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
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
The wave speed to this question is 400 meters
Answer:
30 seconds
Explanation:
A = A02^-(t/hl)
--> ln(A/A0) = -(t/hl)ln2
solving for hl,
hl = -t x ln2 /ln(A/A0)
= -(60 min)xln2/ln(50/200)
= 0.5 min or 30 seconds
Answer:
0.02 m/s^2
Explanation:
change in velocity= 4.5m/s - 2.3m/s = 2.2 m/s
acceleration= change in velocity/change in time
acceleration= 2.2/120= 0.0183
= 0.02 (to 2 significant figures)
