"physical science" is a branch of science that is based on practical tests and explanations of the different phenomena. It is based on scientific evidence and tests/experiments.

Some of the branches that are based on physical science are:
1- Astronomy
2- Electronics
3- Engineering
4- Radiology

5 0

4 years ago

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According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

Greeley [361]

Answer:

16.125 Pa

Explanation:

The Bernoulli equation despising the height changes is:

$\frac{P_{2}-P_{1}}{pg}+(\frac{v_{2}^{2}-v_{1}^{2}}{2g})=0$

The gravity constant can be cancelled.

Applying the equation to the first situation, $v_{1}=0$ , $v_{2}=1$ , $P_{2}-P_{1}=-0.645$

The density 'p' may be calculated because it is the only unknown.

$p=\frac{-2(P_{2}-P_{1})}{v_{2}^{2}}=\frac{-2*-0.645}{1^{2}}=1.29\frac{kg}{m^{3} }$

Applying the equation to the second situation, where the only unknown is the pressure drop ( $P_{2}-P_{1}$ ):

$P_{2}-P_{1}=-p*(\frac{v_{2}^{2}-v_{1}^{2}}{2})$

$P_{2}-P_{1}=-1.29*(\frac{5_{2}^{2}}{2})=-16.125$

In both cases the assumption is $v_{1}=0$ because its supposed that the air is stored.

5 0

4 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun

AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0

3 years ago