If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary
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
Answer:
0.00034 m
Explanation:
Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m
Answer:
280 N
Explanation:
Applying Newton's third second law of motion,
F = m(v-u)/t................... Equation 1
Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.
Note: Let the direction of the initial velocity of the ball be positive
Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s
Substitute into equation 1
F = 4(-4-3)/0.1
F = 4(-7)/0.1
F = -28/0.1
F = -280 N.
Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball
