A string is stretched to a length of 212 cm and both ends are fixed. If the density of the string is 0.02 g/cm, and its tension is 357 N, what is the fundamental frequency? Course hero

Answer:

22.5 m/s

Explanation:

Applying Newton's third law of motion

Momentum of the cannon and cart = momentum of the cannonball

MV = mv..................... Equation 1

Where M = mass of the cannon and the cart, V = Recoil velocity of the cannon and the cart, m = mass of the cannonball, v = velocity of the cannonball

make v the subject of the equation

v = MV/m............. Equation 2

Given: M = 150 kg, V = 1.5 m/s, m = 10 kg

Substitute into equation 2

v = 150(1.5)/10

v = 22.5 m/s

Hence the cannonball leave the cannon with a velocity of 22.5 m/s

Answer:

v2 = 22.5 m/s

Explanation:

Momentum is how hard to stop or turn a moving object . Generally, momentum measures mass in motion. Momentum is a vector quantity. Mathematically,

p = mass ×  velocity

The total momentum of an isolated system of  bodies remains constant.

mometum before = 0

mass of the canon (m1) = 150 kg

mass of the ball (m2) = 10 kg

velocity of the ball (v2) = ?

velocity of the cannon(v1) = 1.5 m/s

momentum after = momentum before

m2v2 + m1v1 = 0

10v2 = 150 × 1.5

10v2 = 225

divide both sides by 10

v2 = 225/10

v2 = 22.5 m/s

Explanation:

Given that,

Number of sodium ions at the negative electrode, [tex]Na^+=2.68\times 10^{16}[/tex]

Number of chloride ions at the positive electrode, [tex]Cl^-=3.92\times 10^{16}[/tex]

(a) The current flowing in the circuit is due to the positive as well as negative charges such that total charge becomes:

[tex]Q=(Na^++Cl^-)e[/tex]

[tex]Q=(2.68\times 10^{16}+3.92\times 10^{16})(1.6\times 10^{-19})[/tex]

Q = 0.01056 C

The current is given by :

[tex]I=\dfrac{Q}{t}[/tex]

[tex]I=\dfrac{0.01056}{1}=10.56\ mA[/tex]

So, the current passing between the electrodes is 10.56 mA.

(b) The direction of electric current is towards negative electrodes.

Explanation:

(a)   First, we will calculate the charge of sodium ions as follows.

              q = ne

                  = [tex]2.68 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]

                  = [tex]4.288 \times 10^{-3} C[/tex]

Now, charge of chlorine ions is calculated as follows.

            q' = ne

                = [tex]3.92 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]

                = [tex]6.272 \times 10^{-3} C[/tex]

Hence, the current will be calculated as follows.

             i = [tex]\frac{q}{t} + \frac{q'}{t}[/tex]

               = [tex]\frac{4.288 \times 10^{-3} C}{1.00} + \frac{6.272 \times 10^{-3} C}{1.00}[/tex]

               = [tex]10.56 \times 10^{-3} A[/tex]

               = 10.56 mA

Therefore, current passing between the electrodes is 10.56 mA.

(b)   Since, positive ions are moving towards the negative electrode. And, current is the flow of ions or electrons therefore, the direction of current is towards the negative electrode.

Answer:

0.0800 is time constant for slope.

Explanation:

See attached pictures for explanation.

To find the time constant from the slope, use the formula RC = -1/slope, where R is the resistance and C is the capacitance. Calculate the slope by taking the ratio of voltage change to time change between two data points.

To determine the time constant from the slope, we can use the formula:

RC = -1/slope

where R is the resistance in the circuit and C is the capacitance of the capacitor.

In this case, since we are only given the voltage and time data, we need to find the slope by taking the ratio of the change in voltage to the change in time between any two points.

Let's take the first and second data points:

Slope = (V2 - V1) / (t2 - t1)

          = (3.0000 V - 1.4790 V) / (0.036894 s - 0.015843 s)

Now, calculate the slope:

Slope = (1.5210 V) / (0.021051 s)

Slope ≈ 72.27 V/s

Once we have the slope, we can plug it into the formula RC = -1/slope to find the time constant:

RC = -1 / (72.27 V/s)

Calculate RC:

RC ≈ -0.0138 s/V

So, the time constant (τ) is approximately 0.0138 seconds per volt (s/V). This value represents the product of resistance and capacitance in the circuit.

https://brainly.com/question/34087350

#SPJ3

Final answer:

The average force exerted by the bat on the ball is 2250 N.

Explanation:

To calculate the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in its momentum. We can calculate the initial momentum by multiplying the mass of the ball by its initial velocity, and similarly, we can calculate the final momentum using the mass and final velocity. By subtracting the initial momentum from the final momentum, we get the change in momentum. Finally, dividing the change in momentum by the time of contact gives us the average force.

Using the given values, the initial momentum of the ball is (0.15 kg) × (-40 m/s) = -6 kg·m/s, and the final momentum is (0.15 kg) × (50 m/s) = 7.5 kg·m/s. The change in momentum is 7.5 kg·m/s - (-6 kg·m/s) = 13.5 kg·m/s. Dividing this by the time of contact, 6.0 × 10^-3 s, gives us an average force of 2250 N.

Answer:

The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s

Explanation:

If we make a comparison we have to:

y = A*(1-e^-(C*x)) + B

If the time remains constant we have to:

t = R*C = 1/C

In this way we calculate the time constant and its uncertainty. this will be equal to:

t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s

Answer:

So, Magnitude of the field (B) = 2.09 x 10^(-2) T.

It's in the negative direction since acceleration is in positive direction.

Explanation:

First of all, since acceleration is in positive x- direction, the magnetic field must be in negative y- direction.

We know that The magnitude of the Lorentz force F is; F = qvB sinθ

So, B = F/(qvsinθ)

F = ma.

Speed(v) = 1.00 x10^(7) m/s

acceleration (a) = 2.00 x10^(13) m/s^(2)

Mass of proton = 1.673 × 10^(-27) kilograms

q(elementary charge of proton) = 1.602×10^(−19)

Since right hand thumb rule, θ= 90°

So;B = [1.673 × 10^(-27) x 2.00 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]

So,B = 2.09 x 10^(-2) T.

It's in the negative direction since acceleration is in positive direction.

The magnitude of the magnetic field that the proton experiences is 2.08 x 10^-4 Tesla. The direction of the magnetic field, determined by the right-hand rule, is in the negative y-direction.

The force experienced by a charged particle moving within a magnetic field, like the proton in your question, can be calculated using the Lorentz force law: F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Because the proton is moving perpendicular to the field, θ = 90°, and sinθ = 1. Therefore, F = qvB. To find the magnetic field, B, we can rearrange this formula to give B = F/qv.

Now, we know from Newton's second law that F = ma, so we can replace F in our formula with the proton's mass (m, which is 1.67 x 10^-27 kg for a proton) times the given acceleration (a), yielding B = ma/qv.

Substitute the given values for acceleration, proton's charge, and velocity into our equation, we get: B = (1.67 x 10^-27 kg)(2 x 10^13 m/s^2) / (1.60 x 10^-19 C)(1 x 10^7 m/s) = 2.08 x 10^-4 Tesla (T).

The direction of the magnetic field is given by the right-hand rule as discussed in Essential Knowledge 2.D.1. If you arrange your right hand such that your thumb points in the direction of the proton's velocity (positive z-direction) and your fingers curl in the direction of the force (positive x-direction), your palm points in the direction of the magnetic field, which would be in the negative y-direction.

https://brainly.com/question/36936308

#SPJ3

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

    [tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]

where [tex]V_{s}[/tex] is the voltage induced in the secondary coil

           [tex]V_{p}[/tex] is the voltage in the primary coil

          [tex]N_{s}[/tex] is the number of turns of secondary coil

         [tex]N_{p}[/tex] is the number of turns of primary coil

From the given question,

    [tex]\frac{487*10^{-3} }{120}[/tex] = [tex]\frac{N_{s} }{2464}[/tex]

⇒    [tex]N_{s}[/tex] = [tex]\frac{2462*487*10^{-3} }{120}[/tex]

            = 9.999733

  ∴   [tex]N_{s}[/tex] = 10 turns

Answer:

607,145 turns

Explanation:

Output voltage, that is secondary voltage,Es = 120 volts

Input voltage, that is primary voltage, Ep = 487/1000 = 0.487 volts

Number of turns in secondary = Ns

Number of turns in primary, Np = 2464

∴ Es/Ep = Ns/Np

Ns = Es * Np/Ep = 1`20 X 2464/0.487 =  607,145 turns ( Step up transformer)

Answer:

Φ = 0.469 Wb

Explanation:

Given,

N = 230 turns

Radius, r = 10 cm

Magnetic field, B = 0.0650 T

Magnetic flux  = ?

now,

Φ = NBA

Φ = 230 x 0.0650 x π x r²

Φ = 230 x 0.0650 x π x 0.1²

Φ = 0.469 Wb

Hence, the magnetic flux is equal to Φ = 0.469 Wb

The magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m². when A circular coil that has N = 230 turns and a radius of r = 10.0 cm.

Given:

N = 230 turns

r = 10.0 cm = 0.1 m

B₀ = 0.0650 T

The magnetic flux through a coil can be calculated using the formula:

ΦB = B₀ × A × N,

The area of the coil, we can use the formula for the area of a circle:

A = π × r²,

Let's calculate the magnetic flux:

A = π × r² = 3.14159 × (0.1)² = 0.0314159 m²

ΦB = B₀ × A × N = 0.0650 × 0.0314159 × 230 = 0.04958735 T·m²

Therefore, the magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m².

To know more about magnetic flux:

https://brainly.com/question/1596988

#SPJ6

Answer:

I₁ = 0.32 A

I₂ = 0.16 A

I₃ = 0.16 A

I₄ = 0.16 A

Explanation:

Part A

The equivalent resistance of the circuit is

Req = R₁ + (R₂||(R₃ + R₄))

Req = 15 + (45||(20 + 25))

Req = 15 + (45||45) = 15 + ((45×45)/(45+45)) = 15 + 22.5 = 37.5 Ω

Part B

From Ohm's law,

V = IR

Ieq = V/(Req) = 12/(37.5) = 0.32 A

Part C

Current through R₁ is the same as Ieq as R₁ is directly in series with the voltage source.

I₁ = 0.32 A

Then, this current flows through the (R₂||(R₃ + R₄)) loop too as the entire loop is in series with R₁

This current is them split into two branches of R₂ and (R₃ + R₄), since these two branches have equal resistances (45 Ω and 45 Ω), 0.32 A is split equally between the R₂ and (R₃ + R₄) branch.

Current through R₂ (using current divider)

I₂ = (45/90) × 0.32 = 0.16 A

Current through (R₃ + R₄) = 0.16 A too.

And because the two resistors are in series, the same current flows through them.

I₃ = I₄ = 0.16 A

First, we resolve the parallel circuits using the formula for resistors in parallel to get the equivalent resistance, Rp. We then substitute this into the series circuits to get the total equivalent resistance, Req. Using Ohm's Law, we can calculate the equivalent current, Ieq.

To solve this, we first need to determine the equivalent resistance of the whole circuit. As it is a combination of parallel and series circuits, we have to start by replacing the resistors in parallel. Using the formula for resistors in parallel, 1/R = 1/R2 + 1/R3, the parallel combination of R2 and R3 can be replaced by one resistor with resistance Rp = 1/ ((1/R2) + (1/R3)). This gives Rp = 13.333Ω. Now we have a simple series circuit with R1, Rp, and R4. The equivalent resistance of the whole circuit is then Req = R1 + Rp + R4 = 15.0Ω + 13.333Ω + 25.0Ω = 53.333Ω.

In part B, we apply Ohm's Law: I = E/R to get the equivalent current through the circuit. Using our values for E (12V) and Req (53.333Ω), we find Ieq = 0.225A.

https://brainly.com/question/14993917

#SPJ11

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

Final answer:

The acceleration at the end of a test tube 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm is calculated to be 15051.2 m/s², demonstrating the significant centrifugal forces generated by such devices.

Explanation:

To calculate the acceleration at the end of a test tube that is 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm, we first need to convert the rotational speed to radians per second. The formula to convert revolutions per minute (rpm) to radians per second (ω in rad/s) is ω = (2π×rpm)/60. Thus, ω = (2π× 3700)/60 = 387.98 rad/s. Next, we use the formula for centripetal acceleration, a = ω²×r, where r is the radius of the circle (distance from the center of rotation to the point of interest) in meters. Given that r = 10 cm = 0.1 m, the acceleration a = (387.98)^2× 0.1 = 15051.2 m/s².

This centripetal acceleration is much larger than Earth's gravitational acceleration, indicating the extreme forces at play in a centrifuge's operation, which is crucial for its role in laboratory settings for sedimentation of materials.

Final answer:

The diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.

Explanation:

To calculate the diameter of the circular contact area, we need to determine the radius of the contact area first. The radius can be found by dividing the diameter of the ball by 2. In this case, the radius is 15 mm.

The area of the circular contact area can be calculated using the formula A = πr^2, where A is the area and r is the radius. The maximum pressure at the center of the contact area can be found by dividing the force applied on the plate by the area of the contact area.

Using these formulas, the diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.

Explanation:

Below is an attachment containing the solution.

Answer:

Explanation:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

(a) 1.2 J

(b) 5 m/s

(c)  6.3 J

Explanation:

(a) Kinetic energy at A

Ek = 1/2mv².................. Equation 1

Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.

Given: m = 0.6 kg, v = 2.0 m/s

Substitute into equation 1

Ek = 1/2(0.6)(2²)

Ek = 0.6(2)

Ek = 1.2 J.

(b)

Speed at point B

Ek' = 1/2mv'²............... Equation 2

Make v' the subject of the equation,

v' = √(2Ek'/m).................. Equation 3

Where, Ek' = kinetic energy at B, v' = velocity at B.

Given: Ek' = 7.5 J, m = 0.6 kg,

Substitute into equation 3

v' = √[(2×7.5)/0.6]

v' =√(15/0.6)

v' = √25

v' = 5 m/s.

(c)

Wt =  Δ kinetic energy from A to B

Where Wt = total work done as the particle moves from A to B.

Wt = 1/2m(v'²-v²)

Wt = 1/2(0.6)(5²-2²)

Wt = 0.3(25-4)

Wt = 0.3(21)

Wt = 6.3 J

Answer:

A three times as large as the initial value

Answer:

Change in momentum =15.01kgm/s

Impulse applied by bat = 15.01Ns

Explanation:

Please see attachment below.

Answer:

V = 2.8cm/s

Explanation:

Please see attachment below.

This problem involves the concept of doppler effect.

Answer:

The speed at which the blood is flowing in cm/s = 1.4 cm/s

Explanation:

emitted frequency ( f ) = 1.1 * 10^6 Hz

detected frequency ( F ) = 21 Hz

speed of sound in tissues ( c ) = 1475 m/s

speed ( V ) = ?

To calculate for speed of blood flowing we apply the detected frequency formula :

F = [tex]\frac{2fV}{c}[/tex]

21 = ( 2* 1100000* V ) / 1475

therefore V = (21 * 1475) / (2 * 1100000)

                V = 30975 / 2200000

                 V = 0.0140 m/s = 1.4 cm/sec

Answer:

Option E is correct.

The two objects rise to the same height.

Explanation:

Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.

Let the mass, velocity and height the object will rise to be m, v and h respectively.

Note that V = v

Using the work energy theorem

The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)

(The final kinetic energy = 0 J because the object comes to rest at the final point)

Work done on the bigger object = work done by all the forces acting on the body

But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH

(-MV²/2) = - MgH

H = (V²/2g)

For the small body,

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh

(-mv²/2) = - mgh

h = (v²/2g)

Since V = v as given in the question (Both bodies have the same speeds)

H = h = (V²/2g) = (v²/2g)

Hope this Helps!!!

Answer: F = 113.4.[tex]10^{-3}[/tex]N

Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is

F = Fe + Fm

Force due to electric field

To determine this force:

Fe = q.E, where q is the charge and E is electric field.

Calculating:

Fe = q.E

Fe = 1.8.[tex]10^{-6}[/tex].4.6.[tex]10^{3}[/tex]

Fe = 8.28.[tex]10^{-3}[/tex]N

Force due to magnetic field: It can only happens when the charge is in movement, so

Fm = q.(v×B), where v represents velocity and B is magnetic field

The cross product indicates that force is perpendicular to the velocity and the field.

Calculating:

Fm = q.v.B.senθ

As θ=90°,

Fm = q.v.B

Fm =  1.8.[tex]10^{-6}[/tex].3.1.[tex]10^{6}[/tex].1.2.[tex]10^{-3}[/tex]

Fm = 6.696.[tex]10^{-3}[/tex]N

F, Fm and Fe make a triangle. So, using Pythagorean theorem:

F = [tex]\sqrt{Fe^{2} + Fm^{2} }[/tex]

F = [tex]\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }[/tex]

F = 113.4.[tex]10^{-3}[/tex]N

The net force acting on the charge is F = 113.4.[tex]10^{-3}[/tex]N

Answer:

[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]

Explanation:

The initial momentum of the racquet ball is:

[tex]||\vec p || = (0.238\,kg)\cdot (12.4\,\frac{m}{s} )[/tex]

[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=mgl[/tex]

[tex]v=\sqrt{2gl}[/tex]

Put the value into the formula

[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]

[tex]u=3.13\ m/s[/tex]

The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]

The initial speed of the block [tex]u_{2}=0[/tex]

(a). We need to calculate the speed of the ball after collision

Using formula of collision

[tex]v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}[/tex]

Put the value into the formula

[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]

[tex]v_{1}=-2.01\ m/s[/tex]

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

[tex]v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}[/tex]

Put the value into the formula

[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]

[tex]v_{2}=1.11\ m/s[/tex]

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Answer with Explanation:

We are given that

Mass of block=m=4.9 kg

Initial velocity, u=0

[tex]F=(2.6-x^2) i N[/tex]

Initial position, x=0

a.We have to find the kinetic energy of the blocks as it passes through x=2.1 m

Work done=Kinetic energy=[tex]\int_{0}^{2.1}(2.6-x^2) dx[/tex]

Kinetic energy of the block=[tex][2.6 x-\frac{x^3}{3}]^{2.1}_{0}[/tex]

Kinetic energy of the block=[tex]2.6\times 2.1-\frac{(2.1)^3}{3}-0=2.373 J[/tex]

Kinetic energy of the block=2.373 J

b.Initial kinetic energy of block=[tex]K_i=\frac{1}{2}(4.9)(0)=0[/tex]

According to work energy theorem

[tex]W=K_f-K_i[/tex]

[tex]2.373 =k_f-0[/tex]

[tex]k_f=2.373 J[/tex]

Hence, the maximum kinetic energy of the block =2.373 J

Answer:

5.51 m

Explanation:

From the question,

The energy used to stretch the spring = the potential energy of the rock.

(1/2)ke²  = mgh ................. Equation 1

Where k = spring constant, e = extension/compression, m = mass of the rock, g = acceleration due to gravity, h = height of the rock above the ground

make h the subject of the equation.

h = ke²/2mg ....................equation 2

Given: k = 5400 N/m, e = 1 m, m = 48 kg.

Constant: g = 9.8 m/s²

Substitute into equation 2

h = 5400(1²)/(2×48×9.8)

h = 5400/940.8

h = 5.51 m.

