what is the mathematical definition of momentum? what is a more conceptual or descriptive definition of momentum?​

Answer:

Endothermic, because the reactants are lower in energy (C)

Explanation:

From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that  when the enthalpy change  Eproducts is gretater than  Ereactants, the reaction is said to be endothermic.

Solution :

The volume rate of flow is given by : R = 5.0 L/s

                                                                 [tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]

The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]

∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]

             = 0.637 meter per second

Then the speed of the water at wider section,

[tex]$R=A_1v_1$[/tex]

Similarly, the speed of water at narrow pipe.

The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m

[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]

             = 2.55 meter per sec

Now from Bernoulli's theorem,

[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]

[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]

    [tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]

    = 50 kPa + 3.05 kPa

    = 53.05 kPa

or 53000 Pa

This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.

The pressure reading in the wide section is "53.05 KPa".

First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.

[tex]V = A_1v_1[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m

A₁ = Area of narrow section = πr₁² = π(0.025 m)²

v₁ = velocity at narrow section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]

Similarly,

[tex]V = A_2v_2[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m

A₂ = Area of wide section = πr₁² = π(0.05 m)²

v₂ = velocity at wide section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]

Now, we will use Bernoulli's Theorem to find out the pressure wide section.

[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]

where,

[tex]\rho[/tex] = density of water = 1000 kg/m³

P₁ = pressure in narrow section = 50 KPa = 50000 Pa

P₂ = pressure in wide section = ?

Therefore,

[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]

P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa

P₂ = 53046.45 Pa = 53.05 KPa

Learn more about Bernoulli's Theorem here:

https://brainly.com/question/13098748?referrer=searchResults

The attached picture shows Bernoulli's Theorem.

Answer:

Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.