A 5L tank of water starts at 20C before a 10cm cube of mild steel at 50C is dropped into the water. When the tank’s contents come to thermal equilibrium (assume an adiabatic exterior), what is the temperature of the steel cube?
20.3°C
22.8°C
24.8°C
27.3°C
31.6°C

The gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa. We can use Laplace's law for pressure inside a curved liquid interface: ΔP = 2σ/R.

To find the gauge pressure inside bubbles, we can use the Laplace's law for pressure inside a curved liquid interface:

ΔP = 2σ/R

where ΔP is the pressure difference across the curved interface, σ is the surface tension of water, and R is the radius of the bubble.

Given:

Surface tension of water (σ) = 0.070 N/m

Radius of the bubble (R) = 10 μm = 10 × 10^(-6) m

Substituting the values into the equation, we have:

ΔP = 2σ/R

= 2 * 0.070 / (10 × 10^(-6))

= 14,000 N/m²

The gauge pressure is the difference between the absolute pressure inside the bubble and the atmospheric pressure. Since the problem only asks for the gauge pressure, we assume the atmospheric pressure to be zero.

Therefore, the gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa.

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a) The speed of the yo-yo at the top of the circular path is given by:

v² = gr [r + h]

Where, v = velocity

g = acceleration due to gravity

r = radius

h = height

Here, r = 0.30m (length of the string)

h = r

  = 0.30m (height of the circle at the top)

g = 9.8 m/s²

Putting these values in the above equation,

v = √(9.8 × 0.6) = 3.4 m/s

The free-body diagram for the yo-yo at the top of the circular path is given below:

b) The speed of the yo-yo at the bottom of the circular path is given by:

v² = gr [r - h]

Where, v = velocity

g = acceleration due to gravity

r = radius

h = height

Here, r = 0.30m (length of the string)

h = r

  = 0.30m (height of the circle at the bottom)

g = 9.8 m/s²

Putting these values in the above equation,

v = √(9.8 × 0.0)

  = 0 m/s

The free-body diagram for the yo-yo at the bottom of the circular path is given below:

c) The maximum tension in the string occurs when the yo-yo is at the bottom of the circular path. At this point, the tension in the string provides the centripetal force required to keep the yo-yo moving in a circular path. The maximum tension in the string is given by:

T = mg + mv² / r

Where, T = tension in the string

m = mass of the yo-yo

v = velocity

r = radius

g = acceleration due to gravity

At the slowest speed, v = 0 and hence, the maximum tension in the string is given by:

T = mg + 0

  = mg

  = 0.050 × 9.8

  = 0.49 N

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The medium for propagation of sound refers to the substance through which sound waves can travel. Sound waves can propagate through various mediums, including gases, liquids, and solids.Sound waves require a medium to travel through, as they are a type of mechanical wave.

The medium through which sound waves travel can have an impact on the speed, direction, and intensity of the sound waves.

Gases: In gases, sound waves can travel through the movement of molecules. These molecules collide with each other, transferring kinetic energy and producing pressure waves that can be detected as sound.

Liquids: In liquids, sound waves can travel through the vibration of molecules. Liquids are more dense than gases, meaning that sound waves can travel faster through liquids. The vibration of molecules transfers energy, producing waves that can be detected as sound.

Solids: In solids, sound waves can travel through the movement of particles. Solids are the most dense medium for sound waves, allowing them to travel even faster than in liquids.

When sound waves move through a solid, the particles move back and forth in the direction of the wave, transmitting energy that produces sound waves.

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The approximate magnitude of the electric field at point Q(<0,2,0>) is 0.015 N/C. The correct option is (B) 0.13 N/C.

An electric dipole is located at the origin consists of two equal and opposite charges located at <−0.01,0,0>m and <0.01,0,0>m.

The electric field at <0,1,0>m has a magnitude of 1 N/C.

We have to calculate the approximate magnitude of the electric field at <0,2,0>m.

Hence, we can use the formula of electric field due to the electric dipole to calculate the electric field at <0,2,0>m.

