Explanation:
The moment of inertia of each disk is:
Idisk = 1/2 MR²
Using parallel axis theorem, the moment of inertia of each rod is:
Irod = 1/2 mr² + m (R − r)²
The total moment of inertia is:
I = 2Idisk + 5Irod
I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]
I = MR² + 5/2 mr² + 5m (R − r)²
Plugging in values:
I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²
I = 23,750 g cm²
Think of a wedge as something you put in between objects, so it is a separates objects
Statement three i do believe
Answer:
The energy dissipated as the puck slides over the rough patch is 1.355 J
Explanation:
Given;
mass of the hockey puck, m = 0.159 kg
initial speed of the puck, u = 4.75 m/s
final speed of the puck, v = 2.35 m/s
The energy dissipated as the puck slides over the rough patch is given by;
ΔE = ¹/₂m(v² - u²)
ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)
ΔE = -1.355 J
the lost energy is 1.355 J
Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J
Answer:
raise it as high as he can
Explanation:
