Fluid flows over a smooth cylinder. The diameter of the cylinder is D and the length normal to the flow direction is L. The drag coefficient is defined as: The drag coefficient is essentially constant with a value of 1.1 in the range of Reynolds numbers of 103 to 105. In this range, at a velocity of 2 m/s the drag force is 3 N. When the velocity is doubled to 4 m/s the drag force is:

Answer:

Ec=53.7×10⁹N/m² =53.7Gpa

Explanation:

To calculate the modulus of elasticity in the longitudinal direction.  This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac

[tex]Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2}  }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa[/tex]

Final answer:

The modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.

Explanation:

To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula:

E = (stress sustained by the fiber phase)/(longitudinal strain)

Given that the stress sustained by the fiber phase is 156 MPa and the total longitudinal strain is 1.25 x 10^-3, we can plug in these values to calculate the modulus of elasticity:

E = 156 MPa / (1.25 x 10^-3) = 124,800 MPa

Therefore, the modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.

Answer:

Consider the following calculations

Explanation:

a )  velocity of the glass is v = distance / time

= 5 X 10-3 / 7 X 365.24 X 24 X 60 X 60

v = 2.263 X 10-11 m/sec

the speed of the light in vaccum is C

C = 3 X 108 m/sec

n = C / v

n = 3 X 108 / 2.263 X 10-11

n = 1.32567 X 1019

b )  given is 40 km/hr

= 40 X 103 / 60 X 60

= 11.11 m/sec

n = C / v

n = 3 X 108 / 11.11

n = 27002700.27

The index of refraction for the 5.00 mm thick slow glass taking light 7.00 years to pass through is approximately 1.33 x 10^19. For a Bose-Einstein condensate where light travels at 40.0 km/hr, the effective index of refraction is about 2.70 x 10^7.

In Bob Shaw's short story 'The Light of Other Days', a fictional material called slow glass is described, which delays the passage of light. To calculate the index of refraction of a 5.00 mm thick piece of slow glass where light takes 7.00 years to pass through, we can use the formula n = c/v, where c is the speed of light in a vacuum (3.00 x 108 m/s), and v is the speed of light through the material.

To find v, we can calculate the total distance light travels in 7.00 years and divide it by the time it takes to travel through the slow glass. Since the slow glass is 5.00 mm thick, which is equivalent to 5.00 x 10-3 m, and one year is 365.24 days, we calculate the speed as follows:

v = distance/time = 5.00 x 10-3 m / (7.00 years x 365.24 days/year x 24 hours/day x 3600 seconds/hour) = 5.00 x 10-3 m / 220,937,280 seconds ≈ 2.263 x 10-11 m/s.

Then, the index of refraction, n, can be calculated as n = c/v ≈ 3.00 x 108 m/s / 2.263 x 10-11 m/s ≈ 1.33 x 1019.

For Lene Hau's experiment using a Bose-Einstein condensate with a light speed of 40.0 km/hr, the index of refraction can also be calculated using n = c/v. Converting 40.0 km/hr to m/s:

v = 40.0 km/hr x (1000 m/km) / (3600 s/hr) = 11.1 m/s.

Using this value for v, we calculate n as n = c/v ≈ 3.00 x 108 m/s / 11.1 m/s ≈ 2.70 x 107.

Answer:

0.358g

Explanation:

Density of Helium = 0.179g/L

ρ=m/v

m=ρv

when the volume was 2L

m1= 0.179*2

m1=0.358g

when the volume increased to 4L

m2= 0.179*4

m2=0.716g

gram of helium added = 0.716g-0.358g

=0.358g

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

Given: W = 12 J

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC  = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

The specific heat of the unknown metal can be determined from the equilibrium of heat transfer in the system. The heat lost by the hot substances is equal to the heat gained by the cooler substances. Solving for the specific heat of the unknown substance involves calculating the heat gained and lost and equating their values.

The specific heat of a substance is a measure of the amount of heat energy required to raise the temperature of a certain mass of the substance by a certain amount. In this case, we're solving for the specific heat (c) of an unknown substance. As the system is in thermal equilibrium, the heat lost by hot substances (copper and unknown metal) is equal to the heat gained by the cooler substances (water and the calorimeter).

The specific heat (c) of the unknown substance can therefore be determined by setting the heat gained (Q_gained = m*c*ΔT) by the cooler substances equal to the heat lost (Q_lost = m*c*ΔT) by the hot substances and solving for the specific heat (c) of the unknown substance. Given that ΔT is the change in temperature, m is the mass, and c is the specific heat, and using the specific heat values for water, aluminum, and copper.

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Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

[tex]E=-\frac{d}{dt}(BACOS\alpha )\\[/tex]

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using [tex]\pi r^{2} \\[/tex] where r is the radius of 5cm or 0.05m

[tex]A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\[/tex]

since we no that the angle is at [tex]0^{0}[/tex]

we determine the magnitude of the magnetic filed

[tex]B=0.5e^{-t} \\t=2s[/tex]

[tex]E=-(0.5e^{-2} * 0.00785)[/tex]

[tex] E=-0.000532v\\[/tex]

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

[tex]V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A[/tex]

[tex]I=2.6mA[/tex]

The magnitude of the induced current in the coil at t = 2s in the given scenario is 2.4 mA. This is calculated using Faraday's law of electromagnetic induction and Ohm's law.

To find the magnitude of the current induced in the coil, we need to consider Faraday's law of electromagnetic induction. This law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

In this situation, we have: B = 0.5 e-0.2t T, and the time derivative of the magnetic field is dB/dt = -0.1 e-0.2t T/s. The area A of the coil is πr²= π(0.05)² m². The induced emf (ε) equals -A dB/dt. Thus, we have ε = -π(0.05)² × -0.1 e-0.2t = 0.0007875 e-0.2t V.

Now, according to Ohm's law, I = ε/R, where R is the resistance of the coil. Substituting the given values, we have I = 0.0007875 e-0.2t / 0.2 = 0.0039375 e-0.2t A. At t=2s, we can substitute into the equation to get I = 0.0039375 e-0.4 = 0.0024 A or 2.4 mA. Therefore, the magnitude of the induced current at t = 2s is 2.4 mA.

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To solve this problem we will use the concepts related to Ohm's law for which voltage, intensity and resistance are related.

Mathematically this relationship is given as

[tex]V = IR \rightarrow R= \frac{V}{I}[/tex]

Where,

V= Voltage

I = Current

R = Resistance

The value of the given voltage is 12V, while the current is 12mA, therefore the resistance would be

[tex]R = \frac{12}{12*10^{-3}}[/tex]

[tex]R = 1000 \Omega[/tex]

Therefore the resistance is [tex]1000\Omega[/tex]

Answer:

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]

Explanation:

The momentum of a photon is:

p = E/c

E = the photon energy

c = the speed of light.

take the time derivative (gives the force)

F = dp/dt = (dE/dt)/c

F = 2(dE/dt)/c (is doubled for complete reflection of the light)

Intensity has the units of energy per unit time per unit area

=  I

then,

Force/unit area = 2I/c

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]

Final answer:

An astronaut cannot walk normally in a space station because there's no frictional force to move forward in the near-weightless environment. To move, astronauts use handholds and walls, pushing against them to create a reaction force.

Explanation:

It is impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally because C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut would indeed "jump" in place due to the lack of friction between their feet and the floor of the space station, which is a result of the near-weightlessness they experience. In space, normal walking is ineffective because walking relies on gravity to pull the body back down to the floor after each step, which isn't present in the same way on a space station in orbit.

In order to move in such an environment, an astronaut must push against a solid object, creating a reaction force in the opposite direction according to Newton's third law of motion. This principle allows the astronaut to propel and steer themselves around the space station using handholds and walls. The environment inside the ISS is similar to that inside a freely falling box where gravity still exists, but occupants do not feel its effects because they are in free fall around Earth, which creates the sensation of weightlessness.

Final answer:

Astronauts cannot walk normally in an orbiting space station due to the lack of gravity and friction. They are in a state of free fall, creating a sensation of weightlessness. Movement can be achieved by utilizing the conservation of momentum and Newton's third law of motion. Therefore option C is the correct answer.

Explanation:

The reason it is impossible for an astronaut inside an orbiting space station to walk from one end to the other by walking normally is C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut cannot walk from one end to the other by walking normally because, in the microgravity environment of an orbiting spacecraft, traditional walking, which relies on the force of gravity and friction between the feet and the ground, does not work. Instead, astronauts move about by pushing off surfaces or floating through the air.

