1) 20 min = 1/3 h
2) distance = speed * time = 12 * 1/3 = 4 km
1.2 x (2.2 x 10⁵) = 264,000 Ω
0.8 x (2.2 x 10⁵) = 176,000 Ω
With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .
Complete Question
An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?
Answer:
The magnetic field strength is 
Explanation:
The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

Making v the subject
Where m is the mass of electron
v is the velocity of electron
q charge on electron
is the potential difference
Substituting values
f

For the electron to move in a circular path the magnetic force[
] must be equal to the centripetal force[
] and this is mathematically represented as

making B the subject

r is the radius with a value = 5.4cm = 
Substituting values


Answer:
m(P4) = 46.175 (grams)
m (KClO3) = 149 (grams)
Explanation:
1) n(P4) = n(P4O10);
m(P4)/M(P4) = m(P4O10)/M(P4O10);
m(P4) = M(P4)*m(P4O10)/M(P4O10)
= 123.90*105.8/283.89
= 46.175 (grams)
2) Analogously, 10n(P4O10) = 3n(KClO3)
m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
= 10*122.55*105.8/283.89/3
= 149 (grams).