Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V
Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
Answer:132.0285
Explanation: Hope this helps!
