Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Potential energy = (weight) x (height)
After the car has been raised 2.5 meters, it has
(11,000) x (2.5) = 27,500 Joules
MORE potential energy than it had before it was lifted.
That's the energy that has to come from the work you do to lift it.
Since no mechanical process is ever 100% efficient, the work required
to accomplish this task is <em>at least 27,500 joules</em>.
Answer:
It is
Explanation:
1 Answer. The volume is 37.0 cm3Au .
Since you are referring to the TI-203 and TI-205, you need to know the actual masses of these two isotopes. TI-203 has 202.9723 amu and TI-205 has 204.9744 amu. Since you are concluding that this Thallium have 29.5% (Ti-203) and 70.5% (Ti-205), you need to multiply the percentage to the actual masses of the isotopes. With that, you should be able to get 204.3833 amu
