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bixtya [17]
4 years ago
6

Identify the generic outer electron configuration for the halogens.

Physics
1 answer:
Delicious77 [7]4 years ago
8 0
<span>The Halogen outer layer is considered to be full with eight electrons. Halogen is also considered to be part of the Noble gases classification. The exchange of electrons in the halogen atom renders a photon energy pulse with produces light. The consentrated light spectrum is produced in the halogen light bulbs and halogen laser products.</span>
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Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)
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Answer:

(a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

P_{1}=\rho g h_{1}

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

P_{1}=\rho gh_{1}.....(I)

Pressure for second pipe,

P_{2}=\rho gh_{2}.....(II)

From equation (I) and (II)

P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

Put the value of P₁ and P₂

\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)

2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2....(III)

We know that,

The continuity equation

v_{1}A_{1}=v_{2}A_{2}

v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})

Put the value of v₂ in equation (III)

2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2

2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2

Here, \dfrac{A_{1}}{A_{2}}=\gamma

So, 2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)

v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Hence, (a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

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Science Question please help
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Answer:

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Explanation:

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Explanation:

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