(a) The ball’s maximum speed over the net is v(max) = √2gh.
(b) The maximum speed of the horizontally moving ball clearing the net is about 27 m/s.
(c) Speed of the ball is independent of its mass.
<h3>
Time of motion of the ball</h3>
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
1 = 0 + ¹/₂(9.8)t²
1 = 4.9t²
t² = 1/4.9
t² = 0.204
t = 0.452 s
<h3>Horizontal speed of the ball</h3>
The horizontal speed of the ball is calculated as follows;
X = vt
v = X/t
v = (12 m)/(0.452)
v = 26.6 m/s ≈ 27 m/s (proved)
<h3>Conservation of energy</h3>
P.E = K.E
mgh = ¹/₂mv²
gh = ¹/₂v²
2gh = v²
√2gh = v(max)
Speed of the ball is independent of its mass.
Learn more about horizontal velocity here: brainly.com/question/24681896
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Answer:
0.2m
The solution is in the picture
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.
The same number there were on the reactant side, before the reaction.
I think B but I would need to see how much forces are applied
