question
light with an energy flux of 25 x 104 wm-2 falls on a perfectly reflecting surface at normal incidence. if the surface area is 15 cm2, the average force exerted on the surface is:



 option 1)
 1.25 x 10-6 n
 
 option 2)
 2.50 x 10-6 n
 
 option 3)
 1.20 x 10-6 n
 
 option 4)
 3.0 x 10-6 n
309
when a ray of light falls on a smooth and shiny surface, the whole of light is sent back in the same direction. it is called reflection. mirrors do not allow even a small amount of light to pass through them. mirrors show regular and complete reflection.when all the parallel rays reflected from a plane are not parallel, the reflection is diffused or irregular. this is not violation of law of reflection, the reason for diffused reflection is due to irregularities in the reflecting surface. we are required to state the effect of one of the laws governing brightness. apparent brightness:b=l4πd2 where b is apparent brightness, l is intrinsic luminosity, and d is the distance from the source. [readmore]as the distance increases, the apparent brightness gets fainter as the distance from the source increases. the fainting brightness is accounted for by the spreading out of photons: as light travels away from its source, the photons get spread out over increasingly greater areas. the more the light is spread out, the fainter it appears in any given patch of area (for example, the small patch of the area covered by the opening in your eye)the correct choice is:d. far from the source, photons spread over a larger area. centripetal force:f=mv2rr is equal to l in this question.1. f = (100)(1602)/2500 = 1024 n2. f = (300)(802)/10000 = 192 n3. f = (400)(802)/2500 = 1024 n[readmore]4. f = (200)(1602)/5000 = 1024 n5. f = (800)(402)/10000 = 128 n6. f = (200)(1202)/5000 = 576 n1 = 3 = 4 > 6 > 2 > 5the order of forces acting on the satelites from the largest to the smallest is 1 = 3 = 4 > 6 > 2 > 5 you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] hi trinity, let's do this one!newton's famous second law states, in math terms: f = ma applying to the first case, we have: 100 = m(4.00) => m = 100/4 = 25 kgnow apply the formula to the second case: f = 25 x 16 = 400 newtons given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n. 1.6 x 10-11 that depends on a lot of things. in particular, you would need to find the electric susceptibility of the paper. the field produced by static charge is inverse-square. the polarization of the paper varies directly with the external field, with a proportionality constant equal to the electric susceptibility. the dipole field created by the polarized paper is inverse-cube. overall, the force between the ruler and the paper will vary with the inverse fifth power of distance. to get any more detailed, you would need to know the geometry of the situation, and the charge distribution.