Answer:
W=76.55 miles.metric tons
Explanation:
Given that
Weight on the earth = 12 tons
So weight on the moon =12/6 = 2 tons
( because at moon g will become g/6)
As we know that

Here x= 1100 miles
F 2 tons

So

We know that
Work = F. dx


![W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7Bx%7D%5Cright%5D_%7B1100%7D%5E%7B1140%7D)
![W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7B1140%7D-%5Cdfrac%7B1%7D%7B1100%7D%5Cright%5D)
W=76.55 miles.metric tons
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana
= 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey
= 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 +
(
R)² +
R² ]w
m/2 × ω₀ = [ m/2 +
(
)² +
]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s