Answer:
As a science fair project, you want to launch an 950g model rocket straight up and hit a horizontally moving target as it passes 33.0m above the launch point. The rocket engine provides a constant thrust of 20.0N . The target is approaching at a speed of 18.0m/s . At what horizontal distance between the target and the rocket should you launch?
= 43.56m
Explanation:
acceleration =
(20 - (0.95 * 9.8) )/ (0.95)
= 10.68 / 0.95
= 11.24 m/s²
we use
s = ut + (1/2) at²
Given that
s= 40
u =0
s = 0 * t + (1/2) (11.24)t²
t = √(66/1.24)
t = √5.87
t = 2.42sec
hence
Horizontal distance = 18 * 2.42
= 43.56m
ANSWER:
250 J
STEP-BY-STEP EXPLANATION:
F = 20N is required to stretch the spring by 4 meters
We know that the force is equal to:
We solve for k (spring constant):
The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:
The work required is 250 joules.
Answer:
O ksm/s
Explanation:
before collision,
Velocity =0
So,momentum of the bottle before collision=mass ×velocity
=mass×0
=0 kgm/s
None of the lenses on that list is a converting lens.
None are converging either.
The motion of the rock is a free-fall motion, which is a uniformly accelerated motion with constant acceleration 
downward, so the distance covered by the rock in a time t can be found by using the formula
By substituting t=3 s into the formula, we find the distance the rock travels after falling for 3 seconds:
