Under the assumption that the three rocks are dropped from the same height, they will hit the ground at the same speed. The gravity of Earth is virtually the same for any object that is small compared to the size of the Earth. The acceleration will change with the distance from the Earth, but this change is so small for the range of heights we work with (consider the range of heights from sea level to the tip of Mount Everest) that we can take the average value and assume it to be constant. This constant value of acceleration due to Earth's gravity is 9.80665m/s²
Because the objects fall under the same constant acceleration, they will hit the ground at the same speed.
Answer
given,
mass of the shuffleboard disk = 0.42 kg
speed of the cue is increased to = 4.2 m/s
acceleration takes over 2 m then acceleration is zero.
the disk additionally slide to 12 m
final speed of disk = 0 m/s
a) increase in thermal energy



b) 
F_f is the frictional force


increase in thermal energy for entire movement of 14 m


c) Work done on the disk by the cue



W = 3.704 + 0.616
W = 4.32 J
What are you asking here?
Answer:
Law of refraction
Explanation:
An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.
We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.
The vertical line is called Normal and all angles must be measured with respect to this reference in optics.
Having these angles and the refractive index of water we can use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
θ₂ =
we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.
the refractive index of air is n1 = 1 and n2= 1.33 (water)
Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J