In orbit, the International Space Station (ISS) and everything inside it, including the astronauts, are in a state of free fall. They are falling around Earth at the same rate as the space station, creating a sensation of weightlessness. This is akin to the sensation of temporary weightlessness one experiences at the topmost point of a roller coaster ride or when an elevator suddenly descends.

Achieving locomotion for an astronaut stranded in the center of the station without contact with any solid surface would necessitate a method that does not rely on gravity or friction. The astronaut would have to utilize the principle of conservation of momentum. For instance, by throwing an object in one direction, the astronaut would move in the opposite direction, as described by Newton's third law of motion: for every action, there is an equal and opposite reaction.

Answer:

a)  Em = 332.8 J , b) # jump = 13, c)   It is reasonable since there are not too many jumps , d) lower the calories consumed

Explanation:

a) Let's use energy conservation

Initial. On the floor

             Em₀ = K = ½ m v²

Final. The highest point

             Emf = U = m g h

Energy is conserved

             Em₀ = Emf

             ½ m v² = m g h

             h = ½ v² / g

            h = ½ 3.2² /9.8

            h = 0.52 m

b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is

            Q = 3500 -2500 = 1000cal / day

Let's reduce this value to the SI system

             Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day

Now the energy in each jump is

               Em = K = ½ m v²

               Em = ½ 65 3.2²

               Em = 332.8 J

They indicate that the body can only use 25% of this energy

              Em effec = 0.25 332.8 J

              Em effec = 83.2 J

This is the energy that burns the body

Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J

              # jump = 1046 J (1 jump / 83.2 J)

              # jump = 12.6 jumps / day

              # jump = 13  

c) It is reasonable since there are not too many jumps

d) That some days consume more vegetables to lower the calories consumed

To solve this problem it is necessary to apply the concepts related to angular resolution, for which it is necessary that the angle is

[tex]\theta = 1.22\frac{\lambda}{nd}[/tex]

Where

d = Diameter of the eye

n = Index of refraction

D = Distance between head lights

[tex]\lambda[/tex]= Wavelength

Replacing with our values we have that

[tex]\theta = 1.22 \frac{(1.22)(575*10{-9})}{1.4(4*10^{-3})}[/tex]

[tex]\theta = 1.252*10^{-4}rad[/tex]

Using the proportion of the arc length we have to

[tex]L = \frac{D}{\theta}[/tex]

Where L is the maximum distance, therefore

[tex]L = \frac{1.6}{1.252*10^{-4}}[/tex]

[tex]L = 12.77km[/tex]

Therefore the maximum distance from the observer that the two headlights can be distinguished is 12.77km

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, [tex]L=9.4\ mH=9.4\times 10^{-3}\ H[/tex]

Resistance, R = 150 ohms

Capacitance, [tex]C=1.9\ \mu F=1.9\times 10^{-6}\ C[/tex]

At resonance, the capacitive reactance is equal to the inductive reactance such that,

[tex]X_C=X_L[/tex]    

[tex]2\pi f_o L=\dfrac{1}{2\pi f_oC}[/tex]

f is the resonant frequency of this circuit  

[tex]f_o=\dfrac{1}{2\pi \sqrt{LC}}[/tex]

[tex]f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}[/tex]

[tex]f_o=1190.91\ Hz[/tex]

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

Answer:

v = 88.76 m / s ,  K = 6.30 10⁶ J

Explanation:

For this exercise the force that is applied is that necessary for the acceleration of the car to be the acceleration of gravity, they do not indicate that there is friction, we look for the final speed

       v² = v₀² + 2 a x

Since the car starts from rest, the initial speed is zero, vo = 0

       v = √ 2 a x

       v = √ (2 9.8 402)

       v = 88.76 m / s

Let's look for kinetic energy

       K = ½ m v²

       K = ½ 160kg 88.76²

       K = 6.30 10⁶ J

Answer:

1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2

2) .w is angular frequency. Answer 2

3) k  is wave number. Answer 1

Explanation:

The electromagnetic wave is given by

         [tex]E_{y}[/tex] = E₀ sin (kx –wt)

This is the equation of a traveling wave on the x axis with the elective field oscillating on the y axis

The terms represent E₀ the maximum amplitude of the electric field,

The wave vector

        k = 2π /λ

Angular velocity

       w = 2π f

To answer the questions let's use the previous definitions

1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2

2) .w is angular frequency. Answer 2

3) k is wave number. Answer 1

Answer:

x = 8.699 10⁻³ m

Explanation:

The proton feels an electric charge that is the opposite direction of speed, let's look for acceleration using Newton's second law

      F = m a

        F = q E

      a = q E / m

      a = 1.6 10⁻¹⁹ 1500 / 1.67 10⁻²⁷

      a = 1,437 10¹¹ m / s²

Now we can use kinematic relationships

      v² = v₀² - 2 a x

When at rest the speed is zero (v = 0)

      x = v₀² / 2 a

Let's calculate

     x = 50,000² / (2 1,437 10¹¹)

     x = 8.699 10⁻³ m

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is [tex]^{0}_{0}\gamma[/tex]. Hence, charge on a gamma particle is also 0.

For example, [tex]^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy[/tex]

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if  a nucleus decays by gamma decay to a daughter nucleus.

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Answer:

The stress S = 1935 [Psi]

Explanation:

This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.

The initial data:

P = internal pressure [Psi] = 90 [Psi]

Di= internal diameter [in] = 22 [in]

t = wall thickness [in] = 0.25 [in]

S = stress = [Psi]

Therefore

ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]

And using the expression to find the stress:

[tex]S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi][/tex]

In the attached image we can see the stress σ1 & σ2 = S acting over the point A.

Answer:

Explanation:

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g       ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.

The maximum height the second rock reach is :

-4 times the distance traveled by the first rock.

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

using Third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g       ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Therefore, the second rock reaches the 4 times the distance traveled by the first rock.

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Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

Answer:

The motion of the parachute = 3.3 m/s²

Explanation:

Weight of the parachute - Resistive force of air = ma

W - Fₐ  = ma.................... Equation 1

making a the subject of formula in equation 1

a = (W- Fₐ)/m.................. Equation 2

Where W = weight of the parachute, Fₐ = resistive force of air, m = mass of the parachute, a = acceleration of the parachute

Constant: g = 9.8 m/s²

Given: Fₐ = 520 N, m = 80 kg

W = mg = 80 × 9.8 = 784 N,

Substituting these values into equation 2

a = (784-520)/80

a = 264/80

a = 3.3 m/s²

Therefore the motion of the parachute = 3.3 m/s²

Answer:

F= 9.45 N

Explanation:

If the mass is fastened to an unstrained horizontal spring, this means that at this position, the spring doesn't exert any force, because it keeps his equilibrium length.

If then the block is given a displacement of +0.113m, this means that the spring has been stretched in the same length.

According to Hooke's Law, the spring exerts a restoring force (trying to return to his equilibrium state) that opposes to the displacement, and which is proportional (in magnitude) to it, being the proportionality constant, a quantity called spring constant, which depends on the type of spring.

We can write the Hooke's Law as follows:

F = - k * Δx

Just before the block is released, we can get the value of F as follows:

⇒ F = 83.6 N/m* 0.113 m = 9.45 N (in magnitude)

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = [tex]\sqrt{\frac{2GM}{R} }[/tex]

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

so

K  - G Mm / 9R = 0

K = G Mm / 9R

Answer:

For this given plane monochromatic electromagnetic wave with wavelength λ=598 nm, the wavenumber is [tex]k=0,0105\ x\ 10^{-9}\ m^{-1}[/tex] .

Explanation:

For a plane electromagnetic wave we have that the electrical and magnetic field are:

[tex]E(r,t)=E_{0}\ cos ( wt-kr)\\\ B(r,t)=B_{0}\ cos(wt-kr)[/tex]

In this case we have the data for the magnetic field. We are told that the magnetic field in a plane electromagnetic wave with wavelength λ=598 nm, propagating in a vacuum in the z direction ([tex]\hat k[/tex]) is described by

         [tex]B=8.7\ x\ 10^{-6}\ T sin(kz-wt) (\hat i+\hat j)[/tex]

([tex]\hat i,\hat j, \hat k[/tex] are the unit vectors in the x,y,z directions respectively)

The wavenumber k is a measure of the spatial frequency of the wave, is defined as the number of radians per unit distance:

          [tex]k=\frac{2\pi}{\lambda}[/tex]

where λ is the wavelength

So we get that

[tex]k=\frac{2\pi}{\lambda} \rightarrow k=\frac{2\pi}{598 nm}  \rightarrow k=0,0105\ x\ 10^{9}\ m^{-1}[/tex]

The wavenumber is

            [tex]k=0,0105\ x\ 10^{9}\ m^{-1}[/tex] .

