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
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
Answer:
2.19 N/m
Explanation:
A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:
T = 2π√(m/K)
Where T is the period, m is the mass (in kg), and K is the damping constant. So:
2.4 = 2π√(0.320/K)
√(0.320/K) = 2.4/2π
√(0.320/K) = 0.38197
(√(0.320/K))² = (0.38197)²
0.320/K = 0.1459
K = 2.19 N/m
