Haven't done one like this in awhile but I see no one is answering so I gave it a try. I think it's right but let me know if you see something fishy...
<span> Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
1 kg = 2.2 lbs, so 2.0 lbs x 1 kg/2.2 lbs = 0.91 kg
The bag would have a weight of 9.8 x 0.91 = 8.9 N
1. 8.9 x 1/6 = 1.5 N
2. 8.9 x 2.64 = 23.5 N
The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight. </span>
Answer: 272.82 drop/tile
Explanation:
Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?
Tiles/ft^2 × drop/tiles = drop/ft^2
Tiles will cancel out. Leaving the answer to be drop/ ft^2
Substitutes all the magnitude of the above units.
17 × drop/tiles = 4638
Make drop/tiles the subject of formula
Drop/tiles = 4638/17
Drop/tiles = 272.82
Therefore, 272.82 drop/tile drops fall on each tile?
Answer:
B.
It will be greater than 10 J.
Explanation:
The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):
E = PE + KE
According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.
The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.
The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.
Here at the beginning, when it is at the top of the roof, the baseball has:
PE = 120 J
KE = 10 J
So the total energy is
E = 120 + 10 = 130 J
As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).
Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.
Answer:
The first
Explanation:
Cause it will continue in motion till another force is applied
