The acceleration of the jet is
Explanation:
Since the jet motion is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the jet in this problem, we have
u = 0 (it starts from rest)
v = 40 m/s (final velocity)
s = 70 m (length of the runway)
Solving for a, we find the acceleration:
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Answer:
101.19keV
Explanation:
r1 = 1.0cm = 0.01m
r2 = 1.90cm = 0.019m
B = 0.050T
q = 1.60*10^-19C
m = 9.11 * 10^-31 kg
Mv /r = qB
v = rqB / m
v1 = (0.01 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31
v1 = 8.78 * 10⁷ m/s
V2 = (0.019 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31
V2 = 1.67 * 10⁸ m/s
Kinetic energy of the system K.E = K.E₁ + K.E₂
K.E₁ = ½mv₁² = ½ * 9.11 *10⁻³¹ * (8.78*10⁷)²
k.e₁ = 3.511 * 10⁻¹⁵ J
k.e₂ = 1/2 mv₂² = 1/2 * 9.11*10⁻³¹ * (1.67*10⁸)²
k.e₂ = 1.27*10⁻¹⁴J
K.E = k.e₁ + k.e₂
K.E = 3.511*10⁻¹⁵ + 1.27*10⁻¹⁴
K.E = 1.6211*10⁻¹⁴J
1eV = 1.602 * 10⁻¹⁹J
xeV = 1.6211*10⁻¹⁴J
x = 101,192.23eV = 101.192keV