Answer:

Explanation:

Given

Velocity after t=4 sec is v=10 m/s downward

assuming u is the initial upward velocity

[tex]v=u+at[/tex]

[tex]-10=u-gt[/tex]

[tex]u=9.8\times 4-10=29.2 m/s[/tex]

[tex]v^2-u^2=2 as[/tex]

[tex](-10)^2-(29.2)^2=2\times (-9.8)\cdot s[/tex]

[tex]s=\frac{29.2^2-10^2}{2\times 9.8}[/tex]

[tex]s=38.4 m[/tex]

i.e. 38.4 m above the initial thrown Position  

To determine the steady state temperature of the wire, one can use the power dissipation formula and the convection heat transfer equation. The time for the wire to reach within 1-degree Celsius of steady state involves transient heat transfer calculations using the given material properties.

The student has asked about the steady state temperature of a 1-meter-long wire with a 1mm diameter submerged in an oil bath at 25 degrees Celsius when a current of 100A flows through it. We also need to calculate how long it takes for the wire to reach within 1-degree Celsius of the steady state temperature. To find the steady state temperature, we use the formula P = I2R, where P is the power, I is the current, and R is the resistance. Given that R = 0.01
Ω/m and I = 100A, we find P = (100A)2 x 0.01
Ω/m = 100W/m. Then, using the convection heat transfer equation Q = hA(Ts - T
bath), where Q is the heat transfer rate, h is the convection coefficient, A is the surface area, Ts is the wire surface temperature, and Tbath is the oil bath temperature, we equate Q to P since the wire is in steady state, and solve for Ts. The time to reach within 1-degree Celsius of steady state temperature requires calculating the transient heat transfer, which involves solving the heat transfer equation with the given material properties such as density, heat capacity, and thermal conductivity.

The steady-state temperature of the wire is approximately [tex]\(343.471 {°C}\)[/tex], and it takes approximately [tex]\(1.539[/tex],  for the wire to reach within 1°C of the steady-state value.

Steady-State Temperature Calculation:

  - Calculate the radius [tex](\(r\))[/tex] of the wire:

   [tex]\[ r = \frac{d}{2} = \frac{0.001 \, \text{m}}{2} = 0.0005 \, \text{m} \][/tex]

  - Calculate the surface area [tex](\(A\))[/tex] of the wire:

   [tex]\[ A = 2\pi r l = 2\pi \times 0.0005 \times 1 = 0.00314 \, \text{m}^2 \][/tex]

  - Calculate the heat transfer rate [tex](\(q\))[/tex]:

   [tex]\[ q = I^2 R = (100)^2 \times 0.01 = 1000 \, \text{W} \][/tex]

  - Calculate the steady-state temperature [tex](\(T_{\text{wire}}\))[/tex]:

    [tex]\[ T_{\text{wire}} = \frac{q}{hA} + T_{\text{fluid}} \][/tex]

    [tex]\[ T_{\text{wire}} \approx \frac{1000}{500 \times 0.00314} + 298.15 \][/tex]

    [tex]\[ T_{\text{wire}} \approx 343.471 \, \text{°C} \][/tex]

Time to Reach Within 1°C of Steady-State:

  - Calculate the volume [tex](\(V\))[/tex] of the wire:

    [tex]\[ V = \pi r^2 l = \pi \times (0.0005)^2 \times 1 = 7.854 \times 10^{-7} \, \text{m}^3 \][/tex]

  - Calculate the thermal time constant [tex](\(\tau\))[/tex]:

    [tex]\[ \tau = \frac{\rho V c}{hA} \][/tex]

   [tex]\[ \tau \approx \frac{8000 \times 7.854 \times 10^{-7} \times 500}{500 \times 0.00314} \][/tex]

    [tex]\[ \tau \approx 0.7854 \, \text{s} \][/tex]

  - Calculate the time [tex](\(t\))[/tex] it takes for the wire to reach within 1°C of the steady-state value:

    [tex]\[ t = \tau \ln\left(\frac{T_{\text{steady}} - T_{\text{initial}}}{T_{\text{steady}} - T_{\text{fluid}}}\right) \][/tex]

