Given Information:
Radius = r = 0.5 m
Magnetic field = 1.0 T
Required Information:
Period = T = ?
Speed = v = ?
Kinetic energy = KE = ?
Answer:
Period = 0.13x10⁻⁶ seconds
speed = 24.16x10⁶ m/s
Kinetic energy = 12.11 MeV
Explanation:
(a) period
The time period of alpha particle is related to its orbital speed as
T = 2πr/v eq. 1
According to newton's law
F = ma
Force due to magnetic field is given by
F = qvB
qvB = ma
qvB = m(v²/r)
qB = mv/r
v = qBr/m eq. 2
substitute the eq. 2 in eq. 1
T = 2πr/qBr/m
r cancels out
T = 2π/qB/m
T = 2πm/qB
T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1
T = 0.13x10⁻⁶ seconds
(b) speed
From equation 1
T = 2πr/v
v = 2πr/T
v = 2π*0.5/0.13x10⁻⁶
v = 24.16x10⁶ m/s
(c) kinetic energy (in electron volts)
Kinetic energy is given by
KE = 0.5mv²
KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²
KE = 1.94x10⁻¹² J
since 1 electron volt has 1.602x10⁻¹⁹ J
KE = 1.94x10⁻¹²/1.602x10⁻¹⁹
KE = 12.11 MeV
The way that scientists ensure that data is reliable Clear Explanation
D is the amount of space object takes up
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
It results change only in it's kinetic energy, it's KE will increase in accord with the work-energy theorem
