Answer:
very hard others will answer it
Explanation:
hard
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv = 
or Eq = 
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
Kinetic energy lost in collision is 10 J.
<u>Explanation:</u>
Given,
Mass,
= 4 kg
Speed,
= 5 m/s
= 1 kg
= 0
Speed after collision = 4 m/s
Kinetic energy lost, K×E = ?
During collision, momentum is conserved.
Before collision, the kinetic energy is

By plugging in the values we get,

K×E = 50 J
Therefore, kinetic energy before collision is 50 J
Kinetic energy after collision:


Since,
Initial Kinetic energy = Final kinetic energy
50 J = 40 J + K×E(lost)
K×E(lost) = 50 J - 40 J
K×E(lost) = 10 J
Therefore, kinetic energy lost in collision is 10 J.
Yes a kg of hydrogen will have more atoms than a kg of lead, because lead has a higher atomic mass, than hydrogen so it will take more atoms of hydrogen to make a kg than lead
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)