Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.

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:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

Answer:

Plss see attached file

Explanation:

Answer:

Using ohms law

The current is found from Ohm's Law.

I = V /R = E /R = Bxy /Rt.

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

=  10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

=  10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

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

Complete question:

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.

Answer:

The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Explanation:

Given;

length of solenoid, L = 2.4 m

radius of solenoid, R = 1.7 cm = 0.017 m

current in the solenoid, I = 0.19 A

number of turns of the solenoid, N = 2100 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length = N/L

I is current in the solenoid

B = μnI = μ(N/L)I

B = 4π x 10⁻⁷(2100 / 2.4)0.19

B = 4π x 10⁻⁷ (875) 0.19

B = 2.089 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

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:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

Answer:

[similar to]

Explanation:

it is the missing word

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

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

Answer: B. The copper wire (resistance 0.1)

Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law

i = [tex]\frac{V}{R}[/tex]

So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.

Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:

P = [tex]i^{2}*R[/tex]

As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.

In conclusion, the copper wire will dissipate more heat when connected in parallel.

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:

The maximum emf induced in the loop is 0.132 Volts

Explanation:

Given;

radius of the circular loop, r = 9.5 cm

intensity of the wave, I = 0.0295 W/m²

wavelength, λ = 6.40 m

The intensity of the wave is given as;

[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]

where;

B₀ is the amplitude of the field

c is the speed of light = 3 x 10⁸ m/s

μ₀ is permeability of free space = 4π x 10⁻⁵ m/A

[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]

Area of the circular loop;

A = πr²

A = π(0.095)²

A = 0.0284 m²

Frequency of the wave;

f = c / λ

f = (3 x 10⁸) / (6.4)

f = 46875000 Hz

Angular velocity of the wave;

ω = 2πf

ω = 2π(46875000)

ω = 294562500 rad/s

The maximum induced emf is calculated as;

emf = B₀Aω

       = (1.572 x 10⁻⁸)(0.0284)(294562500)

       = 0.132 Volts

Therefore, the maximum emf induced in the loop is 0.132 Volts

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:

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:

The rate at which the magnetic field changes is  [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]

Explanation:

From the question we are told that

   The  electric field strength is [tex]E = 4.0 mV/m = 4.0 *10^{-3} V/m[/tex]

   The  radius of the  circular region where the electric field is induced is

   [tex]d = 1.5 \ m[/tex]

Generally the induced electric field is mathematically represented as

     [tex]E = - \frac{r}{2} * \frac{\Delta B }{\Delta t }[/tex]

The  negative sign show that the induced electric field is acting in opposite direction to the change in magnetic field

Where  [tex]\frac{\Delta B }{\Delta t }[/tex] is the change in magnetic field

So  

       [tex]\frac{\Delta B }{\Delta t } = - \frac{2 * E }{r}[/tex]

substituting values

       [tex]\frac{\Delta B }{\Delta t } = - \frac{2 * 4.0 *10^{-3}}{ 1.5 }[/tex]

       [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]

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:

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: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

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:

The factors include age and puberty

Explanation:

Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.

As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.

Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.

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