Lets calculate the initial speed of the block. We know that kinetic energy is given by:
Solving for v₀:
If the speed after it hits a wall is half its original speed then:
v = v₀/2 = 2 m/s / 2 = 1 m/s
Then the kinetic energy at this point is:
The inetic energy of this object at this point is 0.5 J.
Yeah, which one of the three
Answer:
the radius of the protons path is r = 0.85 m.
Explanation:
the force due to magnetic fields lead to the cetripetal force, such that:
F = q×v×B = m×(v^2)/r
q×B = m×v/r
then:
r = m×v/q×B
r = p/q×B
then, the kinetic energy of the proton:
K = 1/2×m×v^2 = p^2/(2×m)
q×B = \sqrt{2×m×K}/r
r = \sqrt{2×m×K}/(q×B)
= \sqrt{2×(1.67×10^-27)×(5.3×1.60×10^-13)}/(1.60×10^-19×0.39)
= 0.85 m
D.as a magnet is being thrust through the coil
Let Mass be m
If its doubled then
Hence force also needs to be doubled.
