Answer:
The the analysis for the free fall part should be done under the constant acceleration.
Explanation:
In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.
Therefore, the the analysis for the free fall part should be done under the constant acceleration.
Answer:
A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape (volume and size), color, texture, flexibility, density, and mass.
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol
Answer:
1.9841256 kg
Explanation:
Given;
Length of the swimming pool = 25.0 ft = 7.62 m ( 1 ft = 0.3048 m )
Width of the swimming pool = 18.5 ft = 5.64 m
Depth of the pool = 9.0 ft =
Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m
now,
Volume of the water in the pool = Length × Width × Depth
or
Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³
also,
1 m³ = 1000 L
thus,
110.2292 m³ = 110229.2 L
also it is given that 18 mg of Cl is added to 1 liter of water
therefore,
In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg
or
= 1.9841256 kg
