Answer:
63.750KeV
Explanation:
We are given that
Initial velocity of second electron,
Radius,
1 m=100 cm
Magnetic field,B=0.0370 T
We have to determine the energy of the incident electron.
Mass of electron,
Charge on an electron,
Velocity,
Using the formula
Speed of electron,
Speed of second electron,
Kinetic energy of incident electron=
Kinetic energy of incident electron=
Kinetic energy of incident electron=
1KeV=1000eV
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2
Find:
Horizontal displacement
Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s
Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
<span>Whatever the focus of the experiment is, plus any others factors that might influence the outcome of the experiment. If you are testing a new cancer drug, the experimental group and the control group must both be people with the same type of cancer, and both be a representative distribution of the population, all races, genders, ages, etc. You want the only difference in the two groups to be what you are studying, i.e. the effects of the drug.</span>
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
B would be the correct answer
