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 heat energy required, Q = 6193.8 J
Explanation:
Given,
The mass of ice cube, m = 18.6 g
The heat of fusion of ice, ΔHₓ = 333 J/g
The heat energy of a substance is equal to the product of the mass and heat of fusion of that substance. It is given by the equation,
<em> Q = m · ΔHₓ joules</em>
Substituting the given values in the above equation
Q = 18.6 g x 333 J/g
= 6193.8 J
Hence, the heat required to melt the ice cube is, Q = 6193.8 J
