Answer:
(a) Region of higher potential
(b) ΔV=1.911V
Explanation:
For Part (a)
Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest
For Part (b)
Potential difference we can calculate bu definition of kinetic energy
ΔK=mv²/2= -qΔV
express in ΔV
ΔV= -mv²/2q
Substitute the given values
So
ΔV= -mv²/2q
ΔV=1.911V
Answer:
Pound
Explanation:
SI Unit stands for the International Standard of Unit. Thus from the options given, metre is a measure of length which is an SI unit. Kilogram is a measure of mass of a goven matter which is also an SI unit. Second is a measure of time which is an SI unit. Pound is a measure of masswhich is not an SI unit but has to be converted to kilograms which an SI unit.
Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J
North and south will apply
Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s
Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?
Applying law of conservation of momentum,
Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s
