Answer:
present
Explanation:
read doesn't change but write is in present tense
Answer:
h f = W + KE
Input energy equals work function plus KE of emitted electron
W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19
W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)
W = (16.6 - 10.1)E-19 = 6.5E-19 J
h f = 6.5E-19 J for electrons to be emitted with zero KE
f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ =
= 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = 
0.005 = 2 x 3.142 x r
r = 0.005/6.284 =
= 0.0079 m =<em> 7.9 mm</em>
Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2
F = 9.00 * 10^9 N.m^2/C^2 * 2.4*10^-8 C * 1.8*10^-6 C / [0.008m]^2 = 38.88 * 10^ -5 N
F = 39 * 10 -5N