Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
Frequency of the wave is 2 per second
Explanation:
- Frequency is the number of times waves pass at a particular point of time. Here, given time period = 3 s and number of waves that pass in 3 seconds = 6. Find the frequency of the wave per second.
- Frequency is given by the formula
f = 1/T, where f is the frequency and T is the time period
- For 3 seconds, frequency is 6
3 seconds ⇒ 6 waves
⇒ 1 second = 6/3 = 2 waves per second
The answer would be false
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