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
Power = work / time = 8000J / 20s = 400W
That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.
Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.