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REY [17]
3 years ago
15

The trajectory an air mass follows, whether cyclonic or anti-cyclonic, greatly affects what characteristics of the air mass?

Physics
1 answer:
Ahat [919]3 years ago
6 0

Answer:

Stability

Explanation:

The Stability of an air mass  is greatly affected by trajectory that air mass follows whether the trajectory is cylonic or anticyclonic.

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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
3 years ago
PLEASE HELP ASAP!! Will give brainliest!!<br><br> Which item is necessary to make an electromagnet?
vichka [17]
To make something electrical you need electricity and the wires will turn a magnet into an electromagnet.
or for a short answer: <span>1) soft iron core 2) coil of insulated wire 3)source of electricity</span>
8 0
3 years ago
A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m
BaLLatris [955]

Answer: a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass


m = son's mass = M/3


V = father's initial speed


v = son's initial speed

(1/2)MV^2 = (1/2)*(1/2)*m v^2


M*V^2 = (1/2)(M/3)v^2


V^2/v^2 = 1/4


V = v/2

Second equation:


(1/2)M*(V + 1.4)^2 = (1/2)m*v^2


= (1/2)*(M/3)*(3V)^2


cancel out the M's and (1/2)'s


(V + 1.4)^2 = 3V^2


V^2 + 2.8V + 1.96 = 3V^2


V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

3 0
3 years ago
Read 2 more answers
Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from
mojhsa [17]

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

8 0
3 years ago
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
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