F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
Sorry im from the UK, ill do my math in metric :)
5) work done = force * distance moved in direction of force
170J = 140N* ?d
170J/140N = 1.21m
F=Ma
140N/9.81ms^-2=
14.27Kg or 31.46Ib
you could do the fast alternative simplified route to this question
as 140/4.45 = 31.46Ib
Answer:15th
Explanation:
I believe it will fall on the 15th stair because 10 times 15 is 150 divided by 10 m/s equals 15
With another person on student A’s side. I’m sorry if I wasn’t able to help. Im trying my best from the information given. :)
Answer:
2.353 × 10⁵ years
Explanation:
Since speed, v = distance/time
time, t = distance/speed
The distance, d = circumference of orbit = 2πr where r = radius of orbit = 8.5 kiloparsecs = 8.5 kiloparsecs × 3.1 × 10¹³ kilometers/parsec = 26.35 × 10¹³ kilometers and velocity of sun, v = 220 km/s
So, the time it takes to complete one orbit, T is
T = 2πr/v
= 2π × 26.35 × 10¹³ km/220 km/s
= 165.562 × 10¹³ km ÷ 220 km/s
= 0.753 × 10¹³ s
= 7.53 × 10¹² s
We now convert T to years
T = 7.53 × 10¹² s ÷ 3.2 × 10⁷ seconds/year
T = 2.353 × 10⁵ years
