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2 years ago

For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.

Physics

1 answer:

Lelechka [254]2 years ago

6 0

Answer:

B. 161.5 J

Explanation:

$n$ = Number of moles = 0.37

$\Delta T$ = Rise in the temperature of the oxygen gas = 15 K

$Q$ = heat added in order to raise the temperature

$c_{p}$ = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

$Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J$

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g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

2 years ago

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by

Marizza181 [45]

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

$f_o$ = Focal length of objective = 0.9 cm

$f_e$ = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

$m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52$

The angular magnification of the compound microscope is -252.52

6 0

3 years ago

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60

german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

$m_1$ = Mass of glider without person = 680-60 kg

$v_1$ = Gliders speed just after the skydiver lets go

$m_2$ = Mass of person = 60 kg

$v_2$ = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

$m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s$

The gliders speed just after the skydiver lets go is 34 m/s

3 0

3 years ago

O How are electricity and magnetism related?

Nataly [62]

Magnetism is the product of a moving charged particle. We can have electricity without magnetism but we can not have magnetism without electricity.An electro magnet is made so that we have a soft metal core and electricity around it. A bar magnet is a normal magnet in bar shape with permanent magnetism.

6 0

3 years ago

Pls help me with this problem!!

Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

$v=\dfrac{2\pi r}{T}$

Putting all the values in the above formula

$v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s$

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0

2 years ago