    [tex]\[ t \approx 0.7854 \times \ln\left(\frac{343.471 - 25}{343.471 - 298.15}\right) \][/tex]

   [tex]\[ t \approx 0.7854 \times \ln\left(\frac{318.471}{45.321}\right) \][/tex]

   [tex]\[ t \approx 0.7854 \times \ln(7.032) \][/tex]

   [tex]\[ t \approx 1.539 \, \text{s} \][/tex]

Answer:

Explanation:

Mass of bullet m = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

Velocity of wooden block + bullet just after impact = √2gH

=√(2 x 9.8 x 0.6)

= 3.43 m / s

Let the launch velocity of bullet be v₁

If v₂ be the velocity with which bullet hits the block

Applying law of conservation of momentum

.03 x v₂ = .530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² - 2 gh

v₁² = v₂² + 2 gh

= 60.6 ² + 2 x 9.8 x 0.4

v₁ = 60.65 m /s this is launch speed.

b )

Initial kinetic energy of bullet

= 1/2 m v²

= .5 x .03 x 3680

= 55 J

Potential energy of bullet + block = 0

Total energy = 5 J

c)

Kinetic energy of bullet block system

1/2 m v²

= .5 x .53 x  3.43

= 3.11 J

d )

Loss of energy in the impact =  Total mechanical energy  lost from beginning to end?

3.11 J  - 5

= 1.89 J

The energy given to each cell during the pulse can be calculated by multiplying the power of the pulse by its duration, and then dividing by the number of cells.

The energy supplied to the cell during the pulse is determined by the power multiplied by the duration of the pulse. In this scenario, the power is 2.46×1012 W and the duration is 3.4 ns (which is 3.4x10-9 s when converted to seconds for mathematical calculations).

We use the formula:
E = P * t
Where,
E is the Energy
P is the Power
t is the time (duration of the pulse)

Substituting the given values into the formula:
E = 2.46x1012 W * 3.4x10-9 s

This gives the total energy supplied. We know the energy is spread uniformly over the faces of 100 cells, so each cell will get 1/100 of the total energy. Using these calculations, we can determine the amount of energy given to each cell during the pulse.

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Answer:

b. 84 minutes

Explanation:

[tex]a_c=g[/tex] = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s[/tex]

Time period is given by

[tex]T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes[/tex]

The time period of Earth’s rotation would be 84.43835 minutes

The new period of the Earth’s rotation is mathematically given as

T=84.43835 min

Question Parameter(s):

The Earth’s radius is 6378.1 kilometers.

g= (9.81 m/s2 ),

Generally, the equation for the   is mathematically given as
[tex]a_c=\dfrac{v^2}{r}[/tex]

Therefore

[tex]v=\sqrt{a_cr}\\\\v=\sqrt{9.81*6378100}[/tex]

v=7910.06706 m/s

In conclusion

[tex]T=\dfrac{2\pi r}{v60}[/tex]

Hence

[tex]T=\dfrac{2\pi 6378.1*10^3}{7910.06706*60}[/tex]

T=84.43835 min

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To solve this problem we will use the parallel axis theorem for which the inertia of a point of an object can be found through the mathematical relation:

[tex]I = I_{cm} +mx^2[/tex]

Where

[tex]I_{cm}[/tex] = Inertia at center of mass

m = mass

x = Displacement of axis.

Our mass is given as 0.71kg,

m = 0.71kg

Para a Stick with length (L) the Moment of Inertia of the stick about and axis passing through the center and perpendicular to stick is

[tex]I_{cm} = \frac{1}{12} mL^2[/tex]

[tex]I_{cm} = \frac{1}{12} (0.71)(1)^2[/tex]

[tex]I_{cm} = 0.05916Kg\cdot m^2[/tex]

The distance between center of mass to the specific location is  

[tex]x = 50cm - 18cm[/tex]

[tex]x = 38cm = 0.38m[/tex]

So, from parallel axis theorem ,

[tex]I = I_{cm} + mx^2[/tex]

[tex]I =0.05916Kg\cdot m^2+ (0.71kg)(0.38m)^2[/tex]

[tex]I = 0.161684Kg\cdot m^2[/tex]

Therefore the rotational inertia is [tex]0.161684Kg\cdot m^2[/tex]