Two insulated wires, each 4.42 m long, are taped together to form a two-wire unit that is 4.42 m long. One wire carries a current of 7.33 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 69.4o relative to a magnetic field whose magnitude is 0.547 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.24 N. What is the current I

Answer:

Explanation:....

The range of the projectile is 4675 m. The maximum height reached by the projectile is 2124 m. The speed of the projectile at impact is 234 m/s.

(a) Range is defined as the maximum horizontal distance covered by a projectile during its motion. It is given by the formula:

[tex]R = \frac{u^2 sin2\theta}{g}[/tex],

where u = initial speed of the projectile = 230 m/s

θ = angle of projection = 60°

g = acceleration due to gravity = 9.8 m/s²

Using these in formulas, we get:

[tex]R = \frac{(230 \hspace{0.8mm} m/s)^2 \hspace{0.8mm} sin2(60)}{9.8 \hspace{0.8mm} m/s^2}[/tex]

or, R = 4674.77 m ≈ 4675 m

(b) The maximum height reached by the projectile is given by the formula:

[tex]H = \frac{u^2 \hspace{0.8mm} sin^2\theta}{2g}[/tex]

using the numeric values, we get:

[tex]H = \frac{(230 \hspace{0.8mm}m/s)^2 sin^2(60)}{2 \times (9.8 \hspace{0.8mm} m/s^2)}[/tex]

or, H = 2024.23 m

However, since the projectile was fired 100 m above the ground, hence, the maximum height would be:

h = 100 m + H = 100 m + 2024.23 m = 2124.23 m ≈ 2124 m

(c) The speed of the projectile comprises of two parts, a horizontal velocity and a vertical velocity.

The horizontal velocity is given as [tex]v_x[/tex]. It is constant throught the motion and its magnitude can be determined by the formula:

[tex]v_x = v cos\theta = 230 \hspace{0.8 mm} m/s \times cos(60)[/tex]

or, [tex]v_x = 115 \hspace{0.8mm} m/s[/tex]

The vertical velocity is given as [tex]v_y[/tex]. It is not constant throught the motion. At the highest point, the magnitude of vertical velocity is zero. Using the kinetic equation,

[tex]v_y^2[/tex] = u² + 2gs, where s = h = 2124 m, we get:

[tex]v_y^2[/tex] = 0 + (2 × 9.8 m/s² × 2124 m)

or, [tex]v_y^2[/tex] = 41630.4 m²/s²

or, [tex]v_y[/tex] = 204.03 m/s ≈ 204 m/s

Now, the magnitude of speed will be equal to the resultant of both the horzontal and vertical velocities.

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

or, [tex]v = \sqrt{204^2 + 115^2} \hspace{0.8 mm} m/s[/tex]

or, v = 234.21 m/s ≈ 234 m/s

Answer:

the answer the correct one is B

Explanation:

The interference phenomenon is described by

           d sin θ = m λ

The spectrum is recorded on a screen, so we can use trigonometry

          tan θ = y / L

In this experiment the angles are very small, so

           tan θ = sin θ /cos θ = sin θ

  We replace

         d y / L = m La m

        y = m Lam L / d

When the slits approach the screen the value of L decreases,

Therefore the value of the separation between the slits must also decrease

When reviewing the answer the correct one is B

Answer:

Explanation:

The system can be modeled as,

Using magnitude relationship for imbalance system.

Check attachment for solution

Answer:

Explanation:

W (Ce) = 2.1 eV

W (Al) = 4.1 eV

W (Be) = 5 eV

W (K) = 2.3 eV

W (Pt) = 6.4 eV

W (Mg) = 3.7 eV

Part A:

Work function is directly proportional to the cut off frequency.

let f denotes the cut off frequency.

So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)

Part B:

Maximum wavelength for the emission is inversely proportional to the cut off frequency

So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)

Part C:

E = 3.10 eV

Let K is the maximum kinetic energy

K = E - W

K (Ce) = 3.1 - 2.1 = 1 eV

K (Al) 3.1 - 4.1 = not possible

K (Be) = 3.1 - 5 = not possible

K (K)  = 3.1 - 2.3 = 0.8 eV

K (Pt) = .1 - 6.4 = not possible

K (Mg) = 3.1 - 3.7 = not possible

So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)

Metals can be ranked based on their work functions, which determine cutoff frequency, the maximum wavelength of light needed to free electrons, and maximum kinetic energy of emitted electrons under specific light. Lower work functions result in higher cutoff frequencies and longer required wavelengths, and more energetic emitted electrons when under 400 nm light.

The ranking of the metals, in terms of cutoff frequency, maximum wavelength of light, and maximum kinetic energy, is based on the work function of each metal, which is the minimum energy required to free an electron from the metal's surface. The cutoff frequency is inversely related to the work function, so the metal with the lowest work function (cesium, 2.1 eV) will have the highest cutoff frequency, and the metal with the highest work function (platinum, 6.4 eV) will have the lowest cutoff frequency.

Conversely, the maximum wavelength of light needed to free electrons is directly related to the work function, so cesium will require the longest wavelength and platinum will require the shortest wavelength. When illuminated with 400 nm (3.10 eV) light, the kinetic energy of the emitted electrons will be the difference between the energy of the light and the work function, so cesium (with the smallest work function) will have the most energetic electrons, and metals with a work function larger than 3.10 eV (like aluminum, beryllium, and platinum) will not emit electrons at all.

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Answer:

The magnetic flux links to its turns = [tex]7.6 \times10^{-3}[/tex] Wb.

Explanation:

Given :

Radius of circular coil = [tex]18 \times 10^{-2}[/tex] m

Number of turns = 25

Magnetic field = [tex]3 \times10^{-3}[/tex] T

Magnetic flux (Φ) is a measure of the magnetic field lines passes through a given area. The unit of magnetic flux is weber (Wb).

We know that,

⇒    Φ = [tex]BA[/tex]

Where [tex]B =[/tex] ext. magnetic field, [tex]A =[/tex] area of loop or coil.

But here given in question, we have turns of wire so our above eq. modified as follows.

⇒   Φ = [tex]NBA[/tex]

Where [tex]N =[/tex] no. of turns.

∴    Φ = [tex]25 \times 3 \times 10^{-3} \pi (18 \times10^{-2} )^{2}[/tex]

     Φ = [tex]7.6 \times 10^{-3} Wb[/tex]

Thus, the magnetic flux links to its turns = [tex]7.6 \times 10^{-3} Wb[/tex]

Answer:

 ΔV = 2k λ ln (r₁ / r₂)

Explanation:

The electric potential is

         ΔV = - ∫ E. ds

The expression for the electric field of a charge line is given

           E = 2k λ / r

Let's replace and integrate

         ΔV = - 2k λ ∫ dr / r

         ΔV = -2 k λ ln r

Let's evaluate

       ΔV = - 2k λ ln r₂- ln r₁

       ΔV = -2k λ ln (r₂ / r₁)

       ΔV = 2k λ ln (r₁ / r₂)

Answer:

A) v/ sq. rt. of 2

Explanation:

The speed of the wave in the string is defined as:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

Where T is the tension on the string and [tex]\mu[/tex] is the linear density, that is, the mass per unit length:

[tex]\mu=\frac{m}{L}[/tex]

Where m is the mass of the string and L its length. We have [tex]m'=2m[/tex], [tex]T'=T[/tex] and [tex]L'=L[/tex]:

[tex]v'=\sqrt\frac{T'}{m'/L'}\\v'=\sqrt\frac{T}{2m/L}\\v'=\frac{1}{\sqrt2}\sqrt\frac{T}{m/L}\\v'=\frac{v}{\sqrt2}[/tex]

Explanation:

Below is an attachment containing the solution.

The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.

The given expression is 6m(2m + 18). First, we distribute 6m across the terms in the parentheses, resulting in 12m² + 108m. So, an expression that uses the variable 'm' and makes the equation true when the given expression is simplified would be 12m² + 108m. This answer is fully simplified, and the variable 'm' is assumed to be an integer.

The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.

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Answer: 1.176×10^-3 s

Explanation: The time constant formulae for an RC circuit is given below as

t =RC

Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F

t = 56×10^-6 × 21

t = 1176×10^-6

t = 1.176×10^-3 s

Given Information:

Internal resistance of battery = 0.460

Resistance = 21.0 Ω

Capacitance = 56.0 µF

Required Information:

time constant = τ = ?

Answer:

τ = 0.0012

Explanation:

The time constant τ provides the information about how long it will take to charge the capacitor up to certain level.

τ = Req*C

Where Req is the equivalent resistance and C is the capacitance.

Req = R + r

Where R is the resistance of the resistor and r is the internal resistance of the battery.

Req = 21.0 + 0.460 = 21.460 Ω

τ = Req*C

τ = 21.46*56x10⁻⁶

τ = 0.0012

A capacitor approximately charges to 63% in one τ and about 99% in 5τ

After 9 seconds, an object dropped on the moon will have fallen 65.61 meters, be traveling at 14.58 meters per second, with an acceleration of 1.62 meters per second squared. The second derivative of the distance function represents the moon's gravitational acceleration.

An object is dropped on the moon from a height of 200 meters, and the distance function of free-fall due to lunar gravity is given by s(t)=0.81t^2. We are asked to determine various properties of the object's motion after 9 seconds.

a) Distance Fallen

To find how far the object has fallen, we plug t=9 into the distance function to get s(9)=0.81(9)^2=65.61 meters. This is the distance the object has fallen, not the distance from its starting height.

b) Velocity

The object's velocity can be determined by finding the derivative of the distance function, which gives us v(t) = 2(0.81)t = 1.62t. Substituting t=9, we find v(9) = 1.62(9) = 14.58 meters per second.

c) Acceleration

The acceleration of the object is constant and equal to twice the coefficient of t^2 in the distance function, which is 1.62 meters per second squared on the moon.

d) Meaning of the Second Derivative

The second derivative of the free-fall distance function represents the acceleration due to gravity on the moon. It's constant at 1.62 meters per second squared, indicating uniform acceleration during free-fall.

Final answer:

The magnitude of force on the wire segment along the y-axis is 3.0N and directed along the negative x-axis, while there's no force on the segment along the x-axis.

Explanation:

To find the magnitude of the force on the wire, we can use the formula F = I * (L x B), where F is the magnetic force, I is the current, L is the length vector of the wire segment, and B is the magnetic field. The force will be perpendicular to both the current direction and the magnetic field direction, according to the right-hand rule.

The wire is in two segments, one along the x-axis and one along the y-axis. First, we calculate the force on the section of the wire along the x-axis: Fx = I * Lx * B, which is 0 since the magnetic field is in the z-direction and Lx is in the x-direction, making the angle between them 90 degrees. So, there's no force for this segment.

Second, we calculate the force on the section of the wire along the y-axis: Fy = I * Ly * B. The force experienced by this segment can be calculated using the formula with the values provided (I=25A, Ly=3.0m, B=40mT). Therefore, Fy = 25A * 3.0m * 40mT = 3000mN or 3.0N, which will be directed along the negative x-axis, under the right-hand rule.

Answer:

MA = 178 Nm

MB = 72 Nm

MR = 249 Nm

For the moments MA the lever arm was taken as r1 + r2 which is the distance of the point of application of the force at point A to point D.

For the moment MB, the distance of the point of application of force is r1 - r2 which is the distance from the outer ring to the inner ring.

The couple moment is given by F × r1. Which is basically a sum of the moments of both forces applied on the wheel.

Explanation:

See the attachment for detail of the calculation.

The moment MA for the force at point A is 124.55 N*m, the moment MB for the force at point B is 53 N*m, and the resulting couple moment MR is 71.55 N*m.

To calculate the moments MA and MB about point D for the forces applied at points A and B, we can use the formula:

M = F * r

Where M is the moment, F is the force, and r is the radius. For point A, the radius is r1 = 0.470 m and the force is F = 265 N. So, MA = 265 N * 0.470 m = 124.55 N*m (counterclockwise).

Similarly, for point B, the radius is r2 = 0.200 m and the force is F = 265 N. So, MB = 265 N * 0.200 m = 53 N*m (clockwise).

To determine the resulting couple moment MR, we can subtract the clockwise moment (MB) from the counterclockwise moment (MA). MR = MA - MB = 124.55 N*m - 53 N*m = 71.55 N*m (counterclockwise).

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Answer:

Option (b)

Explanation:

As the actor swings on a steel cable, the cable stretches due to the weight of the actor. So, the stress produced by the weight of the actor is longitudinal stress and the strain is also longitudinal in nature.

The modulus of elasticity associated to the normal stress and the longitudinal strain is young's modulus.

Answer:

[tex]6.68\times 10^{15}\ kg[/tex]

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of orbit = 45 km

T = Time period = 1.04 days

Mass of Eros would be given by the following equation

[tex]M=\dfrac{4\pi^2R^3}{GT^2}\\\Rightarrow M=\dfrac{4\pi^2\times (45\times 10^3)^3}{6.67\times 10^{-11}\times (1.04\times 24\times 3600)^2}\\\Rightarrow M=6.68\times 10^{15}\ kg[/tex]

The mass of Eros is [tex]6.68\times 10^{15}\ kg[/tex]

Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

Given Information:  

Diameter of solenoid = d = 1.8 cm = 0.018 m

Length of solenoid = L = 37 cm = 0.37 m

Current = I = 4.4 A  

Magnetic field = B = 0.50 T  

Required Information:  

Number of turns = N = ?  

Answer:  

Number of turns ≈ 33,498 or 33,458

Step-by-step explanation:  

The magnetic field at the center of the solenoid is given by

B = μ₀NI/√ (L²+4r²)

N = B√ (L²+4r²)/μ₀I

Where L is the length and r is the radius of the solenoid, N is the number of turns and B is the magnetic field.

r = d/2 = 0.018/2 = 0.009 m

N = 0.50√ (0.37)²+(4*0.009²)/4πx10⁻⁷*4.4

N ≈ 33,498 Turns

Please note that we can also use a more simplified approximate model for this problem since the length of the solenoid is much greater than the radius of the solenoid

L = 0.37 >> r = 0.009

The approximate model is given by

B = μ₀NI/L

N = BL/μ₀I

N = 0.50*0.37/4πx10⁻⁷*4.4

N ≈ 33,458 Turns

As you can notice the results with the approximate model are very close to the exact model.

Answer:

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

Explanation:

The electric field due to electric potential at a distance Δs is given by

E=ΔV/Δs

We have to find the magnitude electric field in the membrane

Ecell= -ΔV/Δs

[tex]E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m[/tex]

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

The magnitude of the electric field inside the membrane, if it were empty, would be approximately [tex]\( 8.75 \times 10^6 \) V/m.[/tex]

To find the magnitude of the electric field inside the membrane, we can use the formula for electric field strength ( E ) due to a uniform field between two parallel plates, which is given by:

[tex]\[ E = \frac{V}{d} \][/tex]

where ( V ) is the potential difference across the plates (in volts) and ( d ) is the separation between the plates (in meters). In this case, the potential difference ( V ) is given as 70 mV, which we need to convert to volts:

[tex]\[ V = 70 \text{ mV} = 70 \times 10^{-3} \text{ V} \][/tex]

The thickness of the membrane ( d ) is given as 8 nm, which we need to convert to meters:

[tex]\[ d = 8 \text{ nm} = 8 \times 10^{-9} \text{ m} \][/tex]

Now we can plug these values into the formula for the electric field:

[tex]\[ E = \frac{70 \times 10^{-3} \text{ V}}{8 \times 10^{-9} \text{ m}} \][/tex]

[tex]\[ E = \frac{70}{8} \times 10^{(3 - (-9))} \text{ V/m} \][/tex]

[tex]\[ E = 8.75 \times 10^{6} \text{ V/m} \][/tex]

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

[tex]I\propto \dfrac{1}{r^2}[/tex]

Hence,

[tex]I_1r_1^2 = I_2r_2^2[/tex]

[tex]r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}[/tex]

From the question, [tex]I_2[/tex] is half of [tex]I_1[/tex]

[tex]r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}[/tex]

[tex]r_2 = r_1\sqrt{2}[/tex]

[tex]r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}[/tex]

Answer:

Distance ,d= 159.81m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, d,from the source.

Using the equation d/do=sqrt2

d= dosqrt2

Where d=113m

d= 113sqrt2

d= 159.81m

Answer:

 V = k Q/l  ln [(l +√(l² + x²)) / x]

Explanation:

The electrical potential for a continuous distribution of charges is

      V = k ∫ dq / r

Let's apply this expression to our case, define a linear charge density for the bar

           λ = dq / dy

          dq = λ dy

The distance from a point on the bar to the x-axis is

            r = √ (x² + y²)

Let's replace

          V = K ∫ λ dy /√ (x² + y²)

We integrate

         V = k λ ln (y + √ (x² + y²))

Let's evaluate between y = 0 and y = l

         V = k λ [ ln (l +√(x² + l²) - ln x]

We substitute the linear density

           V = k Q/l  ln [(l +√(l² + x²)) / x]

Final answer:

To calculate the electric potential at a point on the x-axis due to a uniformly charged rod, we use the concept of charge density and integrate the contributions from each infinitesimal charge element along the rod with respect to the position.

Explanation:

The student is asking about the electric potential due to a uniformly charged rod on the x-axis. To find an expression for the electric potential at a point on the x-axis, we can imagine dividing the rod into infinitesimally small segments of charge dq. Given that the charge Q is evenly distributed along the rod of length L, the linear charge density λ is Q/L. The potential dV due to a small element dq at a distance x from the rod is given by Coulomb's law as dV = k * dq / r, where k is Coulomb's constant and r is the distance from the charge element to the point on the x-axis.

To find the total potential V, we integrate this expression from one end of the rod to the other. The distance r varies along the rod, so we integrate with respect to y, the position along the rod. Setting up the integral, we have V = k * ∫_{-L/2}^{L/2} (dq / √((L/2 - y)^2 + x^2)) where the limits of integration account for the rod's position along the y-axis with one end at the origin and λ = Q/L. After substituting dq with λdy, performing the integration, and simplifying, we obtain the electric potential V(x) as a function of position along the x-axis.

Answer:

The angle through which each bicycle wheel rotates is 35.03 rad.

Explanation:

Given;

the radius of the circular path, R = 9.30 m

the angular displacement of the bicycle, θ = 1.130 rad

the radius of the bicycle wheel, r = 0.3

S = θR

where;

s is the distance of the circular path

S = 1.13 x 9.3 = 10.509 m

The angle (in radians) through which each bicycle wheel of radius 0.300 m rotates is given as;

θr = 10.509 m

θ = 10.509 / 0.3

θ = 35.03 rad.

Therefore, the angle through which each bicycle wheel of radius 0.300 m rotates is 35.03 rad.

The minimum width b that can be used for the beam is 4.32 in

To determine the minimum width (b) of the solid rectangular wood beam, we employ principles of structural engineering.

The maximum bending moment (M) for a simply supported beam with a uniformly distributed load (\(w_0\)) occurs at the center and is given by [tex]\(M = \frac{w_0 L^2}{8}\).[/tex]

The section modulus (S) for a rectangular cross-section is [tex]\(S = \frac{b \times h^2}{6}\).[/tex]

The bending stress (\(σ_b\)) is given by[tex]\(σ_b = \frac{M}{S}\)[/tex].

Setting \(σ_b\) equal to the allowable bending stress [tex](\(1.95 \, \text{ksi}\))[/tex]and using the specified aspect ratio (h/b = 1.75), we can solve for b.

This ensures that the wood beam meets structural safety criteria.

The calculation yields a minimum width b of 4.32 in, ensuring that the beam is structurally sound, withstanding the specified uniformly distributed load and adhering to the aspect ratio constraint in compliance with the allowable bending stress.

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

Answer:

The boat is approaching the dock at a rate of 2.5 ft/s.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, [tex]\frac{dl}{dt}=2\ ft/s[/tex]

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

[tex]l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)[/tex]

Now, differentiating the above equation with time 't', we get:

[tex]2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)[/tex]

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

[tex]b^2=10^2-36\\\\b=\sqrt{64}=8\ ft[/tex]

Now, substituting all the given values in equation (2) and solve for [tex]\frac{db}{dt}[/tex]. This gives,

[tex]\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s[/tex]

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

Answer:

Explanation:

Resonant frequency is 240

4π² x 240² = 1 / LC

230400π² = 1 / LC

Let the required frequency = n

inductive reactance = 2 πn L

capacitative reactance = 1 /  2 π n C

inductive reactance / capacitative reactance

= 4π² x n ² x LC = 5.68

4π² x n ² = 1 / LC x 5.68

= 230400π²  x 5.68

4n ²= 230400 x 5.68

n ²= 57600 x 5.68

n ² = 327168

n = 572 approx

The generator's frequency in this RLC circuit, where the ratio of inductive to capacitive reactance is 5.68 and the resonant frequency is 240 Hz, is approximately 571 Hz.

To solve this problem, we need to use the relationship between inductive and capacitive reactance in an RLC circuit. Given that the ratio of the inductive reactance (XL) to the capacitive reactance (XC) is 5.68,

Reactance is frequency-dependent, with the following formulas for inductive and capacitive reactance:

Given the resonant frequency as 240 Hz, we start by calculating the resonance reactances:

Using the non-resonant frequency, we use the reactance ratio:

Rearranging gives us:

Since 240 Hz is the resonant frequency, substituting f:

Simplifying yields:

Therefore, the generator frequency is approximately 571 Hz.

Answer:

Check Explanation.

Explanation:

Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s

a) Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s

Which is equal to the momentum before collision, hence, momentum is conserved.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s

This is equal to the momentum before collision too, hence, momentum is conserved.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s

This is equal to the momentum before collision, hence, momentum is conserved.

b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.

Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.

Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J

Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J

Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.

In this scenario,

Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J

This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.

But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.

Momentum after collision = (2+2)(1.414) = 5.656 kgm/s

which is not equal to the 4.0 kgm/s obtained before collision.

This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.

Given Information:

frequency = 1240 Hz

width = a = 1.13 m

speed of sound = c = 344 m/s

Required Information:

angle = θ = ?

Answer:

θ = 14.18 rad

Explanation:

We can find out the angle relative to the centerline perpendicular to the doorway by using the following relation

sin(θ) = λ/a

Where λ is the wavelength of the sound wave and a is width

λ = c/f

Where c is the speed of the sound and f is the frequency

λ = 344/1240

λ = 0.277

sin(θ) = λ/a

θ =sin⁻¹(λ/a)

θ =sin⁻¹(0.277/1.13)

θ =sin⁻¹(0.277/1.13)

θ = 14.18 rad

Answer:

W = N!/(n0! * n1!)

Explanation:

Let n0 = number of particles in the lowest energy state

n1 = number of particles in the excited energy state.

Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

Answer:

The conditions are not given in the question. Here is the complete question.

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘

a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?

b) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius ?

Explanation:

Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.

a)

Heat required to convert -15°C ice to 0°C.

  [tex]ms_{ice}[/tex]ΔT = (1.0)(2.100×10³)(15) = 3.150×[tex]10^{4}[/tex]J

Heat required to convert 1.0 kg ice to water.

    [tex]mL_{ice} = 1[/tex]×[tex]3.33[/tex]×[tex]10^{5}[/tex] = 3.33×[tex]10^{5}[/tex]J

Heat required to convert 1.0 kg water at 0°C to 37°C.

   [tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×37 = 1.548×[tex]10^{5}[/tex] J

Total heat required = 3.150×[tex]10^{4}[/tex] + 3.33×[tex]10^{5}[/tex] + 1.548×[tex]10^{5}[/tex]

                              = 5.19×[tex]10^{5}[/tex] J

b)

Heat required to warm 1.0 kg water at 2°C to water at 37°C.

[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×35

                = 1.465×[tex]10^{5}[/tex]J

At the start of meiosis II, the cell exhibits characteristics of a haploid cell preparing for mitosis. It has one set of homologous chromosomes, each with two chromatids, equivalent to a haploid cell in the G₂ phase of interphase.

At the beginning of meiosis II, a cell appears much like a haploid cell preparing to undergo mitosis. After completion of meiosis I, the cell does not duplicate its chromosomes, so at the onset of meiosis II, each dividing cell has only one set of homologous chromosomes, each with two chromatids. This results in half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of chromosomal content, cells at the start of meiosis II are similar to those of haploid cells in the G₂ phase of interphase, where they are preparing for mitosis.

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Explanation:

Given:

m = 1.673 × 10^-27 kg

Q = q = 1.602 × 10^-19 C

r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2

a = 2.45 × 10^17 m/s^2.

Answer:

Explanation:

velocity of electron V = 4 x 10³ m/s

magnetic field B = 1.25 T .

magnetic force = 1.4 x 10⁻¹⁶ N.

If direction of velocity makes angle θ with magnetic field

magnetic force = magnetic field x charge on electron x velocity x sinθ

1.4 x 10⁻¹⁶ = 1.25 x 1.6 x 10⁻¹⁹ x 4 x 10³ x sinθ

1.4 x 10⁻¹⁶ = 8  x 10⁻¹⁶ x sinθ

sinθ  = 1.4 / 8

= .175

θ = 10 degree

or 180 - 10

= 170 degree

because

sinθ = sin (180 - θ)

Answer:

F = 100.0 N

Explanation:

       [tex]F_{c} = m *\frac{v^{2} }{r} = 2.0 kg *\frac{(5.0m/s)^{2} }{0.5m} =100.0 N[/tex]

100.0N

As the ball moves round the circle, a centripetal acceleration which is directed towards the center of the circle will keep the ball from falling off. This acceleration produces a force called centripetal force (F).

Since this is the only force acting on the ball, then the net force acting on the ball to keep it moving round the circle is the centripetal force.

F = m a         [according to Newton's second law of motion]    --------------(i)

Where;

m = mass of the ball.

a = centripetal acceleration. = [tex]\frac{v^2}{r}[/tex]

v = speed of the ball.

r = radius of the circle.

Substitute a = [tex]\frac{v^2}{r}[/tex] into equation (i) as follows;

F = m x [tex]\frac{v^2}{r}[/tex]      --------------------(ii)

From the question;

m = 2.0kg

v = 5.0m/s

r = diameter / 2     [diameter = 1.0m]

r = 1.0 / 2

r = 0.5m

Substitute these values into equation (ii) as follows;

F = 2.0 x [tex]\frac{5.0^{2} }{0.5}[/tex]

F = 2.0 x [tex]\frac{25.0}{0.5}[/tex]

F = 2.0 x 50.0

F = 100.0N

Therefore, the magnitude of the net force on the ball is 100.0N