Answer:
(ω₁ / ω₂) = 1.9079
Explanation:
Given
R₁ = 3.59 cm
R₂ = 7.22 cm
m₁ = m₂ = m
K₁ = K₂
We know that
K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²
if
v₁ = ω₁*R₁
and
I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²
∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² <em>(I)</em>
then
K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²
if
v₂ = ω₂*R₂
and
I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²
∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² <em>(II)</em>
<em>∵ </em>K₁ = K₂
⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²
⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²
⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²
⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)
⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²
⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)
⇒ (ω₁ / ω₂) = 1.9079
Answer:
1:04-1:10 hours
Explanation:
You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:
- In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
- You'll check and keep the letter in that 47 row (S) for future use.
- Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
- Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.
I hope you find this information useful and interesting! Good luck!
Answer:
Explanation:
Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force
T + mg = m v² / r
4 + .25 x 9.8 = .25 x v² / .62
6.45 = .25 v² / .62
v² = 16
v = 4 m /s .
Since we are ignoring air resistance which is a non-conservative force, the potential energy will be completely converted into kinetic energy, resulting in a final kinetic energy of

.