Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
Answer:
The answer to your question is a = -1.85 m/s² the acceleration is negative because it is coming to stop.
Explanation:
Data
vo = 25 m/s
t = 13.5 s
a= ?
vf = 0 m/s
Formula
vf = vo + at
solve for a
a = (vf - vo)/t
Substitution
a = (0 - 25) / 13.5
Simplification
a = -25/13.5
Result
a = -1.85 m/s²
<span>Charge of the glass bead Q = 8.0 x 10^-9 C
Distance d = 2.0 cm = 0.02 m
Coulombs constant K = 8.99 x 10^9 Nm^2/C^2
Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2
E = 71.92 / 0.0004 = 17.98 x 10^4
The electric field is 1.8 x 10^5 N/C</span>
G
has the SI units
m
3
k
g
⋅
s
2
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