(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
2 answers:
Answer:
The work done on the Frisbee is 1.36 J.
Explanation:
Given that,
Mass of Frisbee, m = 115 g = 0.115 kg
Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground
Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,
So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.
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Answer:
(a)
(b)
Explanation:
Given:
- Limiting tension of snapping of the rope, T= 387 N
- Weight of the object to be lifted,
- ∴mass,
- height of letting down the weight, h = 6.1 m
(a)
Now,
The force to be compensated for being on the verge of snapping:
(T-w) = 62 N
Therefore, we need to produce and acceleration equivalent to the above force.
∵
(b)
From the equation of motion ,we have:
....................(2)
where:
u= initial velocity= 0 (here, starting from rest)
v= final velocity = ?
s= displacement =h =6.1 m
Now, putting the values in eq. (2)
is the velocity with which the body will hit the ground in the given conditions.
Answer:
2 m/s
Explanation:
From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.
Let the 10.00 kg mass be and the 12.0 kg mass be
. When they collide and stick, they have a combined mass of
.
Momentum is given by . Set up the following equation:
, where
is the desired final velocity of the masses.
Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.
Solving yields: