i. Refer to the diagram here a side view not to scale. A pipe of radius r connec
ID: 3162456 • Letter: I
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i. Refer to the diagram here a side view not to scale. A pipe of radius r connects a lake to a cylindrical tub of diameter d. At rest in the tub is a cube-shaped Water block of sides s Initially, the tub contains no water (but assume that the pipe is already full of water). Then the pipe's valve is opened for a time At, so that water pours into the tub. At the end of that time interval, the valve is closed; the flow stops. At that moment, the block is 80.0% immersed and the force exerted on it by the bottom of the tub has decreased by 5.00% The water level in the lake has not changed to any measureable extent. s The block is then removed from the tub dried, and transported to the apparatus shown here, which has been at rest, under the conditions shown: A, and A are the areas of mass-less platforms resting on top of cylindrical columns of oil (of density p The initial column heights (measured from the common level shown) are h, and h Column l is open to the atmosphere. Column 2 contains helium (an ideal gas) at an initial pressure P, and a constant temperature. 2 i Now the block is carefully placed onto the platform Ar. and the entire apparatus adjusts to a new equilibrium position (where everything again comes to rest). a. Find At. How long was the valve open b. What is the final height, h2r of column 2? Show how to get an algebraic solution for each The known alue r,d, s, H. p w,.A, A. p h, h, P, g Note: This may involve multiple equations with multiple unknowns (ie.simultaneous equations). You do not actually have to do the algebra to solve them. Simply write them out correctly and completely, identif the unknowns (and so there must be exactly that many equations, of course), then leave it there. One more reminder: For an ideal gas held at a constant temperature, PVisa constantExplanation / Answer
3. The refractive power of the glasses in diaopter = 1/ (focal length in meter)
a) Far point = 35cm = 0.35 m
Refractive power = - (1/0.35) = -2.857
b) In optics, the closest point at which an object can be brought into focus by the eye is called the eye's near point. Here it is 12 cm.
c) The new NP = 25cm = 0.25 m
Refractive power = - (1/0.25) = -4 diopter
d) Now he can hold a book at 17cm from his eye to view properly.
4.
10" is assumed to be the closest distance the human eye can focus for comfortable vision. An object only 1" from your eye would be 10 times larger, but out of focus.
The magnifying power of a lens can be approximated as follows: MP = 10/FL if the focal length is specified in inches. If the focal length is specified in mm, the formula will be MP=250/FL.
a) Here power rating = 250/ 69 = 3.62
b) 25cm is the normal near point.
The peson can use the lens as a 3.6X magnifier.
c)
Focal length of the magnifying glass, f = 69cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v= – d= – 25 cm
According to the lens formula, we have:
1/f = 1/v - 1/u
1/u = 1/v - 1/f = - 1/25 - 1/69 = -0.05
Minimum distance is 0.05cm
Maximum distance for a normal eye is infinity. from lens formula we get
Maximum distance with the lens is 69cm.
Maximum angular magnification = d/ u = 25/ 0.05 = 500
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