Two long, parallel wires are separated by a distance of 2.2 cm. The force per unit length that each wire exerts on the other is 3.6 × 10^-5 N/m, and the wires repel each other. The current in one wire is 0.52 A. What is the magnitude of the current in the second wire? (Give your answer in decimal using "A" (Ampere) as unit)

Answer:

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

Explanation:

We start from:

[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]

Separating variables:

[tex]-50dT=(T-21)dt[/tex]

[tex]-50\frac{dT}{T-21}=dt[/tex]

Integrating with initial conditions:

[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]

[tex]-50ln(\frac{T-21}{91-21})=t[/tex]

[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]

Isolating T:

[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)

The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.

In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).

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Final answer:

To calculate the power converted to thermal energy due to frictional effects, subtract the ideal mechanical power required to lift water to the higher reservoir (13.2435 kW) from the actual pump power (20 kW), yielding 6.7565 kW.

Explanation:

The student is asking to determine the mechanical power that is converted to thermal energy due to frictional effects when water is pumped from a lower reservoir to a higher reservoir. Given the shaft power of the pump is 20 kW, height difference is 45 m, and flow rate is 0.03 m³/s, we can first calculate the ideal mechanical power required to lift the water to that height.

The gravitational potential energy given to the water per second (which is the power) by pumping it to the height is calculated using the formula P = ρghQ, where ρ (rho) is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the height difference (45 m), and Q is the flow rate (0.03 m³/s). This results in P = 1000 kg/m³ * 9.81 m/s² * 45 m * 0.03 m³/s = 13243.5 W or 13.2435 kW.

The difference between the actual power supplied by the pump (20 kW) and the ideal power required (13.2435 kW) is the power lost to thermal energy due to friction. Therefore, the power converted to thermal energy is 20 kW - 13.2435 kW = 6.7565 kW.

The rate of heat generation in the rod is calculated by first determining the resistance using the formula R = ρL/A, then using that resistance value in the power formula P = V²/R. Using the provided values, the rate of heat generation in the rod under a potential difference of 0.50 V is 8.33 Watts.

The subject of this question is about the rate of heat generation in a rod under a potential difference, which is a topic in Physics. To solve for this, we first need to compute for the resistance of the rod using the formula R = ρL/A, where R is the resistance, ρ is the resistivity of the material, L is the length, and A is the cross-sectional area of the rod. Given that ρ = 6.0 × 10−8 Ω ⋅ m, L = 2.0 m, and A = (2.0 mm × 2.0 mm) = 4.0 × 10-6 m², we get R = (6.0 × 10−8 Ω ⋅ m * 2.0 m) / 4.0 × 10-6 m² = 0.03 Ω.

Next, we use the formula P = V²/R to calculate the rate of heat generation (power). Here, P is the power, V is the potential difference, and R is the resistance. With V = 0.50 V and R = 0.03 Ω, after substituting the values we get P = (0.50 V)² / 0.03 Ω = 8.33 Watts. Therefore, the rate at which heat is generated in the rod is 8.33 Watts.

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The rate at which heat is generated in the rod is found by calculating the current and using Joule's Law. The final rate of heat generation is 8.33 Watts.

Heat Generated in a Rod Due to Electric Current

To determine the rate at which heat is generated in the rod, we start by calculating the resistance of the rod. Given:

Length of the rod (L) = 2.0 m

Cross-sectional area (A) = (2.0 mm × 2.0 mm) = (2.0 × [tex]10^{-3}[/tex] m) × (2.0 × [tex]10^{-3}[/tex] m) = 4.0 × [tex]10^{-6}[/tex] m²

Resistivity of the material (ρ) = 6.0 × [tex]10^{-8}[/tex] Ω·m

Potential difference (V) = 0.50 V

The resistance (R) of the rod can be calculated using the formula:

R = ρ(L/A)

Substituting the given values:

R = (6.0 × [tex]10^{-8}[/tex] Ω·m) × (2.0 m / 4.0 × [tex]10^{-6}[/tex] m²) = 3.0 × [tex]10^{-2}[/tex] Ω

Next, we calculate the current (I) flowing through the rod using Ohm's Law:

I = V / R

Substituting the values:

I = 0.50 V / 3.0 × [tex]10^{-2}[/tex] Ω = 16.67 A

The rate of heat generated (P) in the rod is given by Joule's Law:

P = I²R

Substituting the calculated values:

P = (16.67 A)² × 3.0 ×[tex]10^{-2}[/tex] Ω = 8.33 W

Therefore, the rate at which heat is generated in the rod is 8.33 Watts.

Answer:

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

Answer:

B) 206.323

Explanation:

The value of (8 104)2, written with the correct mumber of significant figures is 206.323.

Answer:

B. 206.323

Explanation:

Answer:

Janna levin is a cosmologist and professor at physics.

She is an american by race

She was the presenter of Nova feature Black hole Apocalypse and has writtenany science non-fiction books

www.jannalevin.com is her own page where u get her correct info and bio

Answer:

0.22 m

Explanation:

m = 0.43 kg, K = 561 N/m

Vmax = 8 m/s

Let the amplitude of the oscillations be A.

The formula for the angular frequency of oscillation sis given by

[tex]\omega  = \sqrt{\frac{K}{m}}[/tex]

[tex]\omega  = \sqrt{\frac{561}{0.43}}[/tex]

ω = 36.1 rad/s

The formula for the maximum velocity is given by

Vmax = ω x A

A = Vmax / ω

A = 8 / 36.1 = 0.22 m

Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

[tex]y(t)-h = \frac{1}{2}gt^2[/tex]

where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]

So the bullet has travelled 313.6 m downward.

Answer:

Yes we are right as the force on wire is approx 12500 Lb

Explanation:

Magnetic force on a current carrying bar is given by the equation

[tex]\vec F = i(\vec L \times \vec B)[/tex]

here we know that

L = 1.3 m

B = 12 T

[tex]\theta = 60^0[/tex]

i = 4.1 kA

now from above formula we have

[tex]F = iLBsin60[/tex]

[tex]F = (4.1\times 10^3)(1.3 )(12)sin60[/tex]

[tex]F = 55391 N[/tex]

So this is equivalent to 12500 Lb force

 Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.

"The magnetic force exerted on the conducting bar can be calculated using the formula for the force on a current-carrying conductor in a magnetic field, which is given by [tex]\( F = ILB \sin(\theta) \),[/tex] where [tex]\( I \)[/tex] is the current, \( [tex]L \)[/tex] is the length of the conductor, [tex]\( B \)[/tex]is the magnetic field strength, and [tex]\( \theta \)[/tex] is the angle between the current direction and the magnetic field.

 Given:

[tex]- \( I = 4.1 \times 10^3 \) A (since 4.1 kA = 4.1 \(\times\) 10^3 A) - \( L = 1.3 \) m\\ - \( B = 12 \) T\\ \( \theta = 60^\circ \)\[/tex]

Converting the angle from degrees to radians (since the sine function in physics formulas typically uses radians), we have[tex]\( \theta = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \) radians.[/tex]

 Now, we can calculate the force:

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

[tex]\( F = (4.1 \times 10^3 \text{ A}) \times (1.3 \text{ m}) \times (12 \text{ T}) \times \sin\left(\frac{\pi}{3}\right) \)[/tex]

 Since[tex]\( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex], we can substitute this value into the equation:

[tex]\( F = (4.1 \times 10^3) \times (1.3) \times (12) \times \frac{\sqrt{3}}{2} \)[/tex]

 Calculating the force in newtons:

[tex]\( F = 4.1 \times 10^3 \times 1.3 \times 12 \times \frac{\sqrt{3}}{2} \)[/tex]

[tex]\( F \approx 4.1 \times 1.3 \times 12 \times 0.866 \times 10^3 \)[/tex]

[tex]\( F \approx 5107.2 \times 0.866 \)[/tex]

[tex]\( F \approx 4424.4 \) N[/tex]

 To convert the force from newtons to pounds, we use the conversion factor [tex]\( 1 \text{ N} \approx 0.22481 \text{ lb} \):[/tex]

[tex]\( F \approx 4424.4 \text{ N} \times 0.22481 \frac{\text{lb}}{\text{N}} \)[/tex]

[tex]\( F \approx 994.5 \text{ lb} \)[/tex]

 Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.

 In conclusion, the claim that the magnetic force will exceed 10,000 pounds is incorrect. The actual force is approximately 994.5 pounds, which is less than 10,000 pounds. The conducting bar should still be clamped in place to prevent movement due to the magnetic force, but the initial claim overestimated the force."

Answer:

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Explanation:

As we know that electric field due to long cylinder on a cylindrical Gaussian surface must be constant

so on the Gaussian surface we will have

[tex]\int E. dA = \frac{q_{en}}{\epsilon_0}[/tex]

now the electric field is passing normally through curved surface area of the cylinder

so we will have

[tex]E (2\pi rL) = \frac{q_{en}}{\epsilon_0}[/tex]

here enclosed charge in the cylinder is given as

[tex]q_{en} = \lambda L[/tex]

from above equation

[tex]E(2\pi rL) = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Explanation:

It is given that,

The thinnest soap film appears black when illuminated with light with a wavelength of 535 nm, [tex]\lambda=5.35\times 10^{-7}\ m[/tex]

Refractive index, [tex]\mu=1.33[/tex]

We need to find the thickness of soap film. The soap film appear black means there is an destructive interference. The condition for destructive interference is given by :

[tex]2t=m\dfrac{\lambda}{\mu}[/tex]

t = thickness of film

m = 0,1,2....

[tex]\mu[/tex] = refractive index

[tex]t=m\dfrac{\lambda}{2\mu}[/tex]

[tex]t=\dfrac{\lambda}{2\mu}[/tex]

For thinnest thickness, m = 1

[tex]t=1\times \dfrac{5.35\times 10^{-7}\ m}{2\times 1.33}[/tex]

[tex]t=2.01\times 10^{-7}\ m[/tex]

Hence, this is the required solution.

Final answer:

The thinnest soap film that appears is approximately 50.47 nm due to destructive interference.

Explanation:

The question is asking for the thinnest soap film that appears black when illuminated with light of a specific wavelength, in this case, 535 nm. The phenomenon described is known as thin film interference, which occurs when light waves reflected off the top and bottom surfaces of a film interfere with each other.

The film appears black at the thinnest point where destructive interference occurs, which means the reflected light waves are out of phase and cancel each other out.

To find the thinnest film thickness that appears black, we can use the formula for destructive interference in thin films, taking into account the phase shift that occurs upon reflection from a medium with a lower index of refraction to a higher one. This formula is:
[tex]2nt = (m + \(\frac{1}{2}\))\(\lambda\),[/tex]where n is the index of refraction, t is the thickness of the film, \(\lambda\) is the wavelength of light in vacuum, and m is an integer representing the order of the interference.

For the thinnest film, we use m = 0.

Therefore, the thinnest soap film thickness t is calculated as:

[tex]2nt = \(\frac{1}{2}\)\(\lambda\)2nt = \(\frac{1}{2}\) \\times 535 nm / 1.33t = \(\frac{535 nm}{(2 \\times 1.33 \\times 2)}\)t = 100.94 nm[/tex]

However, regarding physical film thickness, we should only consider the distance that light travels within the film, which is factored by n to give the optical path length. Thus, the actual thickness would be half of 100.94 nm, yielding a value of approximately 50.47 nm.

Answer:

7212.3 N

Explanation:

F = 7.38 x 10^4 N, v = 36.2 m/s

Let b be the strength of magnetic field and charge on the particle is q.

F = q v B Sin theta

Here theta = 90 degree

7.38 x 10^4 = q x 36.2 x B x Sin 90

q B = 2038.7 .....(1)

Now, theta = 17 degree, v = 12.1 m/s

F = q v B Sin theta

F = 2038.7 x 12.1 x Sin 17     ( q v = 2038.7 from equation (1)

F = 7212.3 N

Answer:

Length of wire=2.16 m

Diameter of wire=0.232 mm

Explanation:

m= mass of copper wire= 0.82 g

R= Resistance of copper wire= 0.87 ohms

D= Density of copper= 8.96 g/cm^3

ρ= Resistivity= 1.7×10^-8 Ωm

[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]

[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]

[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

(a) The emf induced in the loop is 0.0171 V.

(b) The current in the loop is determined from the ratio of induced emf to resistance in the loop.

The emf induced in the loop is calculated by applying Faradays law as follows;

emf = dФ/dt

emf = A dB/dt

where;

emf = 0.09 x 0.19

emf = 0.0171 V

The current in the loop is determined by applying Ohm's law;

I = emf/R

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Answer:57.7

Explanation:

Force on a moving charge in  a magnetic field is given by

[tex]F=q\times V\times Bsin\theta [/tex]

Where F=Force experienced by charge

q=charge of particle

V=velocity of particle

B=magnetic field

F=qVBsin(25)-------1

[tex]2F=qVBsin(\theta )------2[/tex]

Divide (1)&(2)

[tex]\frac{1}{2}[/tex]=[tex]\frac{sin(25)}{sin(\theta)}[/tex]

[tex]Sin(\theta )=2\times sin(25)[/tex]

[tex]\theta =57.7 ^{\circ}[/tex]

Given the constraints of the question and the formula for magnetic force on a moving charge, the situation described isn't possible.

The magnetic force experienced by a moving charged particle is given by the formula F = qvBsin(θ), where q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the magnetic field. When the force is F at 25°, to experience a force of 2F, the sin of the new angle must be double that of sin(25°) because all other factors (charge, speed, magnetic field strength) are constant.

In the case given, sin(θ) would need to be 2sin(25°), but since the maximimum value sin(θ) can have is 1 (at 90°), this situation isn't possible within the constraints of the question (angle less than 90°). Hence, the student might have misunderstood the problem or there might be an error the question itself. However, if it is assumed that the question intends to find when the component of the force perpendicular to the field is double, the answer would be when θ is 90 degrees because the sin(90°) = 1. This is because when a charged particle moves perpendicular to a magnetic field, the sin(θ) factor in the force equation is maximized.

To find the magnitude of the induced current in the coil at 0.10 s, we need to calculate the magnitude of the magnetic flux through the coil and use Faraday's law of electromagnetic induction.

To find the magnitude of the induced current in the coil at 0.10 s, we need to first calculate the magnitude of the magnetic flux through the coil. The magnetic flux is given by Φ = B * A, where B is the magnetic field and A is the area of the coil. The area of the coil is calculated as A = π * r^2, where r is the radius of the coil.

Next, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. Mathematically, this can be written as emf = -dΦ/dt. Since the coil has resistance, we can use Ohm's Law to find the magnitude of the induced current, which is given by I = emf/R. Plugging in the values and calculating, we can find the magnitude of the induced current at 0.10 s.

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To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction. The magnetic field B is given as B = 50sin(10πt) mT. At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction, which states that the induced emf (voltage) is equal to the negative rate of change of magnetic flux through the coil.

The magnetic field B is given as B = 50sin(10πt) mT, where t is measured in seconds.

The magnetic flux through the coil is given by Φ = B.A, where A is the area of the coil.

Substituting the given values, we have A = πr^2 = π(0.04^2) = 0.005 cm^2.

At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

We can now calculate the magnetic flux through the coil using Φ = B.A.

Finally, we can use Faraday's Law to find the induced emf and divide it by the resistance of the coil to find the magnitude of the induced current.

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Answer:

EMF = 108.3 Volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

[tex]EMF = N\frac{d\phi}{dt}[/tex]

[tex]EMF = N\frac{BA - 0}{\Delta t}[/tex]

now we know that

N = 50 turns

B = 1.5 T

A = 0.325 m^2

[tex]\Delta t = 0.225 s[/tex]

now we have

[tex]EMF = (50)(\frac{1.5\times 0.325}{0.225})[/tex]

[tex]EMF = 108.3 Volts[/tex]

The magnitude of the average induced emf is 110.12 volts.

The magnitude of the average induced emf can be calculated using Faraday's law. The formula is given as:

emf = -N * A * (dB/dt)

Where:

In this case, the number of turns is 50, the area is 0.325 m², and

the rate of change of the magnetic field is (0 T - 1.5 T) / 0.225 s = -6.67 T/s.

Substituting these values into the formula, we get:

emf = -(50)(0.325)(-6.67) = 110.12 V

The magnitude of the average induced emf is 110.12 volts.

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Answer:

Windmill will complete 0.27 revolutions.

Explanation:

We have equation of motion s = ut + 0.5at²

Initial speed, u = 12.5 rad/s

Time, t = 12 seconds.

Angular acceleration, a = -0.75 rad/s²

Substituting

               s = 12.5 x 12 - 0.5 x 0.75 x 12² = 96 rad

1 revolution = 360 rad

[tex]96rad=\frac{96}{360}=0.27revolution[/tex]

So, windmill will complete 0.27 revolutions.

Answer:

(a) 83.73 rad/s

(b) 251.2 m/s

(c) 3505.4 rad/s^2

(d) 5024 m

Explanation:

R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps

(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s

(b) The relation between linear speed and the angular speed is

V = r w

Here, r = 30 cm = 0.3 m

V = 0.3 x 83.73 = 25.12 m/s

(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2

(d) Time period T = 2 x 3.14 / w

T = 2 x 3.14 / 83.73 = 0.075 sec

In 0.075 second, angle turn = 360 degree

In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree

In 360 degree, the distance traveled = 2 x pi x R

In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360

                                                                     = 5024 m

Answer:

The Poisson's ratio for this material is 0.4370.

Explanation:

Given that,

Diameter of metal = 11.2 mm

Force = 14100 N

Reduction diameter [tex]d=7\times10^{-3}\ mm[/tex]

Elastic modulus = 100 GPa

We need to calculate the change in length

Using formula of modulus elasticity

[tex]E=\dfrac{FL}{A\Delta L}[/tex]

The change in length is

[tex]\Delta L=\dfrac{FL}{AE}[/tex]

[tex]\dfrac{\Delta L}{L}=\dfrac{14100}{\pi\times\dfrac{(11.2\times10^{-3})^2}{4}100\times10^{9}}[/tex]

[tex]\dfrac{\Delta L}{L}=0.00143[/tex]

We need to calculate the Poisson's ratio

Using formula of Poisson's ratio

[tex]\nu=\dfrac{longitudinal\ strain}{Transverse strain}[/tex]

[tex]\nu=\dfrac{-\dfrac{\Delta d}{d}}{-\dfrac{\Delta L}{L}}[/tex]

Put the value into the formula

[tex]\nu=\dfrac{\dfrac{7\times10^{-6}}{11.2\times10^{-3}}}{0.00143}[/tex]

[tex]\nu=0.4370[/tex]

Hence, The Poisson's ratio for this material is 0.4370.

By using similar triangles concept and rates of change in Mathematics, we can find the man's shadow length is decreasing at a rate of 0.5 m/s when he is 4 meters away from the building.

The problem you're describing is typically solved using similar triangles concept in Mathematics and rates of change. As the man walks towards the building, his shadow becomes shorter. We can set up a ratio between the man's height and his distance from the building, and the length of the shadow and the wall. When the man is 4 meters away from the building, a right-triangle is formed with the man's height (2m), his distance from the building (4m), and his shadow's length.

Let s gives the length of the shadow, then the similar triangles imply 12/s = 2/4 which gives us s = 6 meters.

To find the rate at which the man's shadow is decreasing, we can differentiate the similar triangles ratio with respect to time to get -12/s^2 ds/dt =-1/2, sub in s = 6 into the equation to find ds/dt = -0.5 m/s.

So the length of the man's shadow on the building is decreasing at a rate of 0.5 m/s when he is 4 m from the building.

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Final answer:

To find how fast the man's shadow is decreasing, we use similar triangles and related rates in calculus. When the man is 4 m from the building, his 2 m tall shadow is shortening at 3.8 m/s, twice his walking speed of 1.9 m/s.

Explanation:

The question involves determining how fast the length of a man's shadow on a building is decreasing when he walks toward the building and away from the spotlight. This requires the application of similar triangles and the concept of related rates in calculus to solve. If we let the distance between the man and the building be x, and the length of the shadow be y, as the man walks towards the building, both x and y are changing with respect to time.

Given that the man is 2 m tall and the spotlight is 12 m away from the building, we can set up a proportion between the heights of the man and his shadow, and their respective distances from the spotlight. When he is 4 m from the building, the similar triangles can be represented as 2/y = (12-x)/(12). Differentiating both sides with respect to time t yields the related rate of change for the length of the shadow y when x is 4 m and the speed of the man is known, which is 1.9 m/s.

By solving this proportion and applying the derivative, we can find that the rate at which the length of his shadow decreases is exactly double the speed at which the man is walking, because in this scenario, the man and the tip of his shadow form a straight line with the light source. Thus, the length of the shadow decreases at a rate of 3.8 m/s when the man is 4 m from the building.

Answer:

0.24

Explanation:

w=vt

then

w1/v1=w2/v2

0.4/0.5=w2/0.3

w2= 0.3*0.4/0.5=0.24//

The angular acceleration of the bar is 0.75 rad/s², and the acceleration of end A is 0.3 m/s² in the opposite direction to end B. These values were determined using the relationships between linear and angular velocities and accelerations.

To determine the angular acceleration of the bar and the acceleration of end A, let's go through the step-by-step process:

Determine the Angular Velocity

Given that the bar has a length of 0.4 meters and the velocity of end B is 0.5 m/s, we start by calculating the angular velocity ω. The formula relating linear velocity v and angular velocity ω is:

[tex]v = \omega * r[/tex]

where r is the length of the bar. Rearranging for ω :

[tex]\omega = v / r = 0.5 m/s / 0.4 m = 1.25 rad/s[/tex]

Determine the Angular Acceleration

Next, we use the given acceleration of end B, which is 0.3 m/s². To find the angular acceleration α, we use the formula:

[tex]a = \alpha * r[/tex]

Rearranging for α:

[tex]\alpha = a / r = 0.3 m/s^2 / 0.4 m = 0.75 rad/s^2[/tex]

Determine the Acceleration of End A

The acceleration of end A, which is located at the opposite end of the bar, can be computed using:

[tex]a_A = \alpha * r[/tex]

Since end A is on the same bar, r here will be the same:

[tex]a_A = 0.75 rad/s^2* 0.4 m = 0.3 m/s^2[/tex]

Therefore, the acceleration of end A is also 0.3 m/s², but in the opposite direction to end B.

Answer:

N/l = 104

Explanation:

Energy stored in the inductor is given by the formula

[tex]U = \frac{1}{2}Li^2[/tex]

now we have

[tex]6\times 10^{-6} = \frac{1}{2}L(0.400)^2[/tex]

now we have

[tex]L = 7.5 \times 10^{-5}[/tex]

now we have

[tex]L = \frac{\mu_0 N^2 \pi r^2}{l}[/tex]

[tex]7.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} N^2 \pi(0.05)^2}{0.7}[/tex]

[tex]N = 73 turns[/tex]

now winding density is turns per unit length

[tex]N/l = 104[/tex]

The winding density of the given solenoid is 104 turns per meter.

The formula for energy stored in the inductor can be used to determine the inductance of the solenoid as follows.

U = ¹/₂LI²

6 x 10⁻⁶ = ¹/₂ (L) x (0.4)²

6 x 10⁻⁶ = 0.0.8L

L = 7.5 x 10⁻⁵

The number of turns of the solenoid is calculated as follows;

[tex]L = \frac{\mu N^2\pi r^2}{l} \\\\7.5 \times 10^{-5} = \frac{(4\pi \times 10^{-7} ) \times N^2 \times \pi(0.05)^2}{0.7} \\\\N = 73 \ turns[/tex]

The winding density if the number of turns per length

N/l = 73/0.7

N/l = 104

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Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

Answer:

The maximum power delivered by the power supply is 0.81 W.

Explanation:

Given that,

Inductance L= 2.0 H

Resistance R = 100 ohm

Voltage = 9.0 V

We need to calculate the power

Using formula of power

[tex]P = \dfrac{V^2}{R}[/tex]

Where, P = power

V = voltage

R = resistance

Put the value into the formula

[tex]P = \dfrac{(9.0)^2}{100}[/tex]

[tex]P =0.81\ W[/tex]

Hence, The maximum power delivered by the power supply is 0.81 W.

The maximum power delivered by the power supply in a series LR circuit can be calculated using the maximum current, the resistance, and the voltage of the power supply.

In a series LR circuit, the power delivered by the power supply is maximized when the current is at its maximum value. Initially, when the switch is closed, the current rises exponentially with time and eventually reaches its maximum value. The time constant of the circuit is T = L/R, where L is the inductance and R is the resistance. Therefore, the maximum power delivered by the power supply can be calculated using the formula P = (I_max)^2 * R, where I_max is the maximum current.

In this case, the initial current can be calculated using the formula I(0) = V/R, where V is the voltage of the power supply. The time constant can be calculated using the given inductance and negligible internal resistance. Plug these values into the formulas to find the maximum power delivered.

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Answer:

It is actually in both.

Explanation:

The momentum of an isolated system is conserved for both elastic and inelastic collision. OPtion C is correct

Collision is the process by which two bodies come into contact with the release of energy.

Collision can be elastic or inelastic.

Foe elastic collision, both momentum, and energy is conserved while for inelastic collision only momentum is conserved.

From the explanation above, we can see that momentum of an isolated system is conserved for both elastic and inelastic collision

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Answer:

Total displacement traveled = 298

Explanation:

According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).

So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.

Then after a little it can travel up another cm to reach the top.

Therefore, the total displacement traveled = 297 + 1 = 298 cm

"Displacement" means the distance and direction from start to finish, regardless of what happened in between.

The caterpillar's displacement is 99 cm straight up.

Answer:

The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.

Explanation:

Given that,

Angular velocity = 5.40 rad/s

Time t = 0.000 sec

Angular acceleration = 1.50 rad/s^2

(a). We need to calculate the angle at time t = 4.00 s

Using formula for angle

[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t^2[/tex]

Where, [tex]\omega_{0}[/tex]=angular velocity

[tex]\alpha[/tex]=angular acceleration

t = time

Put the value into the formula

[tex]\theta=5.40\times4.00+\dfrac{1}{2}\times1.50\times(4.00)^2[/tex]

[tex]\theta=33.6\ rad[/tex]

[tex]\theta=\dfrac{33.6}{2\pi}\ rad[/tex]

[tex]\theta=5.35\ rev[/tex]

(b). We nee to calculate the angular velocity at 4.00 s

Using formula of angular velocity

[tex]\omega=\omega_{0}+\alpha t[/tex]

[tex]\omega =5.40+1.50\times4.00[/tex]

[tex]\omega=11.4\ rad/s[/tex]

[tex]\omega=\dfrac{11.4}{2\pi}\ rad/s[/tex]

[tex]\omega=1.81\ rev/s[/tex]

Hence, The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.

Answer:

15.7 x 10⁻⁶ A

Explanation:

r = radius of the circular loop = 0.03 m

Area of the loop is given as

A = πr² = (3.14) (0.03)² = 0.002826 m²

R = Resistance of the resistor = 189 Ω

ΔB = Change in magnetic field = 0.25 - 0.35 = - 0.10 T

Δt = time interval = 1 sec

Current is given as

[tex]i = - A\left ( \frac{\Delta B}{\Delta t} \right )[/tex]

[tex]i = \left ( \frac{A}{R} \right )\left ( \frac{\Delta B}{\Delta t} \right )[/tex]

[tex]i = \left ( \frac{-0.002826}{18} \right )\left ( \frac{- 0.10}{1} \right )[/tex]

i = 15.7 x 10⁻⁶ A