Which items are directly related to radioactive decay? electrostatic forces of repulsion e = mc2 electrostatic forces of attract
2 answers:
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Answer:
T = mg/6
Explanation:
Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.
The center rotates about the point where the cords touch the edge. Sum the torques about that point:
∑τ = Iα
mgr = (1/2 mr² + mr²) α
mgr = 3/2 mr² α
g = 3/2 r α
α = 2g / (3r)
(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)
Now, sum of the forces in the y direction:
∑F = ma
2T − mg = m (-a)
2T − mg = -ma
Since a = αr:
2T − mg = -mαr
Substituting expression for α:
2T − mg = -m (2g / (3r)) r
2T − mg = -2/3 mg
2T = 1/3 mg
T = 1/6 mg
The tension in each cord is mg/6.
Answer:
5.625km/h
Explanation:
We are given that
Distance between home and market, d=2.5 km
Speed, v1=5km/h
Speed, v2=7.5 km/h
We have to find the average speed of the man over the interval of time 0 to 40min.
Time,
Using the formula
1 hour =60 min
Distance traveled by man in 10 min with speed 7.5 km/h=
Therefore,
Total distance covered=2.5+2.5/2=3.75 km
Time=40 min=40/60=2/3 hour
Average speed=
Average speed=