Q10. Refer to the Condon table to answer question

1. Where are stars formed?

Ugo [173]

$\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}$

C. A nebula.

thanks
hope it helps

3 0

3 years ago

Read 2 more answers

Algunas fabricas de balones de fútbol ubicadas en la costa inflan los balones que van a ser vendiéndose las ciudades como pasto,

otez555 [7]

Answer:

balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure

Explanation:

Este ejercicio es referente a la mecánica de fluidos, usemos la expresión para la presión

P = ρ g h

En es el caso del balón usemos la presión en la pared extrema, llamemos P la presión por el gas en el interior y P_ext la presión atmosférica del lugar

cuando se llena el valor en una ciudad de baja altura la presión atmosférica es mas alta

P_int1 < P_ext1

por lo cual la pared del balón no se mantiene rígida.

Cuando el balón es trasladado a una ciudad con mayor altura sobre el nivel del mar la presión exterior disminuye

P_ext2 = ρ g h₂ < P_ext1

en promedio la presión disminuye con la altura en 0,029 atm cada 250 m

por lo tanto como la cantidad de gas es constante en el interior la presión interior del globo es constante y en esta ciudades se hace igual o un poco mayor que la presión atmosférica, en consecuencia la pared del globo esta rígida

P_int2 >P_ext2

Traslate

This exercise is related to fluid mechanics, let's use the expression for pressure

P = ρ g h

In the case of the balloon, let's use the pressure on the extreme wall, let's call P the pressure for the gas inside and P_ext the atmospheric pressure of the place

when the value is filled in a low-lying city the atmospheric pressure is higher

P_int1 <P_ext1

therefore the wall of the ball does not remain rigid.

When the ball is transferred to a city with higher altitude above sea level, the external pressure decreases

P_ext2 = ρ g h <P_ext1

on average the pressure decreases with height by 0.029 atm every 250 m

therefore as balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure, therefore the wall of the

Pint 2> Pe

8 0

3 years ago

What would happen to the wavelength of a wave as the velocity is increased and the

slamgirl [31]

Answer: wavelength will increase

Explanation:

Velocity is directly proportional to wavelength,as the velocity is increased, wavelength is also increased

5 0

4 years ago

Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

4 0

3 years ago

mass is 5 lb attached to a rope wound around a pulley. The radius of the pulley is 3 in. If the mass falls at a constant velocit

suter [353]

Answer:

The power transmitted to the pulley is 0.0455 hp.

Explanation:

Given;

mass attached to the rope, m = 5 lb

radius of the pulley, r = 3 in

constant rate of fall of the mass, v = 5 ft/s

acceleration due to gravity, g = 32.2 ft/s²

1 lbf = 32.2 lb.ft/s²

The power transmitted to the pulley is calculated as;

P = Fv

P = (mg)v

$P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2} \ \ = 25 \ \frac{ft.lbf}{s} \\\\$

in horse power, the power transmitted is calculated as;

$P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s} \ \ = 0.0455 \ hp$

Therefore, the power transmitted to the pulley is 0.0455 hp.

3 0

3 years ago