2.5 x 103 kilogram truck with rubber tires moves through a 120 meter radius turn on a dry asphalt surface.

a. Determine the force of friction acting on the truck's tires during the turn.


b. Calculate the maximum speed with which the truck could have made this turn.

(a) At x = 15 m, the tube wall temperature is 432.2 °C, and the quality of the flowing water is 0.23. The heat transfer rate per unit length of the tube is 549.5 W/m.

(b) At a location where single-phase flow of the vapor exists at a mean temperature of 10 bars, the tube wall temperature is 1395.6 °C.

(a) To determine the tube wall temperature and the quality of the flowing water at x=15 m, we need to first calculate the heat transfer rate per unit length of the tube using the given heat flux and tube diameter:

q'' = 7.0 x 10⁴ W/m²

d = 0.1 m

A = pi × \(d^{2/4}\) = 7.85 x 10⁻³ m²

q = q'' × A = 549.5 W/m

Calculate the Reynolds number and the friction factor using the mean velocity and the tube diameter:

\(u_m\) = 0.05 m/s

Re = \(u_m\) × d ÷ nu, where nu is the kinematic viscosity of water at 10 bars.

From the tables, we find nu = 3.3 x 10⁻⁶ m²/s at this pressure.

Re = 1515

Using the Moody chart, we find the friction factor to be f = 0.027.

Now, we can use the energy balance equation to determine the tube wall temperature at x=15 m:

q = m₁ × \(h_{fg\) + m₁ × \(C_{pl\) × (\(T_w-T_4\)) + q'' pid

m₁ = \(rho_l\) × \(Au_m\)

\(rho_l\) = rho₄ = 646.83 kg/m³, the density of saturated liquid water at 10 bars.

\(h_{fg\) = 2230.5 kJ/kg, the enthalpy of vaporization at 10 bars.

\(C_{pl\) = 4.18 kJ/kg.K, the specific heat capacity of liquid water.

T₄ = 179.86 °C, the saturation temperature at 10 bars.

\(T_w\) = 432.2 °C

x = 15.15 m

(b) To determine the tube wall temperature at a location where single phase flow of the vapor exists at a mean temperature at 10 bar, we need to use the energy balance equation again, but this time assuming that the flow is entirely vapor:

q = m₁ × \(C_{pv\)(\(T_w - T_1\))

\(T_w\) = q ÷ (m₁ × \(C_{pv\))

m₁ = \(rho_v\) × \(Au_m\)

\(rho_v\) = 6.09 kg/m³, the density of water vapor at 10 bars and 432.2 °C.

\(C_{pv\) = 1.86 kJ/kg.K, the specific heat capacity of water vapor at 10 bars and 432.2 °C.

\(T_w\) = 1395.6 °C

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The complete question is:

A vertical steel tube carries water at a pressure of 10 bars. Saturated liquid water is pumped into the D = 0.1 m diameter tube at its bottom end (x=0) with a mean velocity of u_m =0.05 m/s. The tube is exposed to combusting pulverized coal, providing a uniform heat flux of q′′ = 7.0 x 10⁴ W/m².

(a) Determine the tube wall temperature and the quality of the flowing water at x=15 m. Assume \(G_{(sf)\) =1.

(b) Determine the tube wall temperature at a location where single-phase flow of the vapor exists at a mean temperature​ of 10 bar.

When the speed of a car is reduced by 12.9 percent, the braking distance decreases by approximately 26.9 percent.

To calculate the percentage decrease in braking distance, we can use the concept of proportionality. Braking distance is directly proportional to the square of the initial speed of the car. Therefore, if the speed is reduced by a certain percentage, the braking distance will decrease by a larger percentage.

Let's assume the initial braking distance is D. When the speed is reduced by 12.9 percent, the new speed becomes 100 percent minus 12.9 percent, which is 87.1 percent of the initial speed. Since the braking distance is directly proportional to the square of the speed, the new braking distance will be (87.1 percent)² of the initial braking distance.

The percentage decrease in braking distance can be calculated as follows:

Percentage decrease = (1 - New distance / Initial distance) * 100

Percentage decrease = (1 - (87.1 percent)²) * 100

Percentage decrease ≈ 1 - 0.7581 ≈ 0.2419 ≈ 24.19 percent

Therefore, the braking distance of the car decreases by approximately 24.19 percent when the speed of the car is reduced by 12.9 percent.

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The average fluid velocity (in m/s) is 3.636 m/s.

The volume of fluid moving past a spot in a given amount of time is measured as the flow rate. Circular and noncircular pipelines both experience fluid flow in daily life.

From the question we have given:

The diameter of the pipe (l) is  3.00 cm or 0.0300 m.

The magnetic field (B) is  0.550 T.

The hall voltage (E) is  60.0 mv or 60 x 10⁻³ v.

  \(v = \frac{E}{Bl}\)

The average fluid velocity can be calculated using the equation E = Blv.

In this instance, the width really corresponds to the diameter:

\(v = \frac{60 \;*\; 10^{-3} \;V}{0.0165\; m} \\\)

\(v = 3.636 \;m/s\)

Thus, the average fluid velocity is 3.636 m/s.

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The orbital period of the satellite circling the Earth at an altitude of 3500 km is 163 minutes

The orbital period for the satellite circling the Earth at an altitude of 3500 km can be obtained as follow:

T² = (4π² / GM) × a³

T² = [(4 × 3.14²) / (6.67×10¯¹¹ × 5.987×10²⁴)] × 9900000³

Take the square root of both sides

T = √[((4 × 3.14²) / (6.67×10¯¹¹ × 5.987×10²⁴)) × 9900000³]

T = 9789.15 s

Divide by 60 to express in minutes

T = 9789.15 / 60

T = 163 minutes

Thus, we can conclude that the orbital period of the satellite is 163 minutes

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Hi there!

We can use the rotational equivalent of Newton's Second Law:

\(\huge\boxed{\Sigma \tau = I \alpha}\)

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\(\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}\)

(A) The wavelength of each of these waves when they are sent through sea water is 0.0126 m and 0.0076 m respectively.

(B) The wavelength of each of these waves when they are sent through freshwater is 0.012 m and 0.0074 m respectively.

(C) The time taken for the echo to return to the ship is 3.97 seconds.

The wavelength of the sound wave in sea water depends on the speed of sound in seawater and frequency of the wave.

The speed of sound in seawater, v = 1,510 m/s

λ = v/f

when the frequency, f = 120 kHz

λ = 1510 / 120,000

λ =  0.0126 m

when the frequency, f = 200 kHz

λ = 1510 / 200,000

λ =  0.0076 m

The speed of sound in freshwater, v = 1481 m/s

when the frequency, f = 120 kHz

λ = 1481 / 120,000

λ =  0.012 m

when the frequency, f = 200 kHz

λ = 1481 / 200,000

λ =  0.0074 m

The time taken for the echo to return is calculated as follows

v = 2d/t

t = 2d/v

t = (2 x 3,000 m) / (1510 m/s)

t = 3.97 s

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B. Wave 1 has a smaller amplitude than wave 3.

Explanation:

The smaller the amplitude is, the more waves it's gonna  have. And you can tell, wave 1 has more waves than wave 3.

The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.

r =T^2/3

r = (11.9)^2/3

r = 5.2 AU

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Answer:

The answer is A. 5.2 AU

Explanation:

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Answer:

Due to intense solar heating near the equator, the warm, moist air is forced up into the atmosphere like a hot air balloon.

Explanation:

Hope this helps you!

Answer:

\(the \: cart's \: velocity \: was : \\ v = \frac{m}{s} = \: \frac{45}{25} = 1.8 \: ms {}^{ - 1} \)

The intensity of sound is inversely proportional to the square of the distance from the source. The intensity of sound at 4 m away from the source is approximately \(0.993W/m^{2}\) (watts per square meter).

The intensity of sound is defined as the power per unit area. According to the inverse square law for sound propagation, the intensity of sound is inversely proportional to the square of the distance from the source. Mathematically, it can be expressed as

\(I=P/(4\pi r^{2})\)

where I is the intensity, P is the power, and r is the distance from the source.

Given that the power of sound at the source outlet is 100 W and the distance from the source is 4 m, we can substitute these values into the formula. Thus, we have

\(I = 100 W / (4\pi (4 m)^2)\)

Simplifying this expression, we find that the intensity of sound at 4 m away from the source is approximately \(0.993W/m^{2}\) (watts per square meter).

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Elastic energy of rubber straps is can be changed to kinetic energy of the capsule which is also changed into gravitational potential energy of a capsule.

Energy can not be created nor destroyed but we could move it from one form to another even though a bit of the energy could be lost as heat during conversion. These are the postulates of the first and second laws of thermodynamics.

As for the energy changes that does take place when there is a capsule ride; elastic energy of rubber straps is can be changed to kinetic energy of the capsule which is also changed into  gravitational potential energy of a capsule.

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Answer:

\(0.361\ \text{m}\)

Explanation:

\(f_s\) = Frequency of source = 375 Hz

\(\Delta f\) = Difference between the maximum and minimum sound frequencies = 2.75 Hz

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

T = Time period = 1.8 s

\(v_m\) = Maximum speed of the microphone

We have the relation

\(\Delta f=2f_s\dfrac{v_m}{v}\\\Rightarrow v_m=\dfrac{\Delta fv}{2f_s}\\\Rightarrow v_m=\dfrac{2.75\times 343}{2\times 375}\\\Rightarrow v_m=1.26\ \text{m/s}\)

Amplitude is given by

\(A=\dfrac{v_mT}{2\pi}\\\Rightarrow A=\dfrac{1.26\times 1.8}{2\pi}\\\Rightarrow A=0.361\ \text{m}\)

The amplitude of the simple harmonic motion is \(0.361\ \text{m}\).

To determine the magnitude of current flowing through a copper wire, we can use the equation that relates current, drift velocity, and the cross-sectional area of the wire.

In Example 20.3 of the textbook, it is stated that the current (I) is given by the equation:

I = n * A * v * q

Where:

I is the current,

n is the number density of charge carriers (electrons in this case),

A is the cross-sectional area of the wire,

v is the drift velocity of the charge carriers,

q is the charge of an electron.

Given that the drift velocity (v) is 1.40 mm/s and the diameter of the wire is 1.700 mm, we can calculate the cross-sectional area (A) of the wire. The diameter of the wire is twice the radius, so the radius (r) is 1.700 mm / 2 = 0.850 mm = 0.850 × 10^(-3) m.

The cross-sectional area (A) of the wire can be calculated using the formula for the area of a circle:

A = π * r^2

Substituting the values, we have:

A = π * (0.850 × 10^(-3))^2

Now we can calculate the current (I) using the known values for the number density (n) and the charge of an electron (q):

I = n * A * v * q

Remember to convert all units to SI units (meters and coulombs) for accurate calculations.

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Answer:

When a substance burns, it produces new substances during a chemical change. Therefore, whether or not a substance is flammable is a chemical property. Knowing which substances are flammable helps you to use them safely. Another chemical property is how compounds react to light

The graph shown in the first option nicely plots the lion's and gazelle's velocities as a function of time, so option A is the correct answer.

Velocity is the rate of change of displacement over time.

As mentioned in the problem of running at a constant speed towards a gazelle with a standing lion as shown above.

So option A is correct.

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The process of heating a cooled cup of coffee from 55°F to 130°F is an example of endothermic reaction. Endothermic reactions absorb energy in the form of heat and result in an increase in temperature. The coffee absorbs the energy from the microwave, causing its temperature to rise.

Heat reactions, also known as thermochemical reactions, are chemical reactions that involve a change in temperature. These reactions can either be endothermic or exothermic. Endothermic reactions absorb heat and result in a decrease in temperature, while exothermic reactions release heat and result in an increase in temperature. Heat reactions play an important role in many industrial processes such as chemical synthesis, power generation, and food preparation. They can also occur in biological systems, where heat energy is used to drive metabolic processes. Understanding heat reactions is essential for developing sustainable energy technologies, controlling environmental pollution, and improving human health.

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Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

The height of the wave is determined by the wind strength and fetch.

The more the strength and the more the fetch size the more will be the height of the wave.

Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.

Answer:

When the circuit switch is off, no electricity will flow and then the circuit is called an open circuit. Electricity will not flow in open circuit.

Explanation:

a) The planar density expression for FCC (100) is 4/a^2.

    The planar density expression for FCC (111) is 2 / [(sqrt(3) / 2) * a^2].

b)  The planar density for the FCC (100) plane is 24.63 atoms/nm^2.

    The planar density for the FCC (111) plane is  12.32  atoms/nm^2.

a) To derive the planar density expression for the FCC (100) and (111) directions in terms of the atomic radius R, we need to consider the arrangement of atoms in these planes.

FCC (100) Plane:

In the FCC crystal structure, there are 4 atoms per unit cell. The (100) plane cuts through the middle of the unit cell, passing through the centers of the atoms at the corners. Since the atoms at the corners are shared with adjacent unit cells, we only count a fraction of these atoms.

For the (100) plane, we have 2 atoms in the plane, located at the corners of the square, and 1/2 atom at each of the 4 face centers. Thus, the total number of atoms in the plane is 2 + (1/2) * 4 = 4 atoms.

The area of the (100) plane is determined by the square formed by the lattice vectors a and a, which gives an area of a^2.

The planar density (PD) is defined as the number of atoms per unit area, so we divide the total number of atoms (4) by the area (a^2):

PD(100) = 4/a^2

FCC (111) Plane:

In the FCC crystal structure, there are 4 atoms per unit cell. The (111) plane passes through the centers of the atoms at the corners and the center of the face. Similarly to the (100) plane, we need to account for the fraction of shared atoms.

For the (111) plane, we have 1 atom in the plane, located at the corner of the equilateral triangle, and 1/3 atom at each of the 3 face centers. Thus, the total number of atoms in the plane is 1 + (1/3) * 3 = 2 atoms.

The area of the (111) plane is determined by the equilateral triangle formed by the lattice vectors a, a, and a, which gives an area of (sqrt(3) / 2) * a^2.

The planar density (PD) is defined as the number of atoms per unit area, so we divide the total number of atoms (2) by the area ((sqrt(3) / 2) * a^2):

PD(111) = 2 / [(sqrt(3) / 2) * a^2]

b) Now, let's compute the planar density values for the FCC (100) and (111) planes using the atomic radius R = 0.143 nm for Aluminum.

For FCC (100) plane:

PD(100) = 4 / a^2

For Aluminum, the lattice constant a is related to the atomic radius R by the formula:

a = 4R / sqrt(2)

Substituting the given value of R = 0.143 nm:

a = 4 * 0.143 nm / sqrt(2) ≈ 0.404 nm

Therefore, the planar density for the FCC (100) plane is:

PD(100) = 4 / (0.404 nm)^2 ≈ 24.63 atoms/nm^2

For FCC (111) plane:

PD(111) = 2 / [(sqrt(3) / 2) * a^2]

Using the calculated value of a = 0.404 nm:

PD(111) = 2 / [(sqrt(3) / 2) * (0.404 nm)^2] ≈ 12.32 atoms/nm^2

Therefore, the planar density for the FCC (111) plane is approximately 12.32 atoms/nm^2

Thus,

a) The planar density expression for FCC (100) is 4/a^2.

    The planar density expression for FCC (111) is 2 / [(sqrt(3) / 2) * a^2].

b)  The planar density for the FCC (100) plane is 24.63 atoms/nm^2.

    The planar density for the FCC (111) plane is  12.32  atoms/nm^2.

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Answer:

C3H6OH + 5O3 = 5CO2 + 6H3O

Explanation:

The energy required to move a charge q across a capacitor with capacitance C and constant voltage V is given by:

E = (1/2)CV^2

Rearranging this formula, we get:

V = sqrt(2E/C)

In this case, the energy required to move a 13.0-mC charge across a 17.0-μF capacitor is 15.2 J. So, we can use this value of energy and the given capacitance to find the voltage across the capacitor:

V = sqrt(2E/C) = sqrt(2 x 15.2 J / 17.0 x 10^-6 F) = 217.3 V

Now that we know the voltage across the capacitor, we can use the formula for capacitance to find the charge on each plate:

C = q/V

Rearranging this formula, we get:

q = CV

Substituting the values of C and V that we found earlier, we get:

q = (17.0 x 10^-6 F) x (217.3 V) = 3.69 x 10^-3 C

Therefore, the charge on each plate of the capacitor is approximately 3.69 milliCoulombs (mC).

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The electrical component that is used for detecting light levels in digital cameras is the light meter.

A light meter is a device used to measure the amount of light. In photography industry, a light meter is used to determine the proper exposure for a photograph.

A digital camera is a camera that captures photographs in digital memory.

Most digital cameras can be grouped into four main types which includes:

Each type of these digital cameras has advantages and disadvantages, and some the types are more expensive than their counterparts.

There are two different kinds of light meters which are:

-An incident light meter measures all the light falling onto a subject. Incident light meters help a camera focus on a subject regardless of how light or dark the surrounding background is.

- Reflective light meters  on the other hand do the opposite by measuring the light reflected by or bouncing off a subject.

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Answer:

Sample Response: The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

A horizontal line, not on the x-axis.

Three sections will lengthen the total by 3 mm; hence, the total length of the two gaps should be equal displacement to or larger than 3 mm, making each gap 1.5 mm or greater.

What does physics mean by displacement?

The term "displacement" refers to a shift in an object's position. It is a vector quantity with a magnitude and direction. The symbol for it is an arrow pointing from the initial location to the ending place.

In physics, how do you express displacement?

Displacement in physics is denoted by the symbol s. The Greek word for "change in" is "delta," which is shaped like a triangle. Spatial position is indicated by the letter "s." It stands for "so"

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Answer:

....

Explanation:

Answer:

What do you mean.

Explanation: