Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb

Answer:

a lower energy than

Explanation:

sorry im a month late but is lower energy than

Answer:

Following are the answer to this question:

Explanation:

Formula:

[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]

Calculating point A:

when the value is [tex]0.38[/tex]

[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]

                   [tex]=2.632[/tex]

[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]

               [tex]=542,802.6[/tex]

Calculating point B:

when the value is [tex]0.75[/tex]

[tex]\to D(PC)=\frac{1}{0.75}[/tex]

                [tex]=1.33[/tex]

[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]

             [tex]=275,020[/tex]

Calculating point C:

when the value is [tex]0.28[/tex]

[tex]\to D(PC)=\frac{1}{0.28}[/tex]

                [tex]=3.571[/tex]

[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]

               [tex]=736660.7[/tex]

Calculating point D:

when the value is [tex]0.42[/tex]

[tex]\to D(PC)=\frac{1}{0.42}[/tex]

                [tex]=2.38[/tex]

[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]

               [tex]=490910.7[/tex]

Calculating point E:

when the value is [tex]0.31[/tex]

[tex]\to D(PC)=\frac{1}{0.31}[/tex]

                [tex]=3.226[/tex]

[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]

               [tex]=665370.97[/tex]

Answer:

Option (c) : 20°C

Explanation:

[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]

T(final) = 500* 10 + 100*70/600 = 20°C

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

1. Electrons transferred from sphere to rod.

Option B is correct.

2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.

Given that initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let us consider that the charge absorbed by the plastic rod  is  q

[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Therefore, electrons transferred from sphere to rod

The charge on one electron is, 1.602 x 10⁻¹⁹ C .

Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]

Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.

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Answer:

she need to use 20 cm thick

Explanation:

given data

wants the attic insulated = R-30

purchased = 10 cm thick

solution

as per given we can say that

10 cm is for the R 15

but she want for R 30

so  

R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]  

R 30 thickness = 20 cm

so she need to use 20 cm thick

Answer:

The cross-sectional area of the larger piston is 392 cm²

Explanation:

Given;

output mass of the piston, m₀ = 2000 kg

input force of the piston, F₁ = 500 N

input area of the piston, A₁ = 10 cm² = 0.001 m²

The output force is given by;

F₀ = m₀g

F₀ = 2000 x 9.8

F₀ = 19600 N

The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]

Therefore, the cross-sectional area of the larger piston is 392 cm²

Answer:

Work = 0.36N

Explanation:

Given

Force = 12N

Distance = 0.03m

Weight = 0.05kg

Required

Determine the work done

Workdone is calculated as thus;

Work = Force * Distance

Substitute 12N for Force and 0.03m for Distance

Work = 12N * 0.03m

Work = 0.36Nm

Using proper S.I units

Work = 0.36N

Hence, work done by the spring on the ball is 0.36N

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

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

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

Answer: Search Results

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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.

Explanation:

Surface waves can exhibit characteristics of both longitudinal waves and transverse waves.

Surface waves are a type of mechanical wave that propagate along the interface between two different mediums, such as the ground and air or the surface of water. These waves combine properties of both longitudinal and transverse waves

Similar to longitudinal waves, surface waves involve particles oscillating in the same direction as the wave propagation. This creates compressions and rarefactions, leading to variations in density or pressure. These compressions and rarefactions are characteristic of longitudinal waves.

However, surface waves also exhibit transverse motion. As the wave propagates along the surface, particles move in a perpendicular direction to the wave's motion. This transverse motion causes particles to displace vertically or horizontally, similar to transverse waves.

By combining both longitudinal and transverse characteristics, surface waves possess a complex motion that allows them to travel along the surface while simultaneously causing particles to oscillate both parallel and perpendicular to the direction of wave propagation.

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Answer:

2 cm.

Explanation:

Data obtained from the question include the following:

Original Length (L₁ ) = 10 m

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Linear expansivity (α) = 5×10¯⁶ /°C

Increase in length (ΔL) =..?

Next, we shall determine the temperature rise (ΔT).

This can be obtained as follow:

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Temperature rise (ΔT) =..?

Temperature rise (ΔT) = T₂ – T₁

Temperature rise (ΔT) = 400 – 0

Temperature rise (ΔT) = 400°C

Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:

Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)

α = ΔL/(L₁ × ΔT)

Original Length (L₁ ) = 10 m

Linear expansivity (α) = 5×10¯⁶ /°C

Temperature rise (ΔT) = 400°C

Increase in length (ΔL) =..?

α = ΔL/(L₁ × ΔT)

5×10¯⁶ = ΔL/(10 × 400)

5×10¯⁶ = ΔL/4000

Cross multiply

ΔL = 5×10¯⁶ × 4000

ΔL = 0.02 m

Converting 0.02 m to cm, we have:

1 m = 100 cm

Therefore, 0.02 m = 0.02 × 100 = 2 cm.

Therefore, the length of the plane will increase by 2 cm.

Answer:

A) Length of an edge = 3.38 × 10^(-9) m

B) 34 times the diameter of a molecule.

C) 11 times the atomic spacing in solids.

Explanation:

A) We will use Avogadro's hypothesis to solve this. It states that 1 mole of gas occupies 22.4 L at STP.

We want to find the volume occupied by 1 mole of gas at 1.06 atm pressure and temperature of 28 °C (= 301 K).

Thus, by the ideal gas equation, we have;

V_mole = (1 × 22.4/273) × (301/1.06) = 23.3 L = 0.0233 m³

Now, since from avogadros number, 1 mole of gas contains 6.02 x 10^(23) molecules, then volume occupied by a molecule is given by;

V_molecule = 0.0233/(6.02 × 10^(23)) m³ = 3.87 x 10^(-26) m³

Thus, length of an edge of the cube = ∛(3.87 × 10^(-26)) = 3.38 × 10^(-9) m

B) We are told that The diameter of a typical molecule is about 10^(-10) m.

Thus, the distance is about;

(3.38 × 10^(-9))/(10^(-10)) ≈ 34 times the diameter of a molecule.

C) We are told that the spacing of atoms is typically are about 0.3 nm apart

Thus;

The separation will be about;

(3.38 × 10^(-9))/(0.3 × 10^(-9)) ≈ 11 times the atomic spacing in solids.

Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

Answer:

The frequency does not change, but the wavelength does

Explanation:

Here are the options

A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.

B. The frequency does change, but the wavelength remains unchanged.

C. Both the frequency and wavelength change.

D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.

E. The frequency does not change, but its wavelength does.

When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.

[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]

where [tex]\lambda_o[/tex] indicates wavelength in vacuum

[tex]\lambda_m[/tex] indicates wavelength in medium

n indicates refractive index

v indicates velocity of light wave

c indicates velocity of light

And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.

Therefore the correct option is E

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Answer:

TOTAL INTERNAL REFLECTION

Explanation:

Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.

For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.

Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

Answer

Work done is 57.9KJ

Explanation

First solve the problem according to work done due to variation in temperature

So W= intergral Cu( 1-Tu/T). at Tu and T

So Given that

C = Heat capacity of the Brick

TEPc= Cold Temperature

TEPh = Hot Temperature

W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)

So

W= (1)-(300-150)-300 (1) ln 2

W= -57.9KJ

Answer:

Well, yes.

We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.

Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.

An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.

Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.

Answer:

It will take. the same distance up as before, but take a longer time

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

Answer:

i) 545.2 N  upwards

ii) 828.2 N  downwards

Explanation:

mass of the roller = 70 kg

force exerted = 200 N

angle the force makes with the ground ∅ = 45°

weight of the roller W = mg

where

m is the mass of the roller

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

weight of the roller = 70 x 9.81 = 686.7 N

The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°

==> F = 200 x 0.707 = 141.5 N

i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards

Resultant vertical force = 686.7 N - 141.5 N = 545.2 N  upwards

ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards

Resultant vertical force = 686.7 N + 141.5 N = 828.2 N  downwards

Answer:

D.

Inverted and reduced

If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.

A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

Hence option B is correct.

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Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

​

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

Answer:

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force

Explanation:

galvanometer is an instrument that can detect and measure small current in an electrical circuit.

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.

The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.

Answer:

1. 0.2m

1. 66m

Explanation:

See attached file

The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;

1) The maximum height when the water leaves the hose is: Δy = 0.20 m

2) The maximum height of the water leaves the nozzle is: Δy = 68.6m

Given parameters

To find

   1. Maximum height of water in hose

  2. Maximum height of water at the nozzle

Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:

           A₁v₁ = A₂.v₂

          P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂

Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.

1, Let's find the exit velocity of the water in the hose.

Let's use subscript 1 for the nozzle and subscript 2 for the hose.

The continuity equation of the flow value that must be constant throughout the system.

      Q = A₁ v₁

      v₁ = [tex]\frac{Q}{A_1 }[/tex]  

The area of ​​a circle is:

     A = π r²

Let's calculate the velocity in the hose.

    A₁ = π (0.94 10⁻²) ²

    A₁ = 2.78 10⁻⁴ m²

    v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]

    v₁ = 1.98 m / s

Let's use Bernoulli's equation.

When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2

      ½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂

      y₂-y₁ = ½  [tex]\frac{v_i^2 - v_2^2}{g}[/tex]  

At the highest point of the trajectory the velocity must be zero.

     y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]

Let's calculate

     y₂-y₁ =  [tex]\frac{1.98^2}{2 \ 9.8}[/tex]  

     Δy = 0.2 m

2.  Let's find the exit velocity of the water at the nozzle

          A₁ = π r²

          A₁ = π (0.22 10⁻²) ²

          A₁ = 0.152 10⁻⁴ m / s

With the continuity and flow equation.

           Q = A v

            v₁ = [tex]\frac{Q}{A}[/tex]  

             v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]  

             v₁ = 36.67 m / s

Using Bernoulli's equation, where the speed of the water at the highest point is zero.

           y₂- y₁ =  [tex]\frac{v^1^2}{g}[/tex]  

Let's calculate.

           Δy =  [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]  

           Δy = 68.6m

In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:

      1) The maximum height when the water leaves the hose is:

          Δy = 0.20 m

      2) The maximum height of the water when it leaves the nozzle is:

          Δy = 68.6 m

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