Answer:
![W=17085KJ](https://tex.z-dn.net/?f=W%3D17085KJ)
Explanation:
From the question we are told that:
Height ![H=16m](https://tex.z-dn.net/?f=H%3D16m)
Radius ![R=3](https://tex.z-dn.net/?f=R%3D3)
Height of water ![H_w=9m](https://tex.z-dn.net/?f=H_w%3D9m)
Gravity ![g=9.8m/s](https://tex.z-dn.net/?f=g%3D9.8m%2Fs)
Density of water ![\rho=1000kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D1000kg%2Fm%5E3)
Generally the equation for Volume of water is mathematically given by
![dv=\pi*r^2dy](https://tex.z-dn.net/?f=dv%3D%5Cpi%2Ar%5E2dy)
![dv=\frac{\piR^2}{H^2}(H-y)^2dy](https://tex.z-dn.net/?f=dv%3D%5Cfrac%7B%5CpiR%5E2%7D%7BH%5E2%7D%28H-y%29%5E2dy)
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
![dw=(pdv)g (H-y)](https://tex.z-dn.net/?f=dw%3D%28pdv%29g%20%28H-y%29)
Substituting dv
![dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)](https://tex.z-dn.net/?f=dw%3D%28p%28%3D%5Cfrac%7B%5CpiR%5E2%7D%7BH%5E2%7D%28H-y%29%5E2dy%29%29g%20%28H-y%29)
![dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy](https://tex.z-dn.net/?f=dw%3D%5Cfrac%7B%5Crho%2Ag%2AR%5E2%7D%7BH%5E2%7D%28H-y%29%5E3dy)
Therefore
![W=\int dw](https://tex.z-dn.net/?f=W%3D%5Cint%20dw)
![W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy](https://tex.z-dn.net/?f=W%3D%5Cint%28%5Cfrac%7B%5Crho%2Ag%2AR%5E2%7D%7BH%5E2%7D%28H-y%29%5E3%29dy)
![W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)](https://tex.z-dn.net/?f=W%3D%5Crho%2Ag%2AR%5E2%7D%7BH%5E2%7D%5Cint%28%28H-y%29%5E3%29dy%29)
![W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1000%2A9.8%2A3.142%2A3%5E2%7D%7B9%5E2%7D%5B%28%289-y%29%5E3%29%7D%5E9_0)
![W=3420.84*0.25[2401-65536]](https://tex.z-dn.net/?f=W%3D3420.84%2A0.25%5B2401-65536%5D)
![W=17084965.5J](https://tex.z-dn.net/?f=W%3D17084965.5J)
![W=17085KJ](https://tex.z-dn.net/?f=W%3D17085KJ)
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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
At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Answer:
at the beginning: ![2.3\cdot 10^{-10} F](https://tex.z-dn.net/?f=2.3%5Ccdot%2010%5E%7B-10%7D%20F)
when the plates are pulled apart: ![1.1\cdot 10^{-10} F](https://tex.z-dn.net/?f=1.1%5Ccdot%2010%5E%7B-10%7D%20F)
Explanation:
The capacitance of a parallel-plate capacitor is given by
![C=k \epsilon_0 \frac{A}{d}](https://tex.z-dn.net/?f=C%3Dk%20%5Cepsilon_0%20%5Cfrac%7BA%7D%7Bd%7D)
where
k is the relative permittivity of the medium (for air, k=1, so we can omit it)
is the permittivity of free space
A is the area of the plates of the capacitor
d is the separation between the plates
In this problem, we have:
is the area of the plates
is the separation between the plates at the beginning
Substituting into the formula, we find
![C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F](https://tex.z-dn.net/?f=C%3D%281%29%288.85%5Ccdot%2010%5E%7B-12%7DF%2Fm%29%5Cfrac%7B0.210%20m%5E2%7D%7B8.15%5Ccdot%2010%5E%7B-3%7D%20m%7D%3D2.3%5Ccdot%2010%5E%7B-10%7D%20F)
Later, the plates are pulled apart to
, so the capacitance becomes
![C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F](https://tex.z-dn.net/?f=C%3D%281%29%288.85%5Ccdot%2010%5E%7B-12%7DF%2Fm%29%5Cfrac%7B0.210%20m%5E2%7D%7B0.0163%20m%7D%3D1.1%5Ccdot%2010%5E%7B-10%7D%20F)