A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull
1 answer:
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Answer
given,
radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m
mass of electron, M = 9.11 x 10⁻³¹ Kg
charge of electron, q₁ = 1.6 x 10⁻¹⁹ C
q₂ = 1.6 x 10⁻¹⁹ C
we know, force between two charges
F = 8.20 x 10⁻⁸ N
b) using newton's second law
F = m a
m a = 8.20 x 10⁻⁸
a = 9 x 10²² m/s²
c) speed of the electron
v² = 4.77 x 10¹²
v = 2.18 x 10⁶ m/s
d) the period of the circular motion.
T = 1.53 x 10⁻¹⁶ s
Answer:
b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
N + T_y -W = 0
N = W -T_y
let's use trigonometry for tension
sin θ = T_y / T
cos θ = Tₓ / T
T_y = T sin θ
Tₓ = T cos θ
we substitute
N = W - T sin 30
we calculate
N = 640 - 160 sin 30
N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
Tₓ - fr = 0
Tₓ = fr
we substitute and calculate
fr = 160 cos 30
fr = 138.56 N
d) the friction force has the formula
fr = μ N
μ = fr / N
we calculate
μ = 138.56 / 560
μ = 0.247
