The heavy stone would produce waves with a higher amplitude, rather than the smaller stone, because since the stone is heaver its going to have a grater impact and displace more water to create a bigger wave.
Answer:
The answer to your question is:
Explanation:
Data
Duane Albert
d = 5 m ; v = 3 m/s v = 4.2 m/s
a) b)
Duane's Albert's
d = 5 + (3)t d = 4.2t
d = 5 + 3t
c) 5 + 3t = 4.2t
4.2t - 3t = 5
1.2t = 5
t = 4.17 s
d)
Duane's
d= 5 + 3(4.17)
d = 17.51 m
Alberts
d = 4.2(4.17)
d = 17.51 m
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is
the vertical velocity.
In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity

, where

- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration

directed downwards, and with initial velocity

. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
(c) +6.67
Explanation:
f1 = 10 cm
f2 = 20 cm
u = Object distance = 15 cm
Distance between lenses = 20 cm
For first lens image distance

Distance from second lens is 10 cm to the right

The final image will appear as +6.67 cm