A stream of air enters a 7.00-cm ID pipe at a velocity of 30.0 m/s at 27.0°C and 1.80 bar (gauge). At a point downstrream, the a
1 answer:
You might be interested in
Answer:
the density of the electrum is 14.30 g/cm³
Explanation:
Given that:
The equilibrium fraction of lattice sites that are vacant in electrum =
Number of vacant atoms =
the atomic mass of the electrum = 146.08 g/mol
Avogadro's number =
The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)
=
× Total number of sites(N)
Total number of sites (N) =
Total number of sites (N) =
From the expression of the total number of sites; we can determine the density of the electrum;
where ;
= Avogadro's Number
density of the electrum
= Atomic mass
Thus; the density of the electrum is 14.30 g/cm³
Answer:
prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S
Explanation:
Given that S = { p q |p, q are prime numbers greater than 0}
E = {0, −2, 2, −4, 4, −6, 6, · · · }
To prove by constructing a bijection from S to E
detailed solution attached below
After the bijection :
<em>prove that | S | = | E |</em> : every element of S there is an Image on E , while not every element on E has an image on S
∴ we can say sets E and S are infinite sets
