Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
Potential difference is the work done In moving a charge from one point to another in a conductor
Answer:
diverging light rays of the bulb are collected by the reflector.
Explanation:
Answer:
80m, assuming g=10m/s^2
Explanation:
40m/s will be reduced to 0m/s in 4 seconds. 4 seconds x 40m/s would be 160m up, but you will only get half of that because you decelerate linearly to 0m/s. This leaves you with 4 x 20 = 80m.
Answer: 2 blocks east if i'm not mistaken.
