A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2, where xo and yo are measured in feet and t is measured in seconds. The angular displacement of a radial line measured from a vertical reference line is θ = 8t4, where θ is measured in radians. Determine the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s.

Answer:

200 nm

Explanation:

Refractive index of gasoline = 1.4

Wavelength = 560 nm

t = Thickness of film

m = Order = 1

Wavelength is given by

[tex]\lambda=\dfrac{560}{1.4}=400\ nm[/tex]

We have the relation

[tex]2t=m\lambda\\\Rightarrow t=\dfrac{m\lambda}{2}\\\Rightarrow t=\dfrac{1\times 400}{2}\\\Rightarrow t=200\ nm[/tex]

The thickness of the film is 200 nm

Answer:

is closest to  100*C  temperature  at the aluminum-copper junction

Explanation:

The expression for calculating the resistance of each rod is given by

[tex]R =\frac{ L}{ kA}[/tex]

Now; for Aluminium

[tex]R_{al} =\frac{ 0.20 }{ 205*0.0004}[/tex]

[tex]R_{al}[/tex] = 2.439

For Copper

[tex]R_{Cu}=\frac{0.70}{385*0.0004}[/tex]

[tex]R_{Cu} = 4.545[/tex]

Total Resistance [tex]R = R_{al} + R_{Cu}[/tex]

= 2.439 + 4.545

= 6.9845

Total temperature  difference = 230*C + 30*C

= 200 *C

The Total rate of heat flow is then determined which is  = [tex]\frac{ total \ temp \ difference}{total \ resistance }[/tex]

=[tex]\frac{200}{ 6.9845 }[/tex]

= 28.635 Watts

However. the temperature difference across the aluminium = Heat flow × Resistance of aluminium

= 28.635 × 2.349

= 69.84 *C

Finally. for as much as one end of the aluminium is = 30 *C , the other end is;

=30*C + 69.84*C  

= 99.84  *C

which is closest to  100*C  temperature  at the aluminum-copper junction

Answer:

a) [tex]\Delta \theta = 617.604\,rad[/tex], b) [tex]\Delta t = 10.392\,s[/tex], c) [tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

Explanation:

a) The final angular speed at the end of the acceleration stage is:

[tex]\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)[/tex]

[tex]\omega = 111.5\,\frac{rad}{s}[/tex]

The angular deceleration is:

[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}[/tex]

[tex]\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}[/tex]

[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

The change in angular position during the acceleration stage is:

[tex]\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}[/tex]

[tex]\theta = 177.604\,rad[/tex]

Finally, the total change in angular position is:

[tex]\Delta \theta = 440\,rad + 177.604\,rad[/tex]

[tex]\Delta \theta = 617.604\,rad[/tex]

b) The time interval of the deceleration interval is:

[tex]\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }[/tex]

[tex]\Delta t = 7.892\,s[/tex]

The time required for the grinding wheel to stop is:

[tex]\Delta t = 2.50\,s + 7.892\,s[/tex]

[tex]\Delta t = 10.392\,s[/tex]

c) The angular deceleration of the grinding wheel is:

[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

Answer:

Ф_T =544

The wheel will stop after 10 s.  

α_z = -11.15 rad/s^2

Explanation:

The angular acceleration is constant. Thus, we will apply the equations of rotation with constant angular acceleration model.

(a) In order to calculate the total angle, we will divide the entire interval from t = 0 to the time the wheel stops into two intervals.  

From t = 0 to t = 2 s:  

Ф-Ф_o =1/2(w_0z+w_z)t                       (1)

We will calculate w_z first:  

w_z = w_0x +α_xt

w_z = 24 + (35)(2.5)

w_z = 111.5

Substitute w_x into Eq.1  

Ф-Ф_o = 1/2(24+111.5)(2)

Ф-Ф_o = 136 rad

We can calculate it directly from the following equation:  

Ф-Ф_o = w_0x*t+1/2a_xt^2

Ф-Ф_o = 24*2.5+1/2*35*2.5

Ф-Ф_o =103.75 rad

Thus, the total angle the wheel turned between t = 0 and the time it stopped:  

Ф_T =103.75 +440

Ф_T =544

(b)  

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.  

w_oz = 111.5 rad/s  

wz = 0 the wheel will stop at the end. 1

Ф-Ф_o =1/2(w_0z+w_z)t    

440 = 1/2(111.5+0)*t

t = 8s

Adding t of the first interval to t of the second interval :  

t_T = 2 + 8  =10

The wheel will stop after 10 s.  

(c)  

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.  

w_oz = 111.5 rad/s  

w_z = 0  

 w_x =w_oz+α_z*t

α_z = -11.15 rad/s^2

Answer:

3 Volts

Explanation:

Using Ohm's Law

         Voltage=current x resistance

              V = I x R

              V = 1.5 x 2

              V = 3 V

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g=[tex]426\times 10^{-3} kg[/tex]

1 kg=1000 g

Radius,r=[tex]\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m[/tex]

1m=100 cm

Height,h=5m

[tex]I=\frac{2}{2}mr^2[/tex]

a.By law of conservation of energy

[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh[/tex]

[tex]\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh[/tex]

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

[tex]gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2[/tex]

[tex]\omega^2=\frac{6}{5r^2}gh[/tex]

[tex]\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s[/tex]

Where [tex]g=9.8m/s^2[/tex]

b.Rotational kinetic energy=[tex]\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J[/tex]

Rotational kinetic energy=8.35 J

Answer:

Explanation:

This question pertains to resonance in air column.  It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

Answer:

They are called absorption lines

Explanation:

Absorption lines are defined as dark lines or lines having reduced intensity, on an ongoing spectrum. A typical example is noticed in the spectra of stars, where gas existing in the outer layers of the star absorbs some of the light from the underlying thermal blackbody spectrum.

A typical stellar spectrum includes dark lines called absorption lines.

A typical stellar spectrum includes a number of deep indentations in which the intensity abruptly falls and then rises. These deep indentations are called absorption lines. Absorption lines are dark lines in a spectrum that correspond to specific wavelengths of light that have been absorbed by elements in the star's outer layers.

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Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = [tex]5.676*10^{-3} \ mm[/tex]

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  [tex]\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}][/tex]

where;

[tex]\epsilon = 0[/tex]

[tex]W_n = \sqrt { \frac{k}{m}}[/tex]

= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]

= 4.0124

replacing them into the above equation and making X the subject of the formula:

[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]

[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]

[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = [tex]5.676*10^{-3} \ mm[/tex]

To solve for the velocity at t=0.40s of a mass in SHM, differentiate the position function with respect to time, and use the resulting velocity function. The spring constant can be calculated using Hooke's law, and the total energy can be found through the energy relationship in SHM.

The student is asking about the characteristics of simple harmonic motion (SHM) related to a mass attached to a spring. To find the velocity at a specific time t for the mass undergoing SHM, we need to differentiate the position function x(t) with respect to time. The provided position function is x(t)=(2.0 cm)cos(10t), implying that the velocity function will be v(t)=-ω A sin(ωt), where ω is the angular frequency, and A is the amplitude. To calculate the spring constant k and total energy E, we will employ both Hooke's law, F = -kx, and the relationship between the total energy in SHM and the displacement from the equilibrium position, given by E = ½ k A².

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The velocity at [tex]t=0.40s[/tex] is [tex]0.1514m/s[/tex].

The spring constant is [tex]6N/m[/tex].

The total energy of the mass is 0.0012J.

Determining Velocity, Spring Constant, and Total Energy in Simple Harmonic Motion

The position of a 60 g oscillating mass is given by the function: [tex]x(t) = (2.0 cm)cos(10t)[/tex], where time t is in seconds. Here's how we solve for the velocity at [tex]t = 0.40 s[/tex], the spring constant k, and the total energy E of the system.

Step-by-Step Solution

Answer:

The maximum charge around the capacitor is 0.03170189C

Explanation:

See attached file

In an RLC series circuit with given values for resistance, inductance, and capacitance, we can calculate the charge on the capacitor five cycles later and fifty cycles later using the equation q(t) = q(0) * e^(-(R/L)t)

In an RLC series circuit with a resistance of 7.092 ohms, an inductance of 10 mH, and a capacitance of 3.0 µF, the charge on the capacitor can be calculated using the equation:

q(t) = q(0) * e^(-(R/L)t)

Given that the initial charge on the capacitor is 8.0 µC, we can calculate the charge five cycles later and fifty cycles later using this equation.

(a) To find the charge five cycles later, we need to find the time period of one cycle. The time period can be calculated using the formula:

T = 2π√(LC)

Once we have the time period, we can calculate the time it takes for five cycles to pass and substitute it in the equation to find the charge.

(b) To find the charge fifty cycles later, we can use the same approach as in part (a), but substituting the time it takes for fifty cycles to pass.

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Answer:

The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules

Explanation

Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.

Gravitational potential energy is expressed as shown:

GPE = mass × acceleration due to gravity × height

Given mass of athlete = 80kg

height covered = 820m

acceleration due to gravity = 9.81m/s

GPE = 80×9.81×820

GPE = 643,536Joules

Answer:

[tex]2.429783984\times 10^{-14}\ A[/tex]

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]

Number of electrons passing per second

[tex]n_e=\dfrac{6020}{0.0476}\\\Rightarrow n_e=126470.588[/tex]

Number of protons passing per second

[tex]n_p=\dfrac{1681}{0.0662}\\\Rightarrow n_p=25392.749[/tex]

Current due to electrons

[tex]I_e=n_ee\\\Rightarrow I_e=126470.588\times 1.6\times 10^{-19}\\\Rightarrow I_e=2.0235\times 10^{-14}\ A[/tex]

Current due to protons

[tex]I_p=n_pe\\\Rightarrow I_p=25392.749\times 1.6\times 10^{-19}\\\Rightarrow I_p=4.06283984\times 10^{-15}\ A[/tex]

Total current

[tex]I=2.0235\times 10^{-14}+4.06283984\times 10^{-15}\\\Rightarrow I=2.429783984\times 10^{-14}\ A[/tex]

The average current is [tex]2.429783984\times 10^{-14}\ A[/tex]

Answer:

false  b) a = v²(radial) / r  and e)

Explanation:

Let's review given statement separately

a) centripetal acceleration

       a = v² / r

linear and angular velocity are related

     v = w r

we substitute

       a = w²r

this acceleration is directed to the center of the circle, so the vector must be negative

        a = - w r2

the bold are vector

True this statement

b) the magnitude is the scalar value

     a = v² / r

where v is the tangential velocity, not the radial velocity, so this statement is

False

c) This is true

      a = v²/ r

this speed is tangential

d) Newton's second law is

         F = m a

if the acceleration is centripetal

         F = m (- v² / r)

we substitute

         F = m (- s² / R)

the statement is true

e) when the car turns to the left, the objects have to follow in a straight line, which is why

you need a force toward the center of the circle to take the curve.

     Consequently there is no outward force

This statement is false

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is [tex]\epsilon_{max}= 26.8 V[/tex]

The emf induced at t = 1.00 s is [tex]\epsilon = 24.1V[/tex]

The maximum rate of change of magnetic flux is   [tex]\frac{d \o}{dt}|_{max} =26.8V[/tex]

Explanation:

    From the question we are told that

        The number of turns is N = 44 turns

          The length of the coil is  [tex]l = 15.0 cm = \frac{15}{100} = 0.15m[/tex]

          The width of the coil is  [tex]w = 8.50 cm =\frac{8.50}{100} =0.085 m[/tex]

          The magnetic field is  [tex]B = 745 \ mT[/tex]

          The angular speed is [tex]w = 64.0 rad/s[/tex]

Generally the induced emf is mathematically represented as

        [tex]\epsilon = \epsilon_{max} sin (wt)[/tex]

 Where [tex]\epsilon_{max}[/tex] is the maximum induced emf and this is mathematically represented as

            [tex]\epsilon_{max} = N\ B\ A\ w[/tex]

Where [tex]\o[/tex] is the magnetic flux

            N is the number of turns

             A is the area of the coil which is mathematically evaluated as

             [tex]A = l *w[/tex]

        Substituting values

           [tex]A = 0.15 * 0.085[/tex]

               [tex]= 0.01275m^2[/tex]

substituting values into the equation for  maximum induced emf

         [tex]\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0[/tex]

                 [tex]\epsilon_{max}= 26.8 V[/tex]

 given that the time t = 1.0sec

substituting values into the equation for induced emf  [tex]\epsilon = \epsilon_{max} sin (wt)[/tex]

      [tex]\epsilon = 26.8 sin (64 * 1)[/tex]

        [tex]\epsilon = 24.1V[/tex]

   The maximum induced emf can also be represented mathematically as

              [tex]\epsilon_{max} = \frac{d \o}{dt}|_{max}[/tex]

  Where  [tex]\o[/tex] is the magnetic flux and [tex]\frac{d \o}{dt}|_{max}[/tex] is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

         [tex]\frac{d \o}{dt}|_{max} =26.8V[/tex]

Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

     n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

      sin θ₂ = n₁ / n₂ sin θ₁

     θ₂ = sin⁻¹ (n₁ /n₂   sin θ₁)

    As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases

Answer is decrease

b) The wave velocity eta related to the wavelength and frequency

      v = λ f

The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.

The speed decreases with the indicated refraction increases and therefore the wavelength decreases

      λ = λ₀ / n

The answer is decrease

c) from the previous analysis the frequency remains unchanged

d) the refractive index is defined by

       n = c / v

So if n increases, the speed must decrease

The answer is decrease

e) the energy of the photon is given by the Planck equation

      E = h f

Since the frequency does not change, the energy does not change either

Answer remains unchanged

a) The Angle te also decreases, b) Decrease, c) Unchanged, d) The speed decrease, e) Unchanged

When a glimmer of light perishes from one translucent material to another, it must capitulate with the law of refraction

n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the happening ray and index 2 is for the refracted ray With this expression let's discuss.

a) They indicate that the refractive index increases,

sin θ₂ = n₁ / n₂ sin θ₁

θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)

As m is more significant than n1 the quantity on the ownership is less than one, the whole quantity in parentheses diminishes so the Angle te also decreases

Solution is decrease

b) The wave velocity eta connected to the wavelength and frequency

v = λ f

The frequency does not modify since the passage from one medium to the other is a procedure of forced oscillation and resonance whereby the frequency in the two mediums is the same.

The speed decreases with the foreshadowed refraction increases and thus the wavelength decreases

So, λ = λ₀ / n

The answer is to decrease

c) from the earlier analysis the frequency remains unchanged

d) When the refractive index is defined by

n is = c / v

Then, if n increases, the speed must decrease

The solution is to decrease

e) When the energy of the photon is given by the Planck equation

E is = hf. Since When the frequency does not transform, the energy does not change either Solution is remains unchanged

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Final answer:

The electric field inside an infinite slab with a constant volume charge density is found using Gauss's law, resulting in an expression where the electric field is directly proportional to both the distance from the midplane and the charge density, and inversely proportional to the permittivity of free space.

Explanation:

Finding the Electric Field Inside an Infinite Slab Using Gauss's Law

To find the electric field strength inside an infinite slab of charge lying in the xy-plane and having a uniform volume charge density (ρ), we employ Gauss's law. Gauss's law states that the net electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). For an infinite slab of thickness 2z0, positioned between z = -z0 and z = +z0, we choose a Gaussian surface in the form of a rectangular box (pillbox) that extends symmetrically above and below the xy-plane. This choice is motivated by the symmetry of the slab and the desire to have an electric field that is either parallel or perpendicular to the faces of the box.

The charge enclosed by the Gaussian surface is given by Q = ρA(2z), where A is the cross-sectional area of the box parallel to the xy-plane and 2z is the extent of the box in the z-direction within the slab. Applying Gauss's law, the electric flux through the box is Ψ = EA = Q/ε0, where E is the magnitude of the electric field inside the slab and is directed along the z-axis. Simplifying, we find E = ρz/ε0. It follows that the electric field strength inside the slab, for -z0 ≤ z ≤ z0, is directly proportional to the distance z from the midplane and to the volume charge density ρ, and inversely proportional to the permittivity of free space ε0.

Using Gauss's law, the electric field strength inside the infinite slab of charge is found to be E = (ρ * z) / ε0.

To find the electric field strength inside an infinite slab of charge with uniform volume charge density ρ, we will use Gauss's law.

The slab extends from z = -z₀ to z = z₀ in the z-direction.

Step-by-Step Solution:

The electric field strength inside the slab, for -z0 ≤ z ≤ z0, is given by:

E = (ρ * z) / ε₀

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z  from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns,  and carry the same current. The currents are in the same sense, and the fields they  produce are in the same direction in the region between them. We place the origin  at the center of the left-hand loop and let x be the coordinate of a point on the axis  between the loops. To calculate the field of the left-hand loop, we set z = x in the  equation above. The chosen point on the axis is a distance s – x from the center of  the right-hand loop. To calculate the field it produces, we put z = s – x in the  equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷²   + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx=  - NμοiR²/2 [3x/ (R²+x²)⁵÷²   + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx=  - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷²   - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

Answer:

0.763 m

Explanation:

Intensity I = power P ÷ area A of exposure (spherical area of propagation)

I = P/A

A = P/I

Power = 10.0 W

Intensity = 1.39 W/m^2

A = 10/1.39 = 7.19 m^2

Area A = 4¶r^2

7.19 = 4 x 3.142 x r^2

7.19 = 12.568r^2

r^2 = 7.19/12.568 = 0.57

r = 0.753 m

Answer:

4117.65 N

Explanation:

Speed of ball, u = 126 km/h = 35 m/s

Mass of ball, m = 160 g = 0.16 kg

Time interval, t = 1.36 ms = 0.00136 s

We can calculate the force as a measure of the momentum of the ball:

F = P/t

Momentum, P, is given as:

P = mv

Therefore:

F = (mv) / t

F = (0.16 * 35) / (0.00136)

F = 4117.65 N

The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.

Answer:

Inductance of first is one-4th that of the second inductor

Explanation:

See attached file

Answer:

The average force on the target is proportional to:

- The number of projectiles hitting the target.

- The mass of the projectiles.

- If the time increases or decreases

Explanation:

This is a problem that applies momentum and the amount of movement. Where this principle can be explained by the following equation:

m*v1 + Imp1_2 = m*v2

where:

∑F *Δt = Imp1_2 = impulse. [N*s]

m*v1  = mass by velocity before the impact [kg*m/s]

m*v2  = mass by velocity after the impact [kg*m/s]

When a problem includes two or more particles, each particle it can be considered separately and equation is written for every particle.

We clear the expression of force in the equation:

∑F *Δt  = m*v2 - m*v1

In this equation if we have a different number of particles, given by the value n, we see that the equation is proportional to the number of particles

∑F *Δt  = n*m*v2 - n*m*v1

Therefore the average force on the target is proportional to the number of projectiles hitting the target.

The Force F is also increased or decreased if the mass of the projectiles is changed. Therefore it is also proportional to the mass of the projectiles.

The Force F also changes if the time increases or decreases.

Using the analytical one-term approximation method, we can determine how long an egg should be kept in boiling water to reach a specific temperature at its center.

In this problem, we are given the initial and final temperatures of an egg, as well as its diameter, the heat transfer coefficient, and the thermal conductivity and diffusivity of water. We are asked to find how long the egg should be kept in boiling water in order for its center temperature to reach a certain value.

To solve this problem, we can use the analytical one-term approximation method, which involves calculating the Biot number and the dimensionless temperature profile. By equating the temperature at the center of the egg to the desired final temperature, we can solve for the time required.

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All objects are described using classical mechanics, with the exception of the buckyball, which requires a quantum mechanical description due to its small mass and high speed that result in a significant de Brogli wavelength.

Whether an object shows particle-wave duality, requiring quantum mechanics to describe, or behaves like everyday objects, described by classical mechanics, is determined by the de Broglie wavelength of the object. Calculating the de Broglie wavelength (λ) can be accomplished using the equation λ = h/mv, where h is Planck's constant and m and v are the object's mass and velocity, respectively.

For the virus, buckyball, mosquito, and turtle, the calculated de Broglie wavelengths are so small compared to their physical dimensions that quantum mechanical effects would be negligible. Hence, all these objects can be described using classical mechanics. The buckyball is the only exception. Owing to its tiny mass and high speed, its de Broglie wavelength becomes significant in relation to its size, requiring quantum mechanical description.

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Objects that are atomic or subatomic in size require quantum mechanics for their description, while larger objects can be adequately explained by classical mechanics. In the provided cases, a virus and a buckyball would require quantum mechanics, while a mosquito and a turtle would be described by classical mechanics.

To determine whether an object would be adequately described by classical mechanics or require quantum mechanics, we need to consider its size, mass, and speed. Under the domain of quantum mechanics are typically objects that are atomic or subatomic in size, meaning they exhibit wave-particle duality due to their small size and significant motion.

1. A virus: Being atomic in size, a virus would require quantum mechanics for its description. Specifically, it would demonstrate both particle and wave properties due to its small size.

2. A buckyball: Given its subatomic mass, size, and significant speed, a buckyball falls into the quantum domain, displaying both particle and wave properties.

3. A mosquito and 4. A turtle : These are larger, macroscopic objects. Their motion and behavior can be adequately described using classical mechanics as they primarily exhibit particle properties, their wave-like characteristics being effectively obscured due to their larger mass and size. Their behaviors conform to the laws defined under classical mechanics and do not exhibit the quantum nature that smaller objects do.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

     [tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]

    [tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]

Explanation:

   From the question we are told that

         The first string has a frequency of   [tex]f_1 = 230 Hz[/tex]

          The period of the beat is  [tex]t_{beat} = 0.99s[/tex]

Generally the frequency of the beat is

             [tex]f_{beat} = \frac{1}{t_{beat}}[/tex]

  Substituting values

            [tex]f_{beat} = \frac{1}{0.99}[/tex]

                   [tex]= 1.01 Hz[/tex]

From the question

        [tex]f_2 - f_1 = f_{beat}[/tex]   for  [tex]f_2[/tex]  having a  higher tension

So

       [tex]f_2 - 230 = 1.01[/tex]

               [tex]f_2 = 231.01Hz[/tex]

 From the question

            [tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]

         [tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]

Substituting values

         [tex]\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}[/tex]

      [tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]

    For [tex]f_2[/tex] having a lower tension

           [tex]f_1 - f_2 = f_{beat}[/tex]

  So

       [tex]230 - f_2 = 1.01[/tex]

            [tex]f_2 = 230 -1.01[/tex]

                  [tex]= 228.99[/tex]

  From the question

            [tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]

         [tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]

    Substituting values

         [tex]\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}[/tex]

      [tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]        

Answer:

1.attraction and repulsion behavior do the north and south poles of a magnet exhibit?A.Like poles repel each other, and opposite poles attract.

2.Explain the left-hand rule for flux direction in conductors. Ans.The left-hand rule is used to determine the direction of rotation of the flux around the conductor

.3.What is the most common source of power for most HVACR systems?Ans.Alternating current is the most common source of power used for most HVACR systems.

4.What is the purpose of transformers?Ans..Transformers are used to increase or decrease incoming voltages to meet the requirements of the load.

5.What is a solenoid valve?Ans.Solenoid valves are electrically operated valves that open or close automatically by energizing or de-energizing a coil of wire.6.Is there an electrical connection between the primary and secondary windings of a transformer?A.There is no electrical connection between the primary and secondary windings.

Explanation:

The poles of a magnet attract and repel based on their polarity. The left-hand rule explains how magnetic fields interact with electric currents. Most HVACR systems use electricity, and transformers convert voltage levels to transmit power efficiently. A solenoid valve controls fluid or gas flow through electromagnetism.

The north and south poles of a magnet exhibit a behavior where like poles repel and opposite poles attract. This means that north pole to north pole or south pole to south pole will repel, while north pole to south pole will attract.

The left-hand rule for flux direction, often used in electromagnetism, states that if you point your thumb in the direction of the current, the direction in which your fingers curl represents the direction of the magnetic field lines. This is significant in understanding the interaction of electric current with magnetic fields in conductors.

The most common source of power for most HVACR (Heating, Ventilation, Air Conditioning, and Refrigeration) systems is electricity. This power is used to run the system's various components including compressors, fans, and controls.

Transformers are used in electric power systems to convert voltages from one level to another. This is mainly used to transmit electricity over long distances with low energy losses and is essential in power grids.

A solenoid valve is an electromechanical device used to control the flow of a fluid or gas. When an electric current is passed through the coil, a magnetic field is created causing the valve to open or close.

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Answer:

a)  x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2

Explanation:

The center of mass of a body is the point where all external forces are applied, it is defined by

      [tex]x_{cm}[/tex] = 1 / M ∑  [tex]x_{i} m_{i}[/tex] = 1 /M ∫ x dm

      [tex]y_{cm}[/tex] = 1 / M ∑ [tex]y_{i} m_{i}[/tex] = 1 / M ∫ y dm

where M is the total body mass

Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis

a) let's start with the reference zero at the left end of the horizontal rod

let's use the concept of linear density

    λ = M / L = dm / dl

since the rod is on the x axis

     dl = dx

    dm = λ dx

let's calculate

      x_{cm} = M ∫ x λ dx = λ / M ∫ x dx

      x_{cm} = λ / M x² / 2

we evaluate between the lower integration limits x = 0 and upper x = L

      x_{cm} = λ / M (L² / 2 - 0)

  we introduce the value of the density that is cosntnate

     x_{cm} = (M / L) L² / 2M

     x_{cm} = L / 2

We repeat the calculation for verilla verilla

     λ = M / L = dm / dy

     y_{cm} = 1 / M ∫ y λ dy

     y_{cm} = λ M y² / 2

     [tex]y_{cm}[/tex] = M/L  1/M (L² - 0)

     y_{cm}= L/2

b) we repeat the calculation for the origin the reference system is top of the vertical rod

     horizontal rod

        x_{cm} = 1 / M ∫λ x dx = λ/M   x² / 2

we evaluate between the lower limits x = 0 and the upper limit x = -L

      x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M

      x_{cm} = L / 2

vertical rod

      y_{cm} = 1 / M ∫y dm

      y_{cm} = λ / M ∫y dy

      y_{cm} = λ / M y2 / 2

we evaluate between the integration limits x = 0 and higher x = -L

      y_{cm} = (M / L) 1 / M ((-L)²/2 -0)

      y_{cm} = L / 2

To find the center of mass of the 'L' structure, use the formal definition and consider the midpoint of each member. The x-coordinate is halfway between the ends and the y-coordinate is halfway between the top and bottom ends.

To determine the location of the center of mass of the 'L' structure, we can use the formal definition of center of mass. Since the thin vertical and horizontal members have the same length and mass, the center of mass of each member is located at its midpoint. The x-coordinate of the center of mass is halfway between the x-coordinate of the left end and the x-coordinate of the right end of the 'L'. The y-coordinate of the center of mass is halfway between the y-coordinate of the bottom end and the y-coordinate of the top end of the 'L'.

(a) Origin at the lower left:

x = -L/4, y = L/4

(b) Origin at the top of the vertical member:

x = L/4, y = -L/4

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Answer:

the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

Explanation:

Given that :

mass of uniform disk M = 1.2 kg

Radius R = 0.11 m

lower radius (b) = 0.14 m

small mass (m) = 0.4 kg

Force (F) = 21 N

time (∆t) =0.2 s

Moment of Inertia of [tex]I_{disk, CM}[/tex] = [tex]\frac{1}{2}MR^2[/tex]

= [tex]\frac{1}{2}*1.2*0.11^2[/tex]

= 0.00726 kgm²

Point mass  [tex]I_{point \ mass}[/tex] = mb²

But since four low rods are attached ; we have :

[tex]I_{point \ mass}[/tex] = 4 × mb²

= 4  × 0.4 (0.14)²

= 0.03136 kgm²

Total moment of Inertia =  [tex]I_{disk, CM}[/tex] + [tex]I_{point \ mass}[/tex]

= (0.00726 + 0.03136) kgm²

= 0.03862 kgm²

Assuming ∝ = angular acceleration = constant;

Then; we can use the following kinematic equations

T = FR

T = 2.1 × 0.11 N

T = 2.31 N

T = I × ∝

2.31 = 0.03862 × ∝

∝ = [tex]\frac{2.31}{0.03862}[/tex]

∝ = 59.81 rad/s²

Using the formula [tex]\omega_f = \omega_i + \alpha \delta T[/tex] to determine the angular speed of the apparatus; we have:

[tex]\omega_f =0 + 59.81*0.2[/tex]         since  [tex]( w_i \ is \ at \ rest ; the n\ w_i = 0 )[/tex]

[tex]\omega_f =11.962 \ rad/s[/tex]

∴ the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

b) Using the formula :

[tex]\theta = \omega_i t + \frac{1}{2}*\alpha*(t)^2\\\\\theta = 0 *0.2 + \frac{1}{2}*59.81*(0.2)^2 \\ \\ \theta = 29.905 *(0.2)^2 \\ \\ \theta = 1.1962 rads \ \ \ ( to \ degree; \ we \ have) \\ \\ \theta = 68.54^0[/tex]

Thus, the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

To solve the problem, compute the angular acceleration using the torque and moment of inertia. Given the angular acceleration and the time, calculate the angular velocity. Finally, use the angular acceleration and time to compute the angle turned through.

The first part of this problem involves calculating angular acceleration, which is the rate of change of angular velocity. Using the formula, angular acceleration (α) = Torque (τ) / Moment of Inertia (I), the torque can be calculated using the formula Torque = Force * Radius, and the moment of inertia is calculated using the given formulas for disk and point masses.

For the disk, I_(disk,CM) = 1/2 M R^2 and for the point masses, I_(point mass) = 4 * m * b^2. Summing these gives the total moment of inertia. The angular acceleration is then obtained by dividing the torque by the total moment of inertia.

Having the angular acceleration and the time, we can calculate the angular velocity (ω) using the formula ω = α * Δt.

For the second part, the angle through which the apparatus turns can be calculated using the formula θ = 0.5 * α * (Δt)^2, as the initial angular velocity was zero.

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Answer:

Explanation:

Moment of inertia of wheel = 1/2 x mR²  , m is mass and R is radius of wheel

= .5 x 9 x .4²

= .72 kg m²

Torque created on wheel by string = T x r , T is tension and r is radius of wheel .

13 x .4 = 5.2 N m

angular acceleration α = torque / moment of inertia

= 5.2 / .72

= 7.222 rad /s²

a ) final angular speed = α x t , α is angular acceleration , t is time.

=  7.222  x .72

= 5.2 rad /s

b )

θ = 1/2 α t² , θ  is angle turned , t is time

= .5 x 7.222 x .72²

= 1.872 rad

average angular speed = θ / t

= 1.872 / .72

= 2.6 rad /s

c )

angle turned = 1.872 rad ( discussed above )

d )

length of string coming off

=  angle rotated x radius of wheel

= 1.872 x .4

= .7488 m .

74.88 cm

The final angular speed of the wheel is 5.2 radians/s. The average angular speed is 0.52 radians/s. The wheel turned through an angle of 0.3478 radians and 0.1391 m of string came off the wheel.

To find the final angular speed, we can use the equation ω = φt + 0.5αt^2, where ω is the final angular speed, φ is the initial angular speed (which is 0 since the wheel starts from rest), α is the angular acceleration, and t is the time. We are given that the force applied to the wheel is 13 N for 0.72 s.

Using the equation F = mα, we can rearrange it as α = F/m, where F is the force and m is the mass of the wheel. Plugging in the values, we get α = 13/9. Now we can substitute the values into the first equation and solve for ω: ω = (0)(0.72) + 0.5(13/9)(0.72)^2 = 5.2 radians/s.

To find the average angular speed, we use the formula ωavg = φ + (1/2)αt, where ωavg is the average angular speed. Since the initial angular speed φ is 0, we can simplify the equation to ωavg = (1/2)αt. Plugging in the values, we get ωavg = (1/2)(13/9)(0.72) = 0.52 radians/s.

To find the angle through which the wheel turned, we can use the equation θ = φt + 0.5αt^2. Since the initial angular speed φ is 0, the equation simplifies to θ = 0.5αt^2. Plugging in the values, we get θ = 0.5(13/9)(0.72)^2 = 0.3478 radians.

To find how much string came off the wheel, we can use the formula s = rθ, where s is the length of string and r is the radius of the wheel. Plugging in the values, we get s = (0.4)(0.3478) = 0.1391 m.

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Answer:

(b) da/√2​

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

Explanation:

Ez = (5.2 V/m)

Magnitude of electric field = 5.2 V/m

Magnitude of magnetic field be B then

Magnitude of electric field  / Magnitude of magnetic field = c , c is velocity of light

5.2 / B = 2.998 x 10⁸

B = 5.2 / 2.998 x 10⁻⁸

= 1.73 x 10⁻⁸ T.

b ) E X B = direction of velocity of light , E is electric vector , B is magnetic V.

E is along z - axis , velocity is along x -axis  , then magnetic field must be in   - Y direction.

c ) E and B  are in phase so when E is in positive z axis , directio of magnetic field must be in  -  y direction.