Conduction, Convection, and in some cases, Radiation.
Answer:
The focal lenth (F) =+10.0cm
Explanation:
The formular for combined focal length (F) is given as;

In this question,
F1 = 20cm
F2 = -30cm
Plugging the values into the formuar above,

![1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017](https://tex.z-dn.net/?f=1%2Ff%20%3D%200.05%20-%200.033%3C%2Fp%3E%3Cp%3E%5Btex%5D1%2Ff%20%3D%20-0.017%3C%2Fp%3E%3Cp%3Ef%20%3D%20%5Btex%5D1%2F%20-0.017)
f = 58.82cm
i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .
Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
![(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]](https://tex.z-dn.net/?f=%280.5%29%5E%7B2%7D-%2810%29%5E%7B2%7D%20%3D%202%2Aa%2A%28100%29%5C%5C0.25-10000%3D200%2Aa%5C%5C200%2Aa%3D-9999.75%5C%5Ca%20%3D-50%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign of acceleration means that the ship slows down its velocity in order to land.
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