/tex]

[tex]\theta_2 = 26.088°[/tex]

Therefore the  angle will the beam make with the normal in the glass is 26°

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

[tex]I = I_0 cos^2\theta[/tex]

Where,

[tex]\theta=[/tex] Angle between the analyzer axis and the polarization axis

[tex]I_0 =[/tex]Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

[tex]I = \frac{I_0}{2}[/tex]

Replacing with our the numerical values we get

[tex]I = \frac{46}{2}[/tex]

[tex]I = 23W/m^2[/tex]

Therefore the  intensity of the light that emerges from the filter is [tex]23W/m^2[/tex]

The correct option can be seen in Option A.

The diagrammatic expression of the question can be seen in the image attached below.

From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.

Mathematically;

[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]

Given that the mass of the woman = 60 kg

[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]

[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]

[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]

Equating both (1) and (2) together, we have:

[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]

Dividing like terms on both side

mass (m) = 60 kg

As such, the correct option can be seen in Option A.

Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.

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When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.

When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).

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.

The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.

The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.

Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.

Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.

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The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.

To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.

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Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii

form i and ii we can write

[tex]v^2= \frac{1}{2} u^2[/tex]

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v

Answer:

Explanation:

The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .

The formula relating to Compton Effect is as follows

[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]

Here [tex]\lambda_i[/tex]  = 3 0 x 10⁻¹¹

For longest [tex]\lambda_f[/tex] θ =180°

[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]

= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]

= .348 nm

For shortest wavelength θ = 0

Putting this value in the given formula

[tex]\lambda_f=\lambda_i[/tex]

[tex]\lambda_f[/tex] = .3 nm

Answer:

-0.31563 m/s

1.16437 m/s

Explanation:

[tex]m_1[/tex] = Mass of girl = 45 kg

[tex]m_2[/tex] = Mass of plank = 166 kg

[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s

[tex]v_2[/tex] = Velocity of the plank relative to ice surface

In this system the linear momentum is conserved

[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]

Velocity of the plank relative to ice surface is -0.31563 m/s

Velocity of the girl relative to the ice surface is

[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]

Answer:

[tex]v_s=27.8m/s[/tex]

Explanation:

If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:

[tex]f=f_0\frac{c}{c+v_s}[/tex]

Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:

[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]

[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]

[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/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: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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Answer:

Current, 7.29 A

Explanation:

It is given that,

Length of the horizontal rod, L = 0.3 m

Magnetic field through a horizontal rod, [tex]B=6.4\times 10^{-2}\ T[/tex]

The magnetic force acting on the rod, F = 0.14 N

Let the current flowing through the rod is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :

[tex]F=ILB\ sin\theta[/tex]

Here, [tex]\theta=90^{\circ}[/tex]

[tex]F=ILB[/tex]

[tex]I=\dfrac{F}{LB}[/tex]

[tex]I=\dfrac{0.14\ N}{0.3\ m\times 6.4\times 10^{-2}\ T}[/tex]

I = 7.29 A

So, the current flowing through the rod 7.29 A.

Final answer:

The current flowing in the rod is approximately 7.292 A when it is subject to a 0.140 N magnetic force within a 6.40 x [tex]10^-^2[/tex] T magnetic field. The calculation uses the formula F = I * L * B, and since the rod is perpendicular to the magnetic field, the angle θ is 90°, which simplifies the calculation.

Explanation:

The question asks for the current flowing through a rod that is experiencing a magnetic force due to an external magnetic field.

The force on a current-carrying conductor in a magnetic field is given by the equation F = I * L * B * sin(θ), where F is the force in newtons, I is the current in amperes, L is the length of the conductor in meters, B is the magnetic field in teslas, and θ is the angle between the direction of the current and the magnetic field. In this case, the rod is perpendicular to the magnetic field, so the angle θ is 90°, making sin(θ) equal to 1.

