Explanation:
Types of light microscope
1. Compound , and 2. Stereo Microscope
Compound microscope has two lens system also called compound lens system. The objective lens and the eyepiece lens. The magnification provided by the objective lens is compounded by the eyepiece lens, the a higher magnification is observed.
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.
There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the elasticity is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.
For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, 
= 926 kg/m³
density of the wood, 
= 974 kg/m³
density of water, 
= 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
![F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm](https://tex.z-dn.net/?f=F_%7Bwood%7D%20-%20F_%7Boil%7D%20-%20F_%7Bwater%7D%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.gh%20-%20%5Crho%20_o%20.g%28h-x%29%20-%20%5Crho_w%20.gx%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%20%5Crho%20_o%20%28h-x%29%20-%20%5Crho_w%20.x%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%5Crho%20_o%20h%20%2B%20%5Crho%20_o%20x%20-%20%5Crho_w%20.x%20%3D0%5C%5C%5C%5Ch%20%28%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%29%20%3D%20x%28%20%5Crho_w%20-%20%5Crho%20_o%29%5C%5C%5C%5Cx%20%3Dh%5B%5Cfrac%7B%20%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%7D%7B%5Crho_w%20-%20%5Crho%20_o%7D%20%5D%5C%5C%5C%5Cx%20%3D%203.69%5C%20cm%20%5Ctimes%20%5B%5Cfrac%7B974%20-%20926%7D%7B1000-926%7D%20%5D%5C%5C%5C%5Cx%20%3D%202.39%20%5C%20cm)
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.
Answer:
I'm pretty sure it's B because I studied this topic and I'm not right I'm sorry.
