The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.
The given parameters;
- time of fall of the first ball, t = 1 s
- time of fall of the second ball, t = 3 s
The distance traveled by each ball is calculated using the second equation of motion as shown below.
The distance traveled by the first ball is calculated as follows;

The distance traveled by the second ball is calculated as follows;

Thus, the second ball traveled a greater distance because it spent more time in motion.
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Answer:to revise or edit
anything that can be made in the non draft one
Explanation:
The image of the object is 8cm to the left of the lens (D)
<h3>
</h3>
What is the image of an object?
The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.
It is calculated thus:
1÷v = 1÷f - 1÷u
<h3>How to calculate the image of an object</h3>
From the formula
1÷v = 1÷f - 1÷u
<h3>
Where </h3>
V = image distance fromthe object
U = object
f = focal length
Substitute the values
1÷v = 1÷8 - 1÷ 4
1÷v = - 1÷8
Make v the subject of formula
v = -8cm
Therefore, the image of the object is 8cm to the left of the lens (D)
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Answer:
The number of people at game are approximately 22909
Explanation:
Given data
When one person shout 
When n number of person shout together 
The sound intensity level during one person shout is given by:

The sound intensity level during n number of person shout is given by:

Since each person generates same sound intensity and hence total number of persons can be determined as

Hence
The number of people at game are approximately 22909
Answer:
block K = 29.39 J and spring #1 Ke = 360 J
Explanation:
In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work
= Ef - E₀
Let's look for the energies
Initial
E₀ = Ke = ½ k₁ x₁²
Final, this is just before starting to compress the spring
Ef = Ke = ½ m v²
The work of the rubbing force is
= -fr x
Let's write Newton's second law the y axis
N-W = 0
N = W
fr = μ N
fr = μ mg
Let's replace
-μ mg x = ½ m v² - ½ k₁ x₁²
v² = 2/m (½ k₁ x1₁² -μ mg x)
v² = 2/6 (½ 2000 0.6²2 - 0.5 6 9.8 1) = 1/3 (360 - 29.4)
v = 3.13 m / s
With this value we calculate the energy of the block
K = ½ m v²
K = ½ 6 3.13²
K = 29.39 J