Electric field due to an electric dipole is given as

E = 1 / 4πε₀ * p / r³

Where, E is the electric field at a point p is the magnitude of electric dipoler is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space

Putting the values in the above formula, we get

E = 1 / 4πε₀ * 2q * d / r³Where,2q is the magnitude of electric dipoled is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space

Thus, the distance of point P(<0,1,0>) from the midpoint of the dipole is

r = √(0.01)² + 1²

r = √(0.0001 + 1)

≈ √(1)

= 1 m

And the distance of point Q(<0,2,0>) from the midpoint of the dipole is

r' = √(0.01)² + 2²r'

= √(0.0001 + 4)

≈ √(4)

= 2 m

We know that the magnitude of electric dipole (p) is given by

p = 2qa

Where, q is the magnitude of the charge and a is the distance between the two charges

Putting the values of q and a in the above formula, we get

p = 2 * 1 * 0.01

p = 0.02 C-m

Thus, the electric field at point P(<0,1,0>) is given by

E = 1 / 4πε₀ * p / r³Putting the values in the above formula, we get

E = 1 / 4πε₀ * 0.02 / 1³

E = 1 / 4πε₀ * 0.02

E = 0.14 N/C

Similarly, the electric field at point Q(<0,2,0>) is given by

E' = 1 / 4πε₀ * p / r'³

Putting the values in the above formula, we get

E' = 1 / 4πε₀ * 0.02 / 2³

E' = 1 / 4πε₀ * 0.02 / 8

E' = 1 / 4πε₀ * 0.0025

E' = 0.015 N/C

The correct option is (B) 0.13 N/C.

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It takes approximately 19.21 seconds for the robot to accelerate the equipment over a horizontal displacement of 140 m.

To determine the time it takes for the robot to accelerate the equipment, we can use the kinematic equation:

v² = u² + 2as

Where:

v is the final velocity

u is the initial velocity (which is 0 m/s since the equipment starts from rest)

a is the acceleration

s is the displacement

In this case, we need to solve for time (t). Rearranging the equation, we have:

t = (v - u) / a

Since the equipment starts from rest (u = 0 m/s), the equation simplifies to:

t = v / a

Substituting the given values:

t = 140 m / (7.29 m/s²)

Calculating:

t ≈ 19.21 seconds

Therefore, it takes approximately 19.21 seconds for the robot to accelerate the equipment over a horizontal displacement of 140 m.

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To calculate the minimum energy required to accelerate the spaceship, we can use Einstein's mass-energy equivalence principle, [tex]\(E = mc^2\)[/tex], where m is the mass and c is the speed of light.

[tex]\[ \text{Kinetic Energy} = E_f - E_i \][/tex]

Given values:

Mass of spaceship (m) = [tex]\(2.35 \times 10^6\)[/tex]kg

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

Final speed ([tex]\(v_f\)[/tex]) = [tex]\(0.850c\)[/tex]

Calculate the final energy ([tex]\(E_f\)[/tex]):

[tex]\[ E_f = mc^2 = (2.35 \times 10^6 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \\\\= 6.615 \times 10^{23} \, \text{J} \][/tex]

The initial energy ([tex]\(E_i\)[/tex]) is the rest energy of the spaceship, which can be calculated using the rest mass-energy equivalence:

[tex]\[ E_i = mc^2 \\\\= (2.35 \times 10^6 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \\\\= 2.115 \times 10^{17} \, \text{J} \][/tex]

Substitute the values to find the kinetic energy required:

[tex]\[ \text{Kinetic Energy} = E_f - E_i \\\\= (6.615 \times 10^{23} \, \text{J}) - (2.115 \times 10^{17} \, \text{J})\\\\ = 6.613 \times 10^{23} \, \text{J} \][/tex]

Part (b): Fuel Required

To find the amount of fuel required, we need to calculate the mass equivalent of the energy required using the mass-energy equivalence ([tex]\(E = mc^2\)[/tex]) and then divide it by the rest energy of the fuel:

[tex]\[ \text{Fuel Mass} = \dfrac{\text{Kinetic Energy}}{c^2} \][/tex]

[tex]\[ \text{Fuel Mass} = \dfrac{2.115 \times 10^{17} \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2} \\\\= 2.35 \times 10^{-10} \, \text{kg} \][/tex]

Thus, it would take approximately [tex]\(2.35 \times 10^{-10}\)[/tex] kg of fuel to provide the energy required for the spaceship's acceleration.

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The rotational inertia of a solid iron disk of mass 37.0 kg, with a thickness of 5.00 cm and radius of 19.0 cm, about an axis through its center and perpendicular to it is 0.6674 kg⋅m².

Rotational inertia is the resistance of a rotating object to any change in its rotational motion. The measurement of an object's rotational inertia is known as the moment of inertia. It is calculated by multiplying the mass of an object by the square of the distance from the axis of rotation to the object's center of mass.

Rotational inertia is important in many fields, including engineering, physics, and sports. Understanding the moment of inertia of an object allows for more efficient and accurate designs of various mechanical systems.

To find the rotational inertia of a solid iron disk about an axis through its center and perpendicular to it, we can use the formula for the rotational inertia of a solid disk:

I = (1/2) * m * r²

Where:

I is the rotational inertia (also known as the moment of inertia),

m is the mass of the disk, and

r is the radius of the disk.

In this case, the mass of the disk is given as 37.0 kg and the radius is 19.0 cm (which is 0.19 m).

Plugging these values into the formula, we have:

I = (1/2) * 37.0 kg * (0.19 m)²

Calculating this expression:

I = 0.5 * 37.0 kg * (0.19 m)²

I = 0.5 * 37.0 kg * 0.0361 m²

I = 0.5 * 1.3347 kg⋅m²

I ≈ 0.6674 kg⋅m²

Therefore, the rotational inertia of the solid iron disk about an axis through its center and perpendicular to it is approximately 0.6674 kg⋅m².

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The siren of an ambulance appears louder and higher in pitch as it moves toward you due to the Doppler effect. This effect is caused by the relative motion between the source of sound (ambulance) and the observer (you), resulting in a change in perceived frequency. As the ambulance moves away, the sound becomes softer and lower in pitch.

The Doppler effect is the change in frequency or pitch of a sound wave due to the relative motion between the source of the sound and the observer. When the ambulance approaches you, its motion compresses the sound waves it emits, causing the wavelength to shorten and the frequency to increase. This increase in frequency makes the sound appear higher in pitch.

As the ambulance moves away from you, the sound waves are stretched out, resulting in a longer wavelength and a decrease in frequency. The decrease in frequency makes the sound appear lower in pitch.

The perceived change in loudness is related to the intensity of the sound waves. As the ambulance approaches, the sound waves are compressed, leading to a more concentrated and intense sound, which makes it appear louder. Conversely, as the ambulance moves away, the sound waves spread out, causing a decrease in intensity and perceived loudness.

Therefore, the combination of the Doppler effect and the change in intensity results in the siren of an ambulance appearing louder and higher in pitch as it approaches and softer and lower in pitch as it moves away from an observer.

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When the natural frequency of a system is 50 Hz, we know that: [tex]$$\omega_n = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s}$$[/tex]

The expression for displacement of a mass on a spring is given by:

[tex]$$x(t) = A\cos (\omega_n t + \phi)$$[/tex]

where A and [tex]$\phi$[/tex] are constants determined by the initial conditions.

To find A and we use the initial conditions.

We know that at

t = 0,

displacement is 0.003m and velocity is 1.0m/s.

[tex]$$\begin{aligned} x(0) &= A\cos \phi = 0.003 \\ \frac{dx}{dt} \bigg|_{t=0} &= -\omega_n A\sin \phi = 1.0 \end{aligned}$$[/tex]

Dividing the second equation by the first, we get:

[tex]$$-\omega_n \tan \phi = \frac{1.0}{0.003}$$$$\tan \phi = - \frac{1}{300 \pi}$$[/tex]

which gives us .

Then we can use the first equation to get A,

which is the amplitude of the motion.

We can also express displacement as a sum of cosine and sine functions.

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Given that the position of a particle is expression as

r= 2t i + t²j+ t³ k,

where r is in meters and t in seconds.  We need to find the scalar tangential components of the acceleration at t=1s and scalar normal components of the acceleration at t = 18.

a) Scalar tangential components of the acceleration at t=1s:We know that, Velocity of the particle is given by the differentiation of the given position of the particle.

r = 2ti + t²j + t³k

Differentiating r with respect to time t, we get

v = dr/dt = 2i + 2tj + 3t²k

Differentiating v with respect to time t, we get

a = dv/dt = 0i + 2j + 6tkAt t = 1s

The acceleration of the particle is given by,Substituting t = 1s in the above equation, we get

a = 0i + 2j + 6k

Therefore, the scalar tangential components of the acceleration at t=1s is given by the dot product of the acceleration vector and the unit tangent vector at t=1s. The unit tangent vector at t=1s is given by,The magnitude of the acceleration is,So, the scalar tangential components of the acceleration at t=1s is given by

aT = a . T= (2.449j + 0.588k).(0.554i + 0.832j)= 1.358

b) Scalar normal components of the acceleration at t = 18:

We know that, Velocity of the particle is given by the differentiation of the given position of the particle.

r = 2ti + t²j + t³k

Differentiating r with respect to time t, we get,

v = dr/dt = 2i + 2tj + 3t²k

Differentiating v with respect to time t, we get

a = dv/dt = 0i + 2j + 6tkAt t = 18s,

The acceleration of the particle is given by,Substituting t = 18s in the above equation, we get

a = 0i + 2j + 6(18)k= 0i + 2j + 108k

Therefore, the scalar normal components of the acceleration at t = 18 is given by the dot product of the acceleration vector and the unit normal vector at t = 18. The unit normal vector at t=18 is given by,The magnitude of the acceleration is,So, the scalar normal components of the acceleration at t = 18 is given by

aT = a . N= (1.928j + 0.296k).(0.830i - 0.558j)= -0.515

Therefore, the scalar normal components of the acceleration at t = 18 is -0.515.

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To determine which light is out on a string of Christmas lights, you can follow these steps  are  Ensure Safety,   Inspect the Bulbs, Replace Bulbs,Check the Light Set,    Wiggle and Inspect, Use a Light Tester.

The following steps are :

By systematically inspecting and replacing bulbs, checking for loose connections, and utilizing a light tester if needed, you can identify and replace the faulty light, allowing your Christmas lights to shine brightly once again.

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The horizontal component of the initial velocity of the ball is 17.8cos(52°) = 10.6m/s and the vertical component is 17.8sin(52°) = 14.0m/s.

When the ball reaches its maximum height, its vertical component of velocity is 0 (at the highest point, the ball has no more upward velocity), so using the formula

v = u + at,

where v is the final velocity,

u is the initial velocity,

a is the acceleration due to gravity and t is the time taken to reach the highest point of the ball's trajectory. We can find t as u = 14.0m/s,

a = -9.8m/s² (negative due to gravity), and

v = 0:0 = 14.0 + (-9.8)t=> t = 1.43 seconds

The time taken for the ball to reach the ground from its highest point is equal to the time it takes for the ball to reach that highest point.

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In a two-slit experiment, when monochromatic coherent light passes through a pair of slits, an interference pattern is formed on a screen located at a certain distance away from the slits. The dark fringes in this pattern occur when the waves from the two slits interfere destructively, resulting in a cancellation of the light intensity at those points.

To find the angle at which the second dark fringe occurs in the given scenario, we can use the formula for the position of dark fringes in a two-slit experiment:

y = (m * λ * L) / d

where:

y is the distance from the centerline to the fringe,

m is the order of the fringe (m = 1 for the first dark fringe, m = 2 for the second dark fringe, and so on),

λ is the wavelength of light,

L is the distance between the slits and the screen, and

d is the separation between the slits.

Given:

λ = 600 nm = 600 * 10^(-9) m

d = 2.20 * 10^(-5) m

m = 2

L is not given.

Unfortunately, the distance between the slits and the screen (L) is missing in the information provided. Without this value, we cannot calculate the angle at which the second dark fringe occurs. Therefore, the correct answer cannot be determined with the given information.

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In order to find out the power consumed machining the material and the coefficient of friction between the chip and the tool using graphical method with given conditions, we need to follow the steps given below.

Step 1: Calculate the cutting velocity vC:Given, cutting speed = 2.5 m/secDiameter of workpiece, D = 120 mmWe know that, cutting velocity vC = (πDN)/1000where, N = rotational speed in revolutions per minute= (1000 x cutting speed) / (πD)Putting the given values in the above formula,[tex]vC = (π x 120 x 1000 x 2.5) / (1000 x π)= 300 m/min , the cutting velocity vC is 300 m/min.[/tex]

Step 2: Calculate the chip thickness (h)The cutting ratio r is given asr = (t - h) / hwhere, t = depth of cut = 2 mmPutting the given values in the above formula,0.555 = (2 - h) / hh = (2 / 1.555)= 1.287 mm, the chip thickness h is 1.287 mm.

Step 3: Calculate the shear angle (φ):We know that, tan φ = (fcosα - tsinα) / (fsinα + tcosα)where, f = cutting force = 900 Nt = thrust force = 600 Nα = tool rake angle = 12° Putting the given values in the above formula,

[tex]tan φ = (900cos12 - 600sin12) / (900sin12 + 600cos12)= 0.2268, the shear angle φ is 12.56°.[/tex]

Step 4: Calculate the shear strain rate:Given, cutting speed = 2.5 m/sec Diameter of workpiece, D = 120 mmThe cutting velocity vC = 300 m/minChip thickness h = 1.287 mm We know that,γ = vC / (h x f) Putting the given values in the above formula,γ = (300 x 10^-3) / (1.287 x 900)= 0.2599 x 10^-3/sec  , the shear strain rate γ is 0.2599 x 10^-3/sec.

Step 5: Calculate the coefficient of friction:We know that,γ = (πD^2)/4 x V x k x cos φwhere, k = coefficient of friction Putting the given values in the above formula,k = γ / [(πD^2)/4 x V x cos φ] Putting the given values in the above formula,[tex]k = (0.2599 x 10^-3)/ [(π x 120^2)/4 x 300 x cos12.56]= 0.33,[/tex] the coefficient of friction is 0.33.

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