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vredina [299]
2 years ago
11

Anyone wanna join whereby? https://whereby.com/dancing-goddess

Physics
1 answer:
AleksandrR [38]2 years ago
6 0
Sure let me download it rn
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Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
2 years ago
Give me 1 example of complete inelastic collision​
Allisa [31]

i hope it helped thanks

4 0
2 years ago
Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h
Sveta_85 [38]
Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0
8 0
3 years ago
A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt
Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m ,  λ₂ =  2.175 10⁻¹⁰ m,    E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

          E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

          Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

          Ef = E₁ + E₂

         E₀ = Ef

         E₀ = E₁ + E₂

The energy can be found with the Planck equation

          E = h f

          c = λ f

          f = c / λ

          E = hc / λ

They indicate that the wavelength of the second photon is

 

           λ₂ =  λ₁ +0.130 10⁻⁹

We replace

           E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

           E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

          3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ +  λ₁) /  λ₁ ( λ₁ + 0.13 10⁻⁹)

          1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹  λ₁) = 2  λ₁ + 0.13 10⁻⁹

           λ₁² + 0.13 10⁻⁹  λ₁ = 1.25 10⁻¹⁰  λ₁ + 8.125 10⁻²¹

            λ₁² + 0.005 10⁻⁹  λ₁ = 8.125 10⁻²¹

            λ₁² + 5 10⁻¹²  λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

            λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

    λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

             λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

            λ₁ = 87.5 10⁻¹² m

             λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

            λ₁ = 87.5 10⁻¹² m

            λ₂ =  λ₁ + 0.13 10⁻⁹

             λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

             λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

            E₂ = E₀ –E₁

            E₂ = E₀ –hc / λ₁

            E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

            E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

             E₂ = 0.93 10⁻¹⁵ J

             E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

             E₂ = 5.8 10³ eV

7 0
3 years ago
Please select the word from the list that best fits the definition
vagabundo [1.1K]

Answer:input work

Explanation:

7 0
2 years ago
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