Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
Answer:
1.2 x 10¹¹ kgm²/s
Explanation:
m = mass of the airplane = 39043.01
r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m
v = speed of airplane = 335 m/s
L = Angular momentum of airplane
Angular momentum of airplane is given as
L = m v r
Inserting the values
L = (39043.01 ) (335) (9200)
L = (39043.01 ) (3082000)
L = 1.2 x 10¹¹ kgm²/s
Gravity is the only one, since there's no air resistance.
For an object that travels at a fixed speed along a circular path, the acceleration of the object is LARGER IN MAGNITUDE THE SMALLER THE RADIUS OF CIRCLE.
