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
Answer:
0.6kg
Explanation:
the unknown here is the mass of the second block
applying the law of the conservation of momentum
m₁v₁ + m₂v₂ = (m₁ + m₂) v₃
where m₁=mass of first block=2.2kg
m₂=mass of colliding block= ?
v₁= velocity of first block=1.2m/s
v₂=velocity of colliding block=4.0m/s
v₃= final velocity of combined block=1.8m/s
applying the formula above
(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8
2.64 + 4m₂ = 3.96 + 1.8m₂
collecting like terms
4m₂ - 1.8m₂ = 3.96 - 2.64
2.2m₂=1.32
divide both sides by 2.2
m₂= 0.6kg
I believe it is green at 1,550
B hey what do u know i took that test to
Diffuse reflection have a great day
