A single slit 1.5 mm wide is illuminated by 420- nm light. Part A What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.5 m away

Answer:

The wavelength is [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]

Explanation:

From the question we are told that

   The wavelength of the first light is  [tex]\lambda _ 1 = 587 \ nm[/tex]

    The order of the first light that is being considered is  [tex]m_1 = 10[/tex]

     The order of the second light that is being considered is  [tex]m_2 = 11[/tex]

Generally the distance between the fringes for the first light is mathematically represented as

      [tex]y_1 = \frac{ m_1 * \lambda_1 * D}{d}[/tex]

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         [tex]y_2 = \frac{ m_2 * \lambda_2 * D}{d}[/tex]

Now given that both of the light are passed through the same double slit

       [tex]\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1[/tex]

=>    [tex]\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1[/tex]

=>     [tex]\lambda_2 = \frac{m_1 * \lambda_1}{m_2}[/tex]

=>    [tex]\lambda_2 = \frac{10 * 587 *10^{-9}}{11}[/tex]

=>   [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]

Answer:

E. The gas absorbs 50 of energy as heat and does 50」ot work

Explanation:

This is following the law of thermodynamics that energy is neither created nor destroyed

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.

Answer:

198 ms-1

Explanation:

According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.

The uncertainty associated with each measurement is given by;

∆x∆p≥h/4π

Where;

∆x = uncertainty in the measurement of position

∆p = uncertainty in the measurement of momentum

h= Plank's constant

But ∆p= mΔv

And;

m= 1.770×10^-27 kg

∆x = 0.15 nm

Making ∆v the subject of the formula;

∆v≥h/m∆x4π

∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142

∆v≥198 ms-1

Answer:

Objects that are closer together have a stronger force of gravity between them.

Explanation:

For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.

Answer:

Explanation:

C. The temperature of the coffee will decrease while the temperature of the table increases.

Answer:

This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.

Explanation:

In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are  taken to the negatively charged plate.

Answer:

The  length is  [tex]l = 8.6 \ m[/tex]

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] ,  [tex]f_2 = 240 \ Hz[/tex]

   The speed of sound in the air is  [tex]v_s = 343 \ m/s[/tex]

Generally the frequency of a given harmonic is mathematically represented as

     [tex]f_n = \frac{n v }{2l}[/tex]

Here  n defines  the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    [tex]220 = \frac{n v}{2l}[/tex]

and  

     [tex]240 = \frac{(n+1) v}{2l}[/tex]

So

   [tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]

=>  [tex]\frac{v}{2l} = 20[/tex]

=>   [tex]l = 8.6 \ m[/tex]

The bullet's vertical velocity at time [tex]t[/tex] is

[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]

(or about 140 s, if you're keeping track of significant figures) after being fired.

Answer:

The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.

Explanation:

Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.

Answer:

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Explanation:

According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:

[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]

[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

Answer:

The magnetic force would be:

[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Explanation:

Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:

F = q v x B

(where the bold represents vectors)

the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:

[tex]F=q\,v\,B\,sin(\theta)[/tex]

Therefore replacing the known quantities for the first case:

[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]

Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:

[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Answer:

[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

Explanation:

From the question we are told that  

     The  number of turns is  [tex]N = 450 \ turns[/tex]

      The  radius is  [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  [tex]e = 8.20 *10^{-6} \ V/m[/tex]

Generally according to Gauss law

        [tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]

=>    [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                [tex]A = \pi r ^2[/tex]

=>      [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]

=>       [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;

Here  [tex]\mu_o[/tex] is the permeability of free space with value

          [tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]

=>     [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]

=>      [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

The value of di/dt from the given values of the solenoid electric field is;

di/dt = 7.415 A/s

We are given;

Number of turns; N = 450 per m

Radius; r = 1.17 cm = 0.0117 m

Electric Field; E = 8.2 × 10⁻⁶ V/m

Position of electric field; r' = 3.45 cm = 0.0345 m

According to Gauss's law of electric field;

∫| E*dl | = |-d∅/dt |

Now, ∅ = BA = μ₀niA

where;

n is number of turns

i is current

A is Area

μ₀ = 4π × 10⁻⁷ H/m

Thus;

E(2πr') = (d/dt)(μ₀niA)  (negative sign is gone from the right hand side because we are dealing with magnitude)

Since we are looking for di/dt, then we have;

E(2πr') = (di/dt)(μ₀nA)

Making di/dt the subject of the formula gives;

di/dt = E(2πr')/(μ₀nA)

Plugging in the relevant values gives us;

di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)

di/dt = 7.415 A/s

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Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

Answer:

v = - 14.08 m / s

Explanation:

The definition of center of mass is

        [tex]x_{cm}[/tex] = 1 /M  ∑sun [tex]x_{i} m_{i}[/tex]

where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.

The velocity of the center of mass can be found by deriving this expression with respect to time

         [tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi

let's find the total mass

          M = m₁ + m₂ + m₃ + m₄

          M = 1.45 + 2.81 +3.89 + 5.03

          m = 13.18 kg

let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0

          0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)

          v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄

let's substitute the given values

v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03

They ask us for the calculations for a time t = 2.83 s

          v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03

          v = - 14.08 m / s

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]

The velocity of the center mass of the particles is calculated as;

[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]

The velocity of particle 1 at time, t = 2.83 s;

[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]

The velocity of particle 2 at time, t = 2.83 s;

[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]

The velocity of particle 3 at time, t = 2.83 s;

[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]

The velocity of the particle 4 at time, t = 2.83 s;

[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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Explanation:

When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur

1. Work done bu the gas = Energy absorbed as heat.

2. Work done by environment = Energy absorbed as heat.

3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.

Hence all option are correct.

Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.

If an ideal gas undergoes a slow isothermal expansion,

To know more about the Ideal gas,

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Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, [tex]N_P[/tex] = 50 turns

number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns

the secondary load resistance, [tex]R_S[/tex] = 250 Ω

Determine the turns ratio;

[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]

Now, determine the reflected resistance in the primary winding;

[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]

Therefore, the reflected resistance in the primary winding is 6250 Ω

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

Answer:

The volume of the block is 306 cm³

The density of the block is 7.07 g/cm³

Explanation:

Given;

weight of block in air, [tex]W_a[/tex] = 21.2 N

Weight of block in water, [tex]W_w[/tex] = 18.2 N

Mass of the block in air;

[tex]W_a = mg[/tex]

21.2 = m x 9.8

m = 21.2 / 9.8

m = 2.163 kg

mass of the block in water;

[tex]W_w = mg[/tex]

18.2 = m x 9.8

m = 18.2 / 9.8

m = 1.857 kg

Apply Archimedes principle

Mass of object in air  - mass of object in water = density of water   x  volume                  of object

2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block

0.306 kg = 1000 kg/m³ x Volume of block

Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]

Volume of the block = 3.06 x 10⁻⁴ m³

Volume of the block = 306 cm³

Determine the density of the block

[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]

0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

Answer:

The correct choice is B & E.  

Explanation:

A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.

Cheers!

Answer:

The wavelength is 2.16 m.

Explanation:

Given the speed of the sound = 343 m/s

Trombone generate the frequency = 159 Hz

Now we have to find the wavelength of the sound. Here, we can find the wavelength by dividing the speed of the sound with frequency.

The wavelength of the sound = Speed of sound/frequency

Wavelength of the sound = 343 / 159 = 2.16 m

Answer:

  λ= 5.4379 10⁻⁷ m = 543.79 nm

Explanation:

The phenomenon of diffraction is described by the expression for destructive diffraction is

           a sin θ = (m + 1/2) λ

           λ = a sin θ / (m + 1/2)

let's reduce the magnitudes to the SI system

           a = 15 um = 15 10⁻⁶ m

           m = 2

           θ = 5.2º

Let's calculate

          λ = 15 10⁻⁶ sin 5.2 / (2 +1/2)

         λ = 5.4379 10⁻⁷ m

Let's reduce to nm

          λ= 5.4379 10⁻⁷ m = 543.79 nm

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

205 V

V[tex]_{R}[/tex] = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

[tex]V_{L} = - IwLsin(wt)[/tex]

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V[tex]_{R}[/tex] = IR

    = (0.044 A) (93 Ω)

V[tex]_{R}[/tex] = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)

Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s

V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V[tex]_{R}[/tex] = 2.05 V

Answer:

3 meters

Explanation:

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]

But heat capacity of object B is twice that of object A

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]

Therefore, the final temperature of both objects is 400 K.

Answer:

The magnification is  [tex]m = -2.15[/tex]

Explanation:

From the question we are told that

   The  radius is  [tex]r = 14.0 \ m[/tex]

    The  focal length eyepiece is  [tex]f_e = 3.25 \ m[/tex]

Generally the objective focal length is mathematically represented as

        [tex]f_o = \frac{r}{2}[/tex]

=>     [tex]f_o = \frac{14}{2}[/tex]

=>     [tex]f_o = 7 \ m[/tex]

The  magnification is mathematically represented as

      [tex]m = - \frac{f_o }{f_e }[/tex]

=>    [tex]m = - \frac{7 }{ 3.25 }[/tex]

=>   [tex]m = -2.15[/tex]