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4 years ago

The expression below was formed by combining different gas laws.

Physics

1 answer:

kap26 [50]4 years ago

7 0

Answer:

Avogadro's law.

Explanation:

Avogadro’s law states that, equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Mathematically,

V n

V = Kn where V = volume in cm3, dm3, ml or L; n = number of moles of gas;

K = mathematical constant.

The ideal gas equation is a combination of Boyle's law, Charles' law and Avogadro’s law.

V 1/P at constant temperature (Boyle’s law)

V T at constant pressure ( Charles’law)

V n at constant temperature and pressure ( Avogadro’s law )

Combining the equations yields,

V nT/P

Introducing a constant,

V = nRT/P

PV = nRT

Where P = pressure in atm, Pa, torr, mmHg or Nm-2; V = volume in cm3, dm3, ml or L; T = temperature in Kelvin; n = number of moles of gas in mol; R = molar gas constant = 0.082 dm3atmK-1mol-1

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The ________ method is duct sizing that considers the each duct section has the same air velocity.

Gennadij [26K]

The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.

How fast is the air moving through a duct?

The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).

What happens when the size of ducts changes to the airflow?

Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.

The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.

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5 0

2 years ago

A series RCL circuit consists of a 50 Ω resistor, 500 Ω capacitor, and a 500 Ω inductor. What is the angle of impedance?

lbvjy [14]

Answer:

Zero

Explanation:

here, the inductive reactance and the capacitive reactance is same, so this is the condition for resonance.

In the condition for resonance,

the circuit and the voltage in the circuit is in the same phase and the impedance of the circuit is minimum which is equal to the resistance of the circuit.

The phase angle is given by

$tan\phi =\frac{X_{L}-X_{C}}{R}$

Ф = 0

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3 years ago

when an electric current is flowing through a wire which of the following is also produced energy domain field force

qaws [65]

When an electric current is flowing through a wire, there is a magnetic <em>field</em> produced. The lines of magnetic force are circles around the wire.

4 0

4 years ago

The velocity of a particle traveling in a straight line is given by v = (6t - 3t²) m/s, where t is in seconds. If s = 0 when t =

qaws [65]

Answer

given,

v = (6 t - 3 t²) m/s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx = (6t - 3 t^2)\int dt$

x = 3 t² - t³

Position of the particle at t= 3 s

x = 3 x 3² - 3³

x = 0 m

now, particle’s deceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}(6t - 3 t^2)$

a = 6 - 6 t

at t= 3 s

a = 6 - 6 x 3

a = -12 m/s²

distance traveled by the particle

x = 3 t² - t³

at t = 0 x = 0

t = 1 s , x = 3 (1)² - 1³ = 2 m

t = 2 s , x = 3(2)² - 2³ = 4 m

t = 3 s , x = 0 m

total distance traveled by the particle

D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s

D = 2 + 4 + 2 = 8 m

average speed of the particle

$v_{avg} = \dfrac{distance}{time}$

$v_{avg} = \dfrac{8}{3}$

$v_{avg} =2.67\ m/s$

8 0

3 years ago

A satellite orbits the Earth in an elliptical orbit. At perigee its distance from the center of the Earth is 22500 km and its sp

aleksley [76]

Answer:

$6.09294\times 10^{24}\ kg$

Explanation:

K = Kinetic energy

$v_p$ = Perigee speed = 4280 m/s

$v_a$ = Apogee speed = 3990 m/s

$r_p$ = Perigee Distance = 22500000 m

$r_a$ = Apogee Distance = 24100000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Earth

m = Mass of satellite

In this system the kinetic and potential energies are conserved

$K_p+P_p=K_a+P_a\\\Rightarrow \frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a}\\\Rightarrow \frac{1}{2}m(v_p^2-v_a^2)+GMm\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=0\\\Rightarrow M=\frac{v_a^2-v_p^2}{2G}\times \left(\frac{1}{r_a}-\frac{1}{r_p}\right)^{-1}\\\Rightarrow M=\frac{3990^2-4280^2}{2\times 6.67\times 10^{-11}}\times \left(\frac{1}{24100000}-\frac{1}{22500000}\right)^{-1}\\\Rightarrow M=6.09294\times 10^{24}\ kg$

The mass of the Earth is $6.09294\times 10^{24}\ kg$

3 0

4 years ago