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
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>
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
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.
Answer:
(a): 
(b): 
(c): 
Explanation:
Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053
m.
Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges 
and 
respectively is given by
where,
= Coulomb's constant = 
= 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, 
The charge on the electron, 
These two are separated by the distance, 
Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by
Part (b):
The gravitational force of attraction between two objects of masses 
and respectively is given by
where,
= Universal Gravitational constant = 
- = distance of separation between the masses.
For the given system,
The mass of proton, 
The mass of the electron, 
Distance between the two,
Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by
The ratio 
:
