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:
(for small oscillations)
Explanation:
The total energy of the pendulum is equal to:

For small oscillations, the equation can be re-arranged into the following form:

Where:
, measured in radians.
If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is
and the total energy of the pendulum is:

The factor of change is:


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1250 decigrams
1 gram = 10 decigrams
Answer:

Given:
Bone density = 2.0 kg/m³
Volume of bone (V) = 0.00027 m³
To Find:
Mass of an adult femur bone (m).
Explanation:
