What is the electric flux Φ3 through the annular ring, surface 3? Express your answer in terms of C , r1, r2, and any constants.

2Q

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = [tex]\frac{1}{2}qV[/tex]            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = [tex]\frac{1}{2} QV[/tex]              ----------------(ii)

Make V subject of the formula in equation (ii)

V = [tex]\frac{2E_{p}}{Q}[/tex]

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = [tex]\frac{1}{2}qV[/tex]

Substitute the value of V into the equation above;

2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])

Solve for q;

[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]

[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]

[tex]q = 2Q[/tex]

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

Answer:

The  self-inductance is  [tex]L = 0.0053 \ H[/tex]

Explanation:

From the question we are told that  

      The number of loops is  [tex]N = 900[/tex]

      The  length of the rod is  [tex]l =6 \ cm = 0.06 \ m[/tex]

      The radius of the rod is  [tex]r = 1 \ cm = 0.01 \ m[/tex]

The  self-inductance for the solenoid is mathematically represented as

        [tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]

Now the cross-sectional of the solenoid is mathematically evaluated as

        [tex]A = \pi r^2[/tex]

substituting values  

         [tex]A =3.142 * 0.01 ^2[/tex]

        [tex]A = 3.142 *10^{-4} \ m^2[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with a value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

    substituting values into above equation

          [tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]

          [tex]L = 0.0053 \ H[/tex]

Answer:

P = 1470980 J

Explanation:

We have,

Mass of the hiker is 79 kg

It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.

It is given by :

[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]

So, the potential energy of 1470980 J is associated with a hiker.

Answer:

v2 = - 16.899 m/s

velocity of ball increases so that the kinetic energy of the ball increases.

Explanation:

given data

mass of elephant, m1 = 5500 kg

mass of ball, m2 = 0.160 kg

initial velocity of elephant, u1 = - 4.70 m/s

initial velocity of ball, u2 = 7.50 m/s

solution

we consider here final velocity of ball = v2

so  collision formula is express as for v2

[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex]      .................1

put here value and we get

[tex]v_{2}=\left ( \frac{2\times 5500}{5500+0.160} \right )(-4.70)+\left ( \frac{0.16-5500}{5500+0.160} \right )(7.50)[/tex]  

solve it we get

v2 = - 16.899 m/s

here negative sign shows that the ball bounces back towards you

and

here we know the velocity of ball increases so that the kinetic energy of the ball increases.

and due to this effect, it will gain in energy is due to the energy from the elephant mass

Air has lower density than water, mineral Water, or salt water. (B)​

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

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

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].

The kinetic energy of an object depends on the mass and velocity with which it moves.

While under free-fall, the mass of an object does not affect the velocity with which it falls.

So, the velocities of both the balls are the same.

Let the mass of ball A is 'm'

So, the mass of ball B is '4m'

The kinetic energy of ball A is given by;

[tex]KE_{A}=\frac{1}{2} mv^2[/tex]

The kinetic energy of ball B is given by;

[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]

Therefore, the ratio of kinetic energies of A and B is,

[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]

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Answer:

B and D

Explanation:

Because

R= resistivity xlenght/ Area

Where R= resistance

Answer:

11.3298W

Explanation:

The rate of heat transfer is determined from the enthalpy of vaporization at the give pressure obtained and the mass flow rate. The mass flow rate is determined from the volume of the boiled water, the given time interval and the specific volume of the saturated liquid.

Given that

1atm as the atmospheric pressure

Internal diameter = 20cm = 0.2m

Time = 15mins = (15×60)secs

Latent heat of vaporization (hevap) = 2256.6

Q = mh(evap)

= m/∆t . hevap

= V/αliq∆t ×h(evap)

D^2π∆h/4αliq ∆t × hevap

= 0.2^2 ×π×0.8×2256.5/4×0.001043×15×60

=0.04×3.142×0.08×2256.6/2.00256

= 22.68876/2.00256

Q = 11.3298W

Answer:

The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

[tex]F=m\,a[/tex]

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]

Answer:

Gas 1

Explanation:

Answer:

The workdone is [tex]W_v = - 20 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the object is [tex]m = 2.5 \ kg[/tex]

     The speed of fall is [tex]v = 2.5 \ m/s[/tex]

     The depth of fall is  [tex]d = 80\ cm = 0.8 \ m[/tex]

Generally according to the work energy theorem

      [tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]

Now here given the that the velocity is  constant  i.e  [tex]v_1 = v_2 = v[/tex] then

We have that

    [tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]  

So in terms of workdone by the potential energy of the object and that of the viscous liquid we have

       [tex]W = W_v - W_p[/tex]

Where  [tex]W_v[/tex] is workdone by viscous liquid

             [tex]W_p[/tex] is the workdone by the object which is mathematically represented as

            [tex]W_p = mgd[/tex]

So  

       [tex]0 = W_v + mgd[/tex]

=>    [tex]W_v = - m * g * d[/tex]

substituting values

       [tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]

      [tex]W_v = - 20 \ J[/tex]

Answer:

a

 [tex]\theta = 0.0022 rad[/tex]

b

 [tex]I = 0.000304 I_o[/tex]

Explanation:

From the question we are told that  

   The  wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]

    The  distance of the slit separation is  [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]

Generally the condition for two slit interference  is  

     [tex]dsin \theta = m \lambda[/tex]

Where m is the order which is given from the question as  m = 2

=>    [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]

 substituting values  

      [tex]\theta = 0.0022 rad[/tex]

Now on the second question  

   The distance of separation of the slit is  

       [tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      [tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]

Where  [tex]\beta[/tex] is mathematically evaluated as

       [tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]

  substituting values

     [tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]

    [tex]\beta = 0.06581[/tex]

So the intensity is  

    [tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]

   [tex]I = 0.000304 I_o[/tex]

Answer:

11.7 m

Explanation:

The radius of the circle is 13 m.

The central angle of the arc is 0.9 radians

The length of an arc is given as:

L = r θ

where θ = central angle in radians = 0.9

=> L = 0.9 * 13 = 11.7 m

Length of the arc will be 11.7 m ≈ 10 m

Arc length refers to the distance between two points along a curve’s section.

Arc length = radius * theta

where

Arc length  = ? to find

given :

radius = 13 m

theta ( central angle) = 0.9 radians

Arc length = 13 m * 0.9 radians

                = 11.7 m ≈ 10 m

length of the arc will be 11.7 m ≈ 10 m

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Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]

Where:

[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.

[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.

[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:

[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]

Roots of the polynomial are, respectively:

[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]

[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.

[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.

[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:

[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]

[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

[tex]v_{1} = v_{1,o} + g \cdot t[/tex]

Second stone

[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]

Where:

[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.

[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.

If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:

First stone

[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]

[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]

Second stone

[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]

[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Answer is given below

Explanation:

given data

we will consider here

Ping-Pong ball weighs = 3.1 g

diameter =  4.2 cm

solution

Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.

Answer:

Explanation:

Using the relationship to calculate the time that elapse 2x = vt

x = distance between the source and the reflector

v is the velocity of sound in air

t = time elapsed

Given x = 30.0m and v = 343m/s

substituting the values into the formula above to get the total time elapsed t;

t = 2x/v

t = 2(30)/343

t= 60/343

t = 0.175sec

If you are seated in the middle row, the time that will elapse between the sound from the stage reaching your ear directly will be;

0.175/2 = 0.0875 secs

Answer:

(a)    vo = 24.98m/s

(b)    t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex]      (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]        

[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]

The time that the ball takes to arrive to the ground is 5.09s

We have that for the Question, it can be said that the speed of  ball and How much longer would it take the ball to reach the ground is

From the question we are told

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.

(a) How fast was the ball originally moving when it was kicked.

(b) How much longer would it take the ball to reach the ground?

a)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]

b)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]

Therefore

[tex]X=5.095-5[/tex]

X=0.095sec

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Answer:

2121.2kg/m^3 is the density of the test liquid on the left

Explanation:

See attached file

Answer:

Explanation:

The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :

D. Quadruple

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.

The stored energy shifts with the square of the electric charge put away within the capacitor.

In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.

Thus, the correct answer is D.

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Answer:

The  tension on the rope  is  T  =  900 N

Explanation:

From the question we are told that  

     The mass of the person on the left is  [tex]m_l = 100 \ kg[/tex]

      The force of the person on the left is  [tex]F_l = 1000 \ N[/tex]

       The mass of the person on the right  is  [tex]m_r = 70 \ kg[/tex]

       The force of the person on the right is  [tex]F_r = 830 \ N[/tex]

Generally the net force is  mathematically represented as

         [tex]F_{Net} = F_l - F_r[/tex]

substituting  values

        [tex]F_{Net} = 1000-830[/tex]

       [tex]F_{Net} = 170 \ N[/tex]

Now the acceleration net acceleration of the rope is mathematically evaluated as

        [tex]a = \frac{F_{net}}{m_I + m_r }[/tex]

substituting  values

     [tex]a = \frac{170}{100 + 70 }[/tex]

     [tex]a = 1 \ m/s ^2[/tex]

The  force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by  a  is  mathematically represented as

        [tex]m_l * a = F_r -T[/tex]

Where T  is  the tension on the rope  

      substituting values

        [tex]100 * 1 = 1000 - T[/tex]

=>      T  =  900 N

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

Answer:39.2J

Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:                  

From above;  the distance between speaker  A and speaker B can be expressed as:

[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]

The path length difference  will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length  will be half the wavelength; which is

= [tex]\dfrac{1}{2}*4 \ m[/tex]

= 2 m

The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.

Given data:

The distance between the speakers is, d = 3 m.

The distance between the observer and speaker is, s = 4 m.

The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then  the distance between speaker  A and speaker B can be expressed as:

[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]

And the path length difference is,

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive

interference for that path length  will be half the wavelength; which is

 [tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]

= 2 m

Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.

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Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

This question involves the concepts of the law of conservation of energy.

The temperature change of the bullet is "84.38°C".

According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.

[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]

where,

Therefore,

[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]

ΔT = 84.38°C

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Answer:

Drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].

Explanation:

Formula for Drift Velocity is:

[tex]u = \dfrac{I}{nAq}[/tex]

Where I is the current

n is the number of electrons in 1 [tex]m^3[/tex] or the electron density

A is the area of cross section and

q is the charge of one electron.

We are given the following:

n = [tex]8.49 \times 10^{28}\ electrons/m^3[/tex]

I = 1 A

A = 0.40 [tex]cm^2[/tex] = 40 [tex]\times 10^{-4}[/tex] [tex]m^{2}[/tex]

We know that q = [tex]1.6\times10^{-19} C[/tex]

Putting all the values to find drift velocity:

[tex]u = \dfrac{1}{8.49 \times 10 ^{28} \times 40 \times 10^{-4}\times 1.6 \times 10^{-19}}\\u = \dfrac{1}{543.36 \times 10 ^{5} }\\u = 1.84 \times 10^{-8}\ m/s[/tex]

So, drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].

Answer:

Net nonzero work is done if the final position of the particle is on options A, C and D

Explanation:

non zero work is done if following will be the final position of the charges :

A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.

C) A line that makes an angle 30° with the dipole moment.

D) A line that makes an angle 45°  with the dipole moment.

Answer:

The maximum number of bright spot is [tex]n_{max} =5001[/tex]

Explanation:

From the question we are told that

     The  slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]

      The  wavelength is  [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]

Generally the condition for interference is  

        [tex]n * \lambda = d * sin \theta[/tex]

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  [tex]\theta = 90[/tex]

=>     [tex]sin( 90 )= 1[/tex]

So

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

substituting values

     [tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]

     [tex]n = 2500[/tex]

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        [tex]n_{max} = 2 * n + 1[/tex]

The  1  here represented the central bright spot

So  

      [tex]n_{max} = 2 * 2500 + 1[/tex]

     [tex]n_{max} =5001[/tex]      

Answer:

The  gauge pressure is  [tex]P = 4.2*10^{5} \ N/m^2[/tex]

Explanation:

From the question we are told that

     The height of the 14th floor from the point where the water entered the building  is  h = 43 m

The  gauge pressure is  mathematically represented as

           [tex]P = mgh[/tex]

Where  the m is the mass of the water which is  mathematically represented as  

      [tex]m = \frac{\rho}{V}[/tex]

Where  [tex]\rho[/tex] is the density of the water which has a constant value of  [tex]\rho = 1000 \ kg/m^3[/tex] and this standard value of density the volume is  [tex]1 m^3[/tex] so

     [tex]m = \frac{1000}{1}[/tex]

     [tex]m = 1000 \ kg[/tex]

Thus  

     [tex]P = 1000 * 9.8 * 43[/tex]

     [tex]P = 4.2*10^{5} \ N/m^2[/tex]