To find the current I, we rearrange the formula to be I = F / (L * B). Substituting the given values:

Force F = 0.140 N

Length L = 0.300 m

Magnetic field B = 6.40×[tex]10^-^2[/tex] T

The current I can thus be calculated as I = 0.140 N / (0.300 m * 6.40×[tex]10^-^2[/tex] T).

Performing the calculation, I equals approximately 7.292 A.

Answer:

1.6 penny

Explanation:

[tex]V_0[/tex] = Original volume = 1 gal (Assumed)

[tex]\Delta T[/tex] = Change in temperature

[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]

Change in volume is given by

[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]

New volume would be

[tex]1+0.008036=1.008036\ gal[/tex]

The amount of money earned extra would be

[tex]0.008036\times 2=0.016072\ \$[/tex]

1.6 penny more would be earned if the temperature is 32°C

Final answer:

By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.

Explanation:

To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.

The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.

The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.

Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:

ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons

To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:

ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters

Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:

Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies

Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.

Answer:

The gauge pressure is 1511.11 psi.

Explanation:

Given that,

Flow rate = 94 ft³/min

Diameter d₁=3.3 inch

Diameter d₂ = 5.2 inch

Pressure P₁= 15 psi

We need to calculate the pressure on other side

Using Bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2[/tex]

We know that,

[tex]V=Av[/tex]

[tex]v=\dfrac{V}{A}[/tex]

Where, V = volume

v = velocity

A = area

Put the value of v into the formula

[tex]P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2[/tex]

Put the value into the formula

[tex]15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=1525.8\ psi[/tex]

We need to calculate the gauge pressure

Using formula of gauge pressure

[tex]P_{g}=P_{ab}-P_{atm}[/tex]

Put the value into the formula

[tex]P_{g}=1525.8-14.69[/tex]

[tex]P_{g}=1511.11\ psi[/tex]

Hence, The gauge pressure is 1511.11 psi.

Answer:

a) 1.13 10-8 T.  b) +y direction

Explanation:

a)

For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.

At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:

E(z,t) = Emax cos (kz-ωt-φ) +x

B(z,t) = Bmax cos (kz-ωt-φ) +y

In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:

Bmax = Emax/c

So, solving for Bmax:

Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.

b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.

To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.

This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.

a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]

b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]

To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.

Part (a): Angle Between the Reflected Beam and the Surface of the Glass

The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).

So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:

[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]

Since the angle of incidence equals the angle of reflection:

[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]

Therefore, the angle between the reflected part of the beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]

So, the angle between the reflected beam and the surface of the glass is:

[tex]\[ 43.5^\circ \][/tex]

Part (b): Angle Between the Refracted Beam and the Surface of the Glass

For the refracted beam, we need to apply Snell's Law, which is:

[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]

Where:

- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],

- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]

- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],

- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.

Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]

[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]

[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]

Calculating [tex]\(\sin(46.5^\circ)\):[/tex]

[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]

So,

[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]

Now we find [tex]\(\theta_t\):[/tex]

[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]

The angle between the refracted beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]

So, the angle between the refracted beam and the surface of the glass is:

[tex]\[ 64.5^\circ \][/tex]

To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.

The flow rate is defined as

[tex]Q = Av[/tex]

Where,

[tex]Q = Discharge (m^3/s)[/tex]

[tex]A = Area (m^2)[/tex]

v = Average speed (m / s)

And also as

[tex]Q = \frac{V}{t}[/tex]

Where,

V = Volume

t = time

Let's start by finding the total volume according to the given dimensions, that is to say

[tex]V = 5*11*2.4[/tex]

[tex]V = 132m^3[/tex]

The entire cycle must be completed in 50 min = 3000s

In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of

[tex]Q = \frac{V}{t}[/tex]

[tex]Q = \frac{132}{3000}[/tex]

[tex]Q = 0.044m^3/s[/tex]

Using the relationship to find the speed we have to

[tex]Q = Av[/tex]

[tex]v = \frac{Q}{A}[/tex]

Replacing with our values,

[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]

[tex]v = 0.488m/s[/tex]

Therefore the air speed in the duct must be 4.88m/s

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

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:

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