Answer: Attach it to clothing, personal flotation device or the person's person.
Explanation:
This particular Question or problem has to do with the rules and regulations or laws governing the use of Personal Watercraft(PWC) in the state of Florida in the United States of America. Hence, one of the rules that is applicable to the use of boats and waterways or the use of personal watercraft in Florida is that in florida, if ones pwc is equipped with an engine cut-off lanyard the person operating it must ATTACH IT TO THE CLOTHING, PERSONAL FLOATATION DEVICE IR THE OPERATORS' PERSON.
Another rule band the use of flammable personal flotation device with Personal Watercraft(PWC) in the state of Florida. All this rules and guildlines are made to limit or minimize hazards.
Im pretty sure its C- Wavelength
Answer:
v = 24 m/s, rightwards
Explanation:
Given that,
The mass of TBT explosive = 5 kg
It explodes into two pieces.
One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.
The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.
Answer:
The final acceleration becomes (1/3) of the initial acceleration.
Explanation:
The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

m = mass
a = acceleration
According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

Let a' is the final acceleration,

m' = 3m



So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.
Answer:
The gplanet is 0.193 m/s^2
Explanation:
The speed of the pulse is:


where
m=mass of the wire=4 g= 4x10^-3 kg
M=mass of the object= 3 kg
Replacing values:
