A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

Answer:

a) rms of electric field =

[tex]E_{rms}[/tex]= 25.97 V/m

b) rms of magnetic field

[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸

[tex]B_{rms}[/tex] = 86.55nT

Explanation:

given

power p = 90.0W

distance d = 2.00m

Intensity = [tex]\frac{power}{area}[/tex]

I = [tex]\frac{p}{A}[/tex]

A = [tex]4\pi d^{2}[/tex]

I = [tex]\frac{p}{4\pi d^{2} }[/tex]

I = [tex]\frac{90}{4\pi(2^{2}) }[/tex]

I = 1.79 W/m²

a) [tex]I_{ave}[/tex] = ε₀ × [tex]E^{2} _{rms}[/tex] × c

where ε₀ is permittivity of free space = 8.85×10⁻¹²,  [tex]E^{2} _{rms}[/tex] is the root mean value and c is speed of light = 3×10⁸m/s

1.79 = 8.85×10⁻¹² × [tex]E^{2} _{rms}[/tex] × 3×10⁸

[tex]E^{2} _{rms}[/tex] = [tex]\frac{1.79}{8.85x10^{-12} x 3x10^{8} }[/tex]

[tex]E^{2} _{rms}[/tex]= 674.1996

[tex]E_{rms}[/tex]= 25.97 V/m

b)for rems magnetic field

[tex]E_{rms}[/tex]= c [tex]B_{rms}[/tex]

[tex]B_{rms}[/tex] = [tex]\frac{E_{rms} }{c}[/tex]

[tex]B_{rms}[/tex] = [tex]\frac{25.97 V/m}{3x10^{8} }[/tex]

[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸

[tex]B_{rms}[/tex] = 86.55nT

Answer:

The quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body

Explanation:

During collision of two bodies, the following quantities are affected;

Kinetic energy of the colliding bodies

change in momentum of the colliding bodies

Force exerted by each body

Two bodies that stick together after collision is inelastic collision.

For inelastic collision, the kinetic energy before collision is greater than kinetic energy after collision.

Change in momentum is zero, that is, momentum before collision is equal to momentum after collision.

According Newton's third law of motion, the force exerted by each body is equal but acts in opposite direction.

Therefore, the quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body

The quantities that always have equal magnitude for both masses during the collision are change in momentum and force exerted by each body

Inelastic collision is a collision in which both bodies stick with each other after collision.

For inelastic collision, the momentum before collision is equal to momentum after collision.

Also, the force exerted by each body is equal but acts in opposite direction.

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Answer:

Rotate a piece of polaroid film about an axis perpendicular to the ray while looking through it in that sky direction.

Explanation:

Polarization involves constraining a transverse wave e.g light waves to vibrate in one phase only. Since unpolarized light vibrates in all direction during propagation. Polarization can be achieved by a polaroid.

A polaroid is a material the make transverse waves to vibrate in one direction after passing through it. It has various applications in sun glasses, wind shield of a car etc.

If the slit of the polaroid is perpendicular to the polarized light from a particular direction in the sky, there would be no propagation of the light. But when it is parallel to the polarized light from the direction, the light would propagate through the polaroid.

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: [tex]n_0[/tex] = 1.52

Wave length of light = λ = 570 nm = [tex]570\times10^{-9}\ m[/tex]

[tex]\text{ Thickness}=\dfrac{\lambda}{4n}[/tex]

[tex]=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}[/tex]

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

Answer:

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

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Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

Answer:

The  acceleration is  [tex]a = 6.2 m/s^2[/tex]

Explanation:

From the question we are told that

    The  angle which the inclined plane make with horizontal is  [tex]\theta = 0.44 \ rad[/tex]

     The frictional coefficients are  [tex]\mu_{\mu s} = 0.61[/tex] and  [tex]\mu_{\mu k} = 0.23[/tex]

The force acting on the  crate  is mathematically represented as  

        [tex]f = F_w + F_N[/tex]

Here f is the net force at which the crate is sliding down the plane which is mathematically represented as

      [tex]f = ma[/tex]

        [tex]F_w[/tex] is the force due to weight which is mathematically represented as

        [tex]F_w = mg sin (\theta)[/tex]

       and  [tex]F_N[/tex] the force due to friction which is mathematically represented as

       [tex]F_N = \mu_{\mu k } * mg cos(\theta )[/tex]

So  

     [tex]ma = mgsin(\theta ) + \mu_{\mu k} mg cos(\theta )[/tex]

      [tex]a = gsin(\theta ) + \mu_{\mu k } * g cos(\theta)[/tex]

substituting values

      [tex]a = 9.8 sin(0.44 ) + 0.23 * 9.8* cos(0.44)[/tex]

      [tex]a = 6.2 m/s^2[/tex]

Answer:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Explanation:

Consider a wire with:

Resistance = R

Length = L

Area = A = πr²

where, r = radius

ρ = resistivity

Then:

R = ρL/A

R = ρL/πr²   --------------- equation 1

Now, the new wire has:

Resistance = R'

Resistivity = ρ

Length = L' = 2 L

Radius = r' = 2r

Area = πr'² = π(2r)² = 4πr²

Therefore,

R' = ρL'/πr'²

R' = ρ(2 L)/4πr²

R' = (1/2)(ρL/πr²)

using equation 1:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Answer:

A.21.3T

B.1.8x 10^6J/m^3

C.0.27x10^2J

D.6.6x10^-3H

Explanation:

Pls see attached file

Answer:

The new acceleration would be 9 m/s².

Explanation:

Acceleration of an object is 6 m/s²

Net force is equal to the product of mass and acceleration i.e.

F = ma

[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]

If the net force was tripled and the mass were doubled, it means,

F' = 3F

m' = 2m

Let a' is new acceleration. So,

[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]

So, the new acceleration would be 9 m/s².

Answer:

The  wavelength is [tex]\lambda = 612nm[/tex] and the color is  Orange

Explanation:

from the question we are told that  

     The thickness is [tex]D = 115 nm = 115 *10^{-9} \ m[/tex]

      The refractive index of water is  [tex]n_w = 1.33[/tex]

Generally the condition for constrictive interference is  

         [tex]2 * D = \frac{\lambda _n}{2}[/tex]

Where  [tex]\lambda _n[/tex] is the wavelength of light in a particular medium

  Now considering the medium of water(soap bubble )

 The  wavelength of light in this medium is mathematically represented as

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

So  

    [tex]2 * D = \frac{\frac{\lambda }{n} }{2}[/tex]

     [tex]2 * D = \frac{\lambda }{2 * n }[/tex]

=>    [tex]\lambda = 4 *n * D[/tex]

substituting values  

      [tex]\lambda = 4 *1.33 * 115*10^{-9}[/tex]

      [tex]\lambda = 6.118 *10^{-7} \ m[/tex]

     [tex]\lambda = 612nm[/tex]

The color is orange because the wavelength range of yellow is  

       590–625 nm

Answer:

Explanation:

a )

current in the wire = potential diff / resistance

= 23 / (15 x 10⁻³ )

= 1.533 x 10³ A .

b )

For resistance of a wire , the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area of wire

putting the given values

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ ohm. m

= 10.6 x 10⁻⁸ ohm m

The metal wire appears to be platinum .

(a) The current in the wire is 1.533 x 10³ A

(b) The resistivity of the wire material is 10.6 x 10⁻⁸ Ωm

(c) The material of the wire is Platinum

(a) According to the Ohm's Law:

V = IR

where V is the potential difference

I is the current

and R is the resistance

So,

I = V/R

I = 23 / (15 x 10⁻³ )

I = 1.533 x 10³ A

(b) The resistance of a wire is expressed as:

R = ρ L / A

where ρ is the resistivity,

L is length

and A is the cross-sectional area

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ Ωm

ρ = 10.6 x 10⁻⁸ Ωm

The metal from which the wire is made is platinum.

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Complete question is;

Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun

Answer:

Minimum distance = 147,099,713.4 km

Maximum distance = 152,096,286.6 km

Explanation:

The formula for the eccentricity of an ellipse is given by;

e = c/a

where;

c is distance from the center of the ellipse to the focus of the ellipse.

a is distance from the center of ellipse to a vertex.

From the question, we are given;

a = 149,598,000

e = 0.0167

Thus;

0.0167 = c/149,598,000

c = 0.0167 × 149,598,000

c = 2,498,286.6

Now, formula for the minimum distance (perihelion) is;

Minimum distance = a - c

Minimum distance = 149598000 - 2498286.6

Thus;

Minimum distance = 147,099,713.4 km

Similarly, formula for the maximum distance (aphelion) is;

Max distance = a + c

Max distance = 149598000 + 2498286.6

Maximum distance = 152,096,286.6 km

Answer

fringe separation l distance between maxima decreases

Explanation:

Because the wavelength of blue light is smaller than that if red light

Answer:

1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

2)If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum

3) must be able to see the well-collimated light emission source

Explanation:

1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.

These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.

The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.

2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.

If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.

3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.

The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.

Answer:

1

  [tex]R = 1.692*10^{6} \Omega[/tex]

2

  [tex]C = 2.837 *10^{-6} \ F[/tex]

Explanation:

From the question we are told that

    The voltage is  [tex]E = 110 \ V[/tex]

     The current is  [tex]I = 6.5 *10^{-5} \ A[/tex]

     The time constant is  [tex]\tau = 4.8 \ s[/tex]

The resistance of resistor is mathematically evaluated as

      [tex]R = \frac{E}{I}[/tex]

substituting values

      [tex]R = \frac{ 110 }{ 6.5*10^{-5}}[/tex]

      [tex]R = 1.692*10^{6} \Omega[/tex]

The  capacitance of the capacitor is mathematically represented as

        [tex]C = \frac{\tau}{R}[/tex]

substituting values

       [tex]C = \frac{ 4.8}{ 1.692*10^{6}}[/tex]

       [tex]C = 2.837 *10^{-6} \ F[/tex]

one of the answers that i found was   5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.

Answer:

The positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

Explanation:

Given;

distance between the conducting planes, d = 80 cm

frequency of the electromagnetic wave, f = 750 MHz

speed of light, c = 3 x 10⁸ m/s²

Determine the wavelength

λ = C/f

where;

λ is the wavelength

C is the speed of light

f is the frequency

λ = C/f

λ = (3 x 10⁸) / (750 x 10⁶)

λ = 0.4 m = 40 cm

One complete cycle = one wavelength = 40 cm

half of the wavelength ( λ / 2) = 20 cm

one wavelength + half wavelength (3λ / 2) = 60 cm

The positions of the wave at zero amplitude (between 0 and 80 cm) = 20 cm, 40 cm, 60 cm

Thus, the positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

Answer:

a) Height reached before coming to rest is 42.86 m

b) Energy lost to friction is 17902.45 J

Explanation:

mass of the motorcycle = 185 kg

speed of the towards the hill = 29 m/s

The wheels weigh 12 kg each

Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m

a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height

the kinetic energy of the motorcycle is given as

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

where m is the mass of the motorcycle

v is the velocity of the motorcycle

[tex]KE[/tex]  = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J

This will be converted to potential energy

The potential energy up the hill will be

[tex]PE[/tex] = mgh

where m is the mass

g is acceleration due to gravity 9.81 m/s^2

h is the height reached before coming to rest

[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h

equating the  kinetic energy to the potential energy for energy conservation, we'll have

77792.5 = 1814.85h

height reached before coming to rest  = 77792.5/1814.85 = 42.86 m

b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is

[tex]PE[/tex] = mgh

[tex]PE[/tex]  = 185 x 9.81 x 33 = 59890.05 J

The energy lost to friction will be the kinetic energy minus this potential energy.

energy lost = 77792.5 - 59890.05 = 17902.45 J

A) The motorcycle can coast up the hill by ; 42.86m  

B) The amount of energy lost to friction :  17902.45 J

A) Determine how high the motorcycle can coast up the hill when friction is neglected

apply the formula for kinetic and potential energies

K.E = 1/2 mv²  ---- ( 1 )

P.E = mgH  ---- ( 2 )

As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.

∴ K.E = P.E

1/2 * mv² = mgH

∴ H = ( 1/2 * mv² ) / mg  ---- ( 3 )

where ; m = 185 kg ,  v = 29 m/s ,  g = 9.81

Insert values into equation ( 3 )

H ( height travelled by motorcycle neglecting friction ) =  42.86m  

B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest  

P.E = mgh = 185 * 9.81 * 33  = 59890.05 J

where : h = 33 m , g = 9.81

K.E = 1/2 * mv² = 77792.5 J   ( question A )

∴ Energy lost ( ΔE ) =  ( 77792.5  - 59890.05 ) = 17902.45 J

Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction :  17902.45 J.

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(a) The drift speed of electrons in the wire is 2.22 x 10⁻⁴ m/s.

(b) The potential difference between two points in the wire is 0.013 V.

The given parameters;

The drift speed of electrons in the wire is calculated as follows;

[tex]v_d = \frac{I}{qn A} \\\\but , \ \frac{I}{A} = \frac{E}{\rho} \\\\v_d = \frac{E}{qn \rho}[/tex]

where;

[tex]v_d = \frac{0.052}{1.602 \times 10^{-19} \times 8.5\times 10^{28} \times 1.72 \times 10^{-8}} \\\\v_d = 2.22 \times 10^{-4} \ m/s[/tex]

The potential difference between two points in the wire, separated by 25 cm;

V = Ed

where;

V = 0.052 x 0.25

V = 0.013 V

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Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

Answer:

The approximate displacement of the object is  23  m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]

= 23 m

Answer:

Vx' = (Vx - u) / (1 - Vx *u / c^2)      velocity transformation formula

In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship

VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c

VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c

In both cases an observer on earth will observe the light traveling at speed c.

Answer:

The drag coefficient of the car is 0.189

Explanation:

mass of the car = 2250 kg

Frontal area of the car  = 2.35 m^2

initial speed of the car = 72 km/hr = (72 x 1000)/3600 = 20 m/s

final speed of the car = 54 km/hr = (54 x 1000)/3600 = 15 m/s

time taken by the car to slow down = 105 sec

We'll assume that the value of the drag coefficient is constant throughout the deceleration.

The car decelerates from 20 m/s to 15 m/s in 105 seconds, the deceleration is calculated from

[tex]a = \frac{v-u}{t}[/tex]

where a is the deceleration

v is the final speed of the car

u is the initial speed of the car

t is the time taken to decelerate.

imputing values, we'll have

[tex]a = \frac{15-20}{105}[/tex] = -0.0476 m/s^2  (the -ve sign indicates a deceleration, which is a negative acceleration)

we can safely ignore the -ve sign in other calculations that follows

The force (drag force) with which the air around the decelerates the car is equal to..

[tex]F_{D} = ma[/tex]

where [tex]F_{D}[/tex] is the drag force

m is the mass of the car

a is the deceleration of the car

imputing values, we'll have

[tex]F_{D} = 2250*0.0476[/tex] = 107.1 N

equation for drag force is

[tex]F_{D} = \frac{1}{2}pAC_{D} v^{2}[/tex]

where p is the air density ≅ 1.225 kg/m³

A is the frontal area of the car

[tex]C_{D}[/tex] is drag coefficient of the car

v is the relative velocity of air and the car, and will be taken as the initial velocity of the car before starting to decelerate.

imputing these values, we'll have

[tex]107.1 = \frac{1}{2}*1.225*2.35*C_{D}*20^{2}[/tex] = 575.75[tex]C_{D}[/tex]

[tex]C_{D}[/tex] = 107.1/575.75 = 0.189

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

A) geometric optics, B) geometric optics , c) geometric optics ,

e) geometric optics, f)  geometric optics

Explanation:

For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing

                λ> = a

Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact

let's analyze each situation

a) let's find the wavelength

                v = λ f

                λ= v / f

                λ= 343/1000

                λ = 0.343 m

                 0.343 << 1m

therefore the description of the geometric optics of

b) red light passes through the pupil of the eye

red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me

         λ = 700 10⁻⁹ m = 7 10⁻⁷ m

         a = 8 mm 10⁻³

longitudinal is much less therefore the geometric optics is correct

c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m

again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient

d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz

let's find the wavelength

          c = λf

          λ = c / f

           λ= 3 108 / 1,400 106

          λ= 2.14 102 m

in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon

e) X-rays have wavelength lam = 10-10 m

the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient

Answer:

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

Answer:

Explanation:

Here potential energy lost = mgh

h = L ( 1 - cos 46 ) where L is length of rope

= 2.6 x ( 1 - cos 46 )

PE lost = 20 x 9.8 x 2.6 ( 1 - cos 46 )

= 155.6 J

gain of kinetic energy = loss of PE

1/2 m v² = 155.6

.5 x 20 x v² = 155.6

v² = 15.56

v = 3.94 m /s