Hence the height of the rock = 5.51 m

Answer:

a) 2 x10^7 eV

b) 2 x10^4 keV

c) 20 MeV

d) 0.02 Gev

e) 3.2 x 10^-12J

Explanation:

The potential difference = 20 x 10^6 V

The charge on the proton = 1.6 x10^-19

The work done to move the proton will be basically the proton will acquire if it accelerates.

Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19

                                                 =3.2 x 10^-12J or 2 x10^7 eV

2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev

Explanation:

Below is an attachment containing the solution.

Answer:

Since at 20V most of the photons released are red then when the voltage keeps increasing the hotter the filament will be, therefore the color of light will be bright red.

Explanation:

The higher the energy the more the electrons in the molecules of the object will be excited, and when they de-excite to their ground states they release energy in the form of infrared light. The increase in voltage and higher temperatures make the object release brighter color and sometimes at the highest temperatures +1400 degrees Celsius, the color glows hot white.

Answer:

0.82 mm

Explanation:

The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:

[tex]y_n=\frac{n \lambda D}{d}[/tex]

so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:

[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]

where;

n = 2

[tex]\lambda_1[/tex] = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]

[tex]y_2[/tex] = 0.00276 m

[tex]y_2[/tex] = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:

[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]

[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

[tex]\delta y = y_2-y^{'}_2[/tex]

[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)

[tex]\delta y[/tex] = 10 ⁻³ (0.82)

[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m

[tex]\delta y[/tex] =  0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]

[tex]\delta y[/tex] = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

Answer:

A plane of infinite length and width.

Explanation:

For the case of the sphere, we should consider spherical coordinates.

You can move around the sphere without changing your distance from the center of the sphere, however you can alter your azimuthal angle and polar angle, since they are symmetric in spherical coordinate system.

r --> Cannot change

Φ --> Free to change

θ --> Free to change

For the case of the infinite cylinder, we should consider cylindrical coordinates.

You can change your height and angular coordinate, but you cannot change your distance from the axis of the cylinder.

r --> Cannot change

θ --> Free to change

z --> Free to change

For the case of the infinite plane, we should consider cartesian coordinates.

Since the length and width of the plane is infinite, we cannot recognize whether we are getting closer or further away.

x --> Free to move

y --> Free to move

z --> Free to move

Therefore, in the case of infinite plane you have the most freedom to move about without your view of the object changing.

Final answer:

An infinite plane allows the most freedom of movement without changing the view, due to its infinite symmetry. In contrast, a sphere and a cylinder have more limited symmetrical properties, resulting in a restricted freedom of movement to keep the view unchanged.

Explanation:

When comparing the freedom to move above a sphere, an infinitely long cylinder, and an infinite plane, the plane provides the most freedom without changing your view of the object. On a sphere, any move results in a different orientation or distance from the surface. With the cylinder, movement along the axis does not change the view, but any other direction does. However, the infinite plane allows movement in any direction on a two-dimensional plane without changing the viewed geometry or orientation.

From a mathematical perspective, this question revolves around symmetry and geometric invariance under transformation. The infinite plane exhibits infinite symmetry; therefore, no matter how you move parallel to its surface, the view remains unchanged. Conversely, a sphere has rotational symmetry around its center, and a cylinder along its axis, limiting the directions in which one can move without altering the perceived distance or orientation.

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

The apparent depth of a swimming pool is measured by considering the water's refractive index. With the pool completely filled, the bottom appears to be 2.25m deep. When halfway filled, the pool appears to be 2.625m deep.

When light travels from a medium with a high refractive index to one with a lower refractive index, the light is refracted, or bent, making objects appear closer than they actually are. We can calculate this apparent depth by using the formula d' = d / n, where d' is the apparent depth, d is the actual depth, and n is the refractive index.

(a) If the pool is completely filled with water, for a person looking from above ground, the bottom of the pool appears to be closer than it actually is. Substituting the given values into the formula, we get the apparent depth: d' = 3.00m / 1.333 = 2.25 m below ground level.

(b) If the pool is halfway filled with water, the apparent depth of the water is calculated in the same way. However, the depth beneath the water is below the refractive index border and is not subject to refraction. Therefore, the apparent total depth of the partially filled pool is the sum of the actual depth of the air part (1.50m) and the apparent depth of the water part (1.50m / 1.333 = 1.125m). This gives us 2.625m.

https://brainly.com/question/32255407

#SPJ3

To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,

[tex]E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}[/tex]

Here,

[tex]k_e[/tex] = Coulomb constant

q = Charge

x = Distance to the axis

r = Distance between the charges

Our values are given as,

[tex]r = 10.0cm = 10*10^{-2} m[/tex]

[tex]q = 75\mu C = 75*10^{-6} C[/tex]

[tex]x_a = 1.00cm[/tex]

[tex]x_b = 5.00cm[/tex]

[tex]x_c = 30cm[/tex]

[tex]x_d = 100cm[/tex]

So applying this to our 4 distances, we have

PART A)

[tex]E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_a = 6.64*10^6N/C \hat{i}[/tex]

PART B)

[tex]E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_b = 24.1*10^6N/C \hat{i}[/tex]

PART C)

[tex]E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_c = 6.39*10^6N/C \hat{i}[/tex]

PART D)

[tex]E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_d = 0.664*10^6N/C \hat{i}[/tex]

Answer:

Parallel, R2, R1, series

Explanation:

From Ohm's law, V=IR hence making current the subject of the formula then [tex]I=\frac {V}{R}[/tex]

Since R1>R2 eg 4>2 then current for individual connection of R2 will be greater than that for R1 for example, assume V is 8 then if R2 we will have 8/2=4 A but for R1 we shall have 8/4=2 A.

When in parallel, the equivalent resistance will be given by [tex]\frac {1}{R1}+\frac {1}{R2}[/tex] for example here it will be 1/4+1/2=3/4. Still taking V of 8 then I= 8/(3/4)=10.667 A

When in series connection, equivalent resistance is given by adding R1 and R2 hence using the same figures we shall have 4+2=6 hence I=8/6=1.33A

We can conclude that the arrangement of current from greatest will be parallel, R2, R1, series

According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".

By using Ohm's law

→ Voltage (V) = Current (I) × Resistance (R)

or,

→ I = [tex]\frac{V}{R}[/tex]

Since R1 > R2

Let, Current (V) = 8 then the resistance will be:

R2 = [tex]\frac{8}{2}[/tex] = 4 A

and,

R1 = [tex]\frac{8}{2}[/tex] = 4 A

When the resistance is in parallel connection,

= [tex]\frac{1}{R1} + \frac{1}{R2}[/tex]

= [tex]\frac{1}{4} +\frac{1}{2}[/tex]

By taking L.C.M, we get

= [tex]\frac{1+2}{4}[/tex] = [tex]\frac{3}{4}[/tex] and,

The current be:

→ I = [tex]\frac{8}{\frac{3}{4} }[/tex] = 10.667 A

When the resistance is in series connection,

= R1 + R2

= 4 + 2

= 6 and,

The current be:

→ I = [tex]\frac{8}{6}[/tex] = 1.33 A

Thus the response above is appropriate.

Find out more information about resistor here:

https://brainly.com/question/24858512

Answer:

The stream function for this potential flow is -4696.8

Explanation:

Given that,

Circulation = 8000 m²/s

Radius = 40 m

Pressure = 2 kPa

Suppose the determine the stream function for this potential flow

We need to calculate the stream function

Using formula of stream function

[tex]\Psi=-(\dfrac{\Gamma}{2\pi})ln(r)[/tex]

Where, Γ = circulation

r = radius

Put the value into the formula

[tex]\Psi=-\dfrac{8000}{2\pi}\times ln(40)[/tex]

[tex]\Psi=-4696.8[/tex]

Hence, The stream function for this potential flow is -4696.8