what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s

Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

For the same reason that we need units of length, instead of just "long" or "short". And units of distance instead of "near" and "far". And units of time instead of "early" and "late" or "old" and "new". And units of weight instead of "light" and "heavy". And units of sound-pressure instead of "loud" and "soft".

"Hot" and "cold" mean different things to different people, and may even mean different things to the same person at different times.

A person who grew up in Panama, and comes to visit Chicago in July, says it's cold.

A person who lives in Chicago, and goes to visit Jamaica in January, says it's hot.

A professional chef, following a recipe, can't just cook the steak until it's "hot inside". He needs a number, so he can cook it the same every time.

A technician in a Chemistry lab may have two solutions, and he's supposed to pour half of the cooler one into the warmer one. One of them is 25°C and the other one is 22°C. He's got a problem. He can't tell the difference. He never learned temperature scales. All he knows is "hot" and "cold", and they both feel luke-warm to him. He doesn't even have a way to measure them, because temperature scales were never invented. He's stumped. And while he's standing there scratching his head, both solutions drift to the same temperature, and the lab goes up in flames. The technician is so petrified, he becomes overwhelmed with shame and regret, and makes himself sick and feverish. His forehead feels hot but nobody can measure his temperature, so nobody knows how sick he is.

All because Franz Fahrenheit and Sven Celsius had planned to invent measurable scales in their lab, but decided to go fishing that day.

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

momentum p= m x v = 15 x -4 = -60 N.s

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

Answer:

the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Explanation:

For an object to have a static equilibrium, it must meet two relationships

             ∑ F = 0

             ∑ τ =0              

force acting on a body fulfills the relation of

         sum F = F - F = 0

when two forces do not move from position.

To find the torque we assume that the counterclockwise rotations are positive

        Σ τ = - F r - F r

        Στ = -2 Fr <> 0

consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Answer:

147 Newtons

Explanation:

To find tension, you can use the formula Tension = (mass)(gravity)

*Gravity's acceleration = 9.8 m/s^2 because of Newton's law of universal gravitation*

T = (15kg)(9.8m/s^2)

  = 147 Newtons

Hope this helps! Best of luck <3

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs

Answer:

53.8Joule

Explanation:

hope it is helpful

Answer:

approximate 153.1J

Explanation:

W= 1/2k(x^2) = 1/2x725x(0.650)^2 = 153.15625 (J)

Answer:

148.5 N

Explanation:

Given that,

The mass of a bungee jumper, m = 55 kg

The downward acceleration, a = 7.1 m/s²

We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :

T = m(g-a)

Put all the values,

T = 55(9.8 - 7.1)

= 148.5 N

So, the force exerted on the bungee cord is 148.5  N.

Answer:

The downward force is 148.5 N.

Explanation:

mass, m = 55 kg

downwards acceleration, a = 7.1 m/s^2

Let the force is F.

According to the newton's second law

m g - F = m a

F = m (g - a)

F = 55 (9.8 - 7.1)

F = 148.5 N

Answer:

it will be 10x

Explanation:

workdone(potential energy before it hits the wall)= horizontal force × distance

=10× x = 10x joules

A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is  10x.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,

Work done (potential energy before it hits the wall)

= horizontal force × distance

=10× x = 10x joules

The potential energy of a particle due to this force is  10x.

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ2

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]

Speed of rocket 1 with respect to rocket 2 :

[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]

[tex]$u' = \frac{0.7 c}{1.12}$[/tex]

[tex]u'=0.625 c[/tex]

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

Answer:

Parte B : 31.18º , Parte C: 31.17º

Explanation:

Parte B: The angle between the glass and the water before it enters the water is going to be equal to the value of the angle when it enters the glass , 27.13º.

Using the formula n1 senθ1 = n2 senθ2 , where n1=1.51 , θ1=27.13º, n2=1.33 , it gives us θ2=31.18º.

Parte C: n1= 1 , θ1=43.5º, n2=1.33

Using the same formula : n1 senθ1 = n2 senθ2 , it gives us θ2= 31.17º.

Answer:

a) Dielectric constant ( λ ) = 750 * 10^-9 m

   minimum thickness of coating ( d )  = 187.5 nm

b) 3.6%

Explanation:

Given data:

wavelength of red light in air = 750 nm

εr = 4

μr = 1,  σ = 0

a) Determine the required dielectric constant and min thickness of coating used

Refractive index of coating ( n ) = √εr * μr =  √4*1 = 2

the refractive index of glass( ng)  = 1.5 which is < 2

λ = 750 * 10^-9 m

Dielectric constant ( λ ) = 750 * 10^-9 m

To determine the minimum thickness we will apply the formula below

d = m λ/2n ;  where  m = 1

∴ d = 750 nm / 2 ( 2 )

      = 187.5 nm

minimum thickness of coating ( d )  = 187.5 nm

b) Determine the percentage of the incident power that will be reflected

R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2

   = [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2

   = 0.03628 =  3.6%

Answer:

A. When moving towards a high pressure center the pressure values ​​increase in the equipment

B. This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

Answer:

P = density (p) * g * h

P = 1000  kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m +  pressure 80 m) for liquid of uniform density

Answer:

The answer is "80 cm".

Explanation:

The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]

[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]

   [tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer: hertz

Explanation:

Wavelength is also measured in metres ( ) - it is a length after all. The frequency ( ) of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz ( ), since one hertz is equal to one wave per second.

Answer:

The number of waves that pass a particular point in a unit of time.

Explanation:

The frequency of a wave describes the number of waves that pass a particular point in a unit of time. The frequency is inversely proportional to the period of the wave, which is the time required for one full wave cycle to complete.

PF

Explanation:

Given that,

Maximum potential, V = 4. mV

Distance, d = 0.350 m

Frequency of the wave, f = 100 Hz

(a) The maximum electric field strength created is given by:

[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]

(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]

(c) The wavelength of the electromagnetic wave can be calculated as :

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]

So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]

Answer:

Explanation:

A bullet being shot out of a gun tends to leave tiny amounts of the bullet behind due to friction between the bullet and the gun barrel.

A roller skater pushing requires the conversion of food chemical energy to muscle contraction energy. This conversion increases the body temperature and sweat is excreted to counteract the heat increase. The evaporation of the sweat causes a slight decrease in body mass.

A jet plane taking off consumes some of the fuel carried onboard to provide thrust. The products of combustion become part of the exhaust stream leaving the airplane rearward providing forward thrust.

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.