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Natasha_Volkova [10]
2 years ago
10

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

w on a planet where the acceleration of gravity varies with height so that g=g0−cz, where g0 and c are constants. Integrate "F=ma" along a streamline to obtain the equivalent of the Bernoulli equation for this flow.
Physics
1 answer:
irina1246 [14]2 years ago
4 0

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

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Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to
Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate m=100 kg

Crate slows down in s=1.5 m

initial speed u=1.77 m/s

inclination \theta =30^{\circ}

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2

W_{gravity}=mg(0-h)=mgs\sin \theta

W_{gravity}=-mgs\sin \theta

W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N

change in kinetic energy=\frac{1}{2}\times 100\times 1.77^2=156.64 J

W_{friction}=156.64+735=891.645

(b)Coefficient of sliding friction

f_r\cdot s=W_{friciton}

891.645=f_r\times 1.5

f_r=594.43 N

and f_r=\mu mg\cos \theta

\mu 100\times 9.8\times \cos 30=594.43

\mu =0.7

5 0
2 years ago
Which statement about electrons and atomic orbitals is NOT true?
balandron [24]

<em>An electron has the same amount of energy in all orbitals  is not true</em>

\boxed{\boxed{\bold{Further~explanation}}}

In an atom there are levels of energy in the skin and sub skin.

  • This energy level is expressed in the form of electron configurations.

Writing electron configurations starts from the lowest to the highest sub-shell energy level.

So electrons that occupy the orbitals in the lowest sub-skin have the lowest energy level

  • In the principle of Pauli's prohibition it was stated that there are no two electrons in one atom that can have the same four quantum numbers.

So suppose that there are two electrons occupying one orbital can have the same main quantum number (n), azimuth (l) and magnetic (m), then the last quantum number that is the quantum spin number (s) must be different.

So that the two electrons are different from just the quantum spin number, even though the other quantum numbers are the same.

So in one orbital only a maximum of 2 electrons is occupied, because if there is a third electron, this third electron will have the same quantum spin number as the previous electron

  • The electron cloud is a visual representation of the location of electrons in an atom.

Orbital is the place around the nucleus where electrons may be found

Electron clouds show the state of electrons in their orbitals

So electron clouds can show the condition of all orbitals in an atom

The lowest energy level of an electron occupies a sub-skin of 1s which has only one orbital

Charging electrons in the sub skin uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Statement about electrons and atomic orbitals is not true is An electron has the same amount of energy in all orbitals

the electron configuration for barium (Ba) in noble-gas notation brainly.com/question/11147367

the formation of a bond.

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quantum number

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Keywords: the electron configuration, orbitals, atoms, energy, skin, sub skin, electron clouds

8 0
2 years ago
Read 2 more answers
The wavelength of green light is 550 nm.
SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

    velocity of electron is 1323.88 m/s

   momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

  velocity of electron is 1654.85 m/s

  momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s

velocity of electron is given by;

P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s

velocity of electron is given by;

v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s

momentum of the electron is given by;

p = mv

p =  (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0
2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
2 years ago
You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0
2 years ago
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