A gymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =
109 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?N
2)What is the force the right support exerts on the beam?N
3)How much extra mass could the gymnast hold before the beam begins to tip? kg
4)Now the gymnast (not holding any additional mass) walks directly above the right support. What is the force the left support exerts on the beam?N
5)What is the force the right support exerts on the beam?N
6)At what location does the gymnast need to stand to maximize the force on the right support?
a. at the center of the beam
b. at the right support
c. at the right edge of the beam
1)What is the force the left support exerts on the beam?N
2)What is the force the right support exerts on the beam?N
3)How much extra mass could the gymnast hold before the beam begins to tip? kg
4)Now the gymnast (not holding any additional mass) walks directly above the right support. What is the force the left support exerts on the beam?N
5)What is the force the right support exerts on the beam?N
6)At what location does the gymnast need to stand to maximize the force on the right support?
a. at the center of the beam
b. at the right support
c. at the right edge of the beam
1 answer:
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Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
,
- Initial and final speeds of the shell, measured in meters per second.
If we know that ,
,
and
, then the explosive force experienced by the shell inside the barrel is:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Answer:
See the explanation below.
Explanation:
The force is a vector therefore we can decompose the force into components x & y. as we need the horizontal component of the force, we must use the cosine function of the angle.