A sharp edged orifice with a 50 mm diameter opening in the vertical side of a large tank discharges under a head of 5m. If the coefficient of contraction is 0.62 and the coefficientof velocity is 0.98, what is thedischarge?

Answer:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

Learn more about refraction here: https://brainly.com/question/10729741

Explanation:

ω = 45 rev/min × (2π rad/rev) × (1 min / 60 s) = 4.71 rad/s

r = 10 cm = 0.10 m

1) The linear speed is:

v = ωr

v = (4.71 rad/s) (0.10 m)

v = 0.471 m/s

2) The linear acceleration in the tangential direction is 0.

The linear acceleration in the radial direction is:

a = v² / r

a = (0.471 m/s)² / (0.10 m)

a = 2.22 m/s²

Answer:

Yes. Something similar occurs here on Earth.

Explanation:

Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.

Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine

Answer:

To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.  

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]

Where:

[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.

[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.

[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]

If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:

[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]

[tex]v = 0.1\,\frac{m}{s}[/tex]

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]

[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]

Where:

[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.

[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]

Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:

[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]

[tex]Q_{disp} = 200\,kJ[/tex]

The energy coverted to heat is 200 kilojoules.

Answer:

Poynting vector oscillate at a frequency of 2omega

Explanation:

This is because The poynting vector is proportional to the cross product of electric and magnetic field vectors. So Because both fields oscillate sinusoidally with frequency w, trigonometric identities show that their product is a sinusoidal function of frequency of 2w.

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = [tex]\pi r^{2}[/tex]

==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

[tex]I[/tex] = E/R

where R is the resistor

[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = [tex]I[/tex]E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

Answer:

Δx = 9 x 10⁻³ m = 9 mm

Explanation:

The formula for fringe spacing in Young's Double Slit Experiment is given as follows:

Δx = λL/d

where,

Δx = fringe spacing = ?

λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m

L = Distance between slits and screen = 2 m

d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m

Therefore,

Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)

Δx = 9 x 10⁻³ m = 9 mm

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

g' = 13.5 m/s²

Answer:

The fraction of light that escapes the water surface as the water moves deeper into the water will decrease.

Explanation:

The speed of light in water is small compared to the speed of light in air, and a larger part of the light energy is absorbed in water than in air. When the bulb is immersed in water, some of the light energy is absorbed by the mass of water. When the light bulb is further moved deeper into the water, the fraction of light that escapes decreases, because more mass of water is made available to absorb more of the light energy from the bulb.

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Answer:

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

Answer:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

Answer:

C) the rope tension is less than the objects weight

Explanation:

According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.

In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.

Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:

C) the rope tension is less than the objects weight

Answer:

(a) D.  Zero.

(b) C.  Equal to the battery's terminal voltage.

Explanation:

The question is incomplete, see the complete question for your reference and information.

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage

across its terminals. At the moment contact is made with the battery

(a) the voltage across the capacitor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero.

(b) the voltage across the resistor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero

A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.

basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R

Answer:

The angle between the polarizing axes of the two polarizers is 54°

Explanation:

Given;

intensity of unpolarized light, I₀ = 54.0 W/m²

intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²

The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.

I₁ = I₀Cos²θ

Cos²θ = I₁ / I₀

Cos²θ = 19 / 54

Cos²θ =0.3519

Cos θ = √0.3519

Cos θ = 0.5932

θ = Cos⁻¹(0.5932)

θ = 53.6°

θ = 54°

Therefore, the angle between the polarizing axes of the two polarizers is 54°

Answer:

The Independent variable in this experiment is the time taken by the ball to roll down each distance.

The dependent variable is the distance  through which the ball rolls

The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.

Explanation:

The complete question is

Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?

Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.

A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.

The relationship between the dependent and independent variables in an experiment is given as

y = f(x)

where y is the output or the dependent variable,

and x is the independent variable.

Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.

Answer:

Option C

Explanation:

The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.

Answer:

C

Explanation:

In an endothermic reaction, the energy-time graph shows reactants are at a lower energy level than the products.

Answer:

31.55° and 148.45°

Explanation:

Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;

F = qvBsinθ where;

F is the magnetic force experienced by the proton

q is the charge on the proton

v is the velocity of the proton

B is the magnetic field

θ is the angle between the proton's velocity and the field (Required)

Given parameters

F =  7.00 * 10⁻¹³N

q = 1.602*10⁻¹⁹C

v = 4.80 * 10⁶ m/s

B = 1.74 T

θ  =?

From the formula F = qvBsinθ;

sinθ = F/qvB

sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74

sinθ =  7.00 * 10⁻¹³/13.38*10⁻¹³

sinθ = 0.5231689 * 10⁰

sinθ = 0.5231689

θ = sin⁻¹0.5231689

θ = 31.55°

The following are the positive values of the angle between 0° and 360°

Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.

Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°

Answer:

Fish C would float to the top & Fish A would sink to the bottom

Explanation:

A would sink to the bottom.Since B just remains where it is shown, its mass must balance the buoyant force and its weight must be equal to the weight of the water displaced. Since A is heavier, its weight will be greater than its buoyant force, it will sink. Since C is lighter, its buoyant force will be greater thanits weight and it will float to the top.

Answer and Explanation:

When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.

Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.

Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:

P = IV                 -------------(i)

Where;

P = Power rating = 1000W

I = Current used

V = Voltage rating = 240V

Substituting these values into equation (i) gives;

1000 = I x 240

I = [tex]\frac{1000}{240}[/tex] = 4.17A

Therefore, the current used by the stove is 4.17A

To get the resistance R of the stove, we can use the relation;

P = [tex]\frac{V^2}{R}[/tex]

R = [tex]\frac{V^2}{P}[/tex]

R = [tex]\frac{240^2}{1000}[/tex]

R = 57.6Ω

Therefore, the resistance of the stove is 57.6Ω

Answer:

a)    a = 17.1 m / s², b)    F = 3.04 N

Explanation:

This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities

* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components

* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components

We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is

                θ = 10 -30 = -20º

The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force

              sin (-20) = F_{y} / F

              cos (-20) = Fₓ / F

              F_{y} = F sin (-20)

              Fₓ = F cos (-20)

              F_y = 18 sin (-20) = -6.16 N

              Fₓ = 18 cos (-20) = 16.9 N

The decomposition of the weight is the customary

               sin 30 = Wₓ / W

               cos 30 = W_y / W

               Wₓ = W sin 30 = mg sin 30

                W_y = W cos 30 = m g cos 30

                Wₓ = 0.8 9.8 sin 30 = 3.92 N

                 W_y = 0.8 9.8 cos 30 = 6.79 N

Notice that in the case  the angle is measured with respect to the axis y perpendicular to the plane

Now we can write Newton's second law for each axis

X axis

      Fₓ - fr = m a

Y Axis  

      N - [tex]F_{y}[/tex] - Wy = 0

      N =F_{y} + Wy

      N = 6.16 + 6.79

They both go to the negative side of the axis and

      N = 12.95 N

The friction force has the formula

        fr = μ N

we substitute

        Fₓ - μ N = m a

        a = (Fₓ - μ N) / m

we calculate

       a = (16.9 - 0.25 12.95) / 0.8

       a = 17.1 m / s²

b) now the block slides down with constant speed, therefore the acceleration is zero

ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced

       Newton's law for the x axis

              Fₓ -fr = 0

              Fₓ = fr

              F cos 20 = μ N

              F = μ N / cos 20

we calculate

              F = 0.25 12.95 / cos 20

              F = 3.04 N

this is the force applied at an angle of 10º to the horizontal

Answer:

F = 213.75 N

Explanation:

First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.

ωf = ωi + αt

where,

ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s

ωi =initial angular velocity = 0 rad/s

t = time = 12 s

α = angular acceleration = ?

Therefore,

0.58 rad/s = 0 rad/s + α(12 s)

α = (0.58 rad/s)/(12 s)

α = 0.05 rad/s²

Now, we shall find the linear acceleration of the merry-go-round:

a = rα

where,

a = linear acceleration = ?

r = radius = 4.5 m

Therefore,

a = (4.5 m)(0.05 rad/s²)

a = 0.225 m/s²

Now, the force is given by Newton;s 2nd Law:

F = ma

where,

F = Force = ?

m = mass pf merry-go-round = 950 kg

Therefore,

F = (950 kg)(0.225 m/s²)

F = 213.75 N

Answer:

C. The spinning of an object on its axis.

Explanation:

The definition of rotation answers this question.

A good way to remember is that when you rotate, you turn, and when you revolve, you move around.

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]

angular frequency ω = 2π / T

= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 (  ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ  = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

That is how:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

Answer:

[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]

Explanation:

[tex]\mathrm{force=mass \times acceleration}[/tex]

The force is given 0 newtons.

[tex]\mathrm{force=0 \: N}[/tex]

Plug force as 0.

[tex]\mathrm{0=mass \times acceleration}[/tex]

Divide both sides by mass.

[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]

[tex]\mathrm{0 =acceleration}[/tex]

[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]

Answer:

The  wavelength is  [tex]\lambda = 1029 nm[/tex]

Explanation:

From the question we are told that

    The  wavelength of the left light is  [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]

      The  voltage across A  and  B is  [tex]V_{AB } = 1.2775 \ V[/tex]

Let the stopping potential  at A  be [tex]V_A[/tex] and the electric potential at B  be  [tex]V_B[/tex]

The voltage across A and B is mathematically represented as

      [tex]V_{AB} = V_A - V_B[/tex]

Now  According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

        [tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]

Where  W is the work function of the metal

             h is the Planck constant with values  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             c  is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

          [tex]eV_B = \frac{h * c}{\lambda } - W[/tex]

So  

      [tex]eV_{AB} = eV_A - eV_B[/tex]

=>    [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]

=>   [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]

=>   [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]

=>  [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]

Where e is the charge on an electron with the value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>   [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]      

=>  [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]  

=>   [tex]\lambda = 1.029 *10^{-6} \ m[/tex]

=>   [tex]\lambda = 1029 nm[/tex]

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz