CI. Design challenge: Your goal is to build a compound microscope than can at least double the size of the object. We will const
rain ourselves to values that will work later in a simulation, which has the unrealistic- for microscopes - focal lengths in meters. (Imagine instead that those values are in cm) You need to choose two converging lenses. You have available lenses with focal lengths of 2, 4, 8 and 12 m. First choose an objective lens focal length, and a distance to place the object from the lens. Then choose an eyepiece lens focal length and a distance to place the lens from the objective lens. C2. Draw your microscope design including a ray diagram. Your diagram should include: 1. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the objective. 2. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the eyepiece. C3. Calculate the magnification of your microscope
1 answer:
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Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s