Answer:
-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing
and
0.023 m right of diverging lens
Explanation:
given data
focal length f2 = 14 cm = -0.14 m
Separation s = 12 cm = 0.12 m
focal length f1 = 21 cm = 0.21 m
distance u1 = 38 cm
to find out
final image be located and Where will the image
solution
we find find image location i.e v2
so by lens formula v1 is
1/f = 1/u + 1/v ...............1
v1 = 1/(1/f1 - 1/u1)
v1 = 1/( 1/0.21 - 1/0.38)
v1 = 0.47 m
and
u2 = s - v1
u2 = 0.12 - 0.47
u2 = -0.35
so from equation 1
v2 = 1/(1/f2 - 1/u2)
v2 = 1/(-1/0.14 + 1/0.35)
v2 = -0.233 m
so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing
and
for Separation s = 45 cm = 0.45 m
v1 = 1/(1/f1 - 1/u1)
v1 =0.47 m
and
u2 = s - v1
u2 = 0.45 - 0.47 =- 0.02 m
so
v2 = 1/(1/f2 - 1/u2)
v2 = 1/(-1/0.14 + 1/0.02)
v2 = 0.023
so here 0.023 m right of diverging lens
Answer:
-4.4 m/s
Explanation:
Since the ball is thrown upward, we can ignore the horizontal velocity. This means that the vertical velocity is 25 m/s. The vertical acceleration is -9.8 m/s^2 (g). This means that after 3 seconds, the velocity would've decreased by 9.8 m/s^2 * 3 s. This is 29.4 m/s.
This should be subtracted from the initial velocity:
25 m/s - 29.4 m/s = -4.4 m/s
Hope this helps!
