The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect
Answer:
In fact, carving letters into a tree probably won't hurt it. ... In general, the tree will compartmentalize the wound and it will heal over. The initials that remain visible are essentially scar tissue, permanent scar tissue.
Explanation:
Unfortunately, when carving into the trunk of a tree the blade of a knife often penetrates the outer bark and cuts into the inner bark. ... In cases that the phloem is damaged all the way around the trunk (in a ring for example), the tree will slowly and eventually starve to death.
add my s n a p
luke_raines19
Answer:
Bow Line
Explanation:
If the wind or current is pushing your boat away from the dock, bow line should be secured first.
1- We should cast off the bow and stern lines.
2-With the help of an oar or boat hook, keep the boat clear of the dock.
3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.
4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.
1) The total mechanical energy of the rock is:

where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to

where

is the mass,

is the gravitational acceleration and

is the height.
Putting the numbers in, we find the potential energy

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:

where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
Answer:
36 N
Explanation:
Velocity of a standing wave in a stretched string is:
v = √(T/ρ),
where T is the tension and ρ is the mass per unit length.
300 m/s = √(T / 4×10⁻⁴ kg/m)
T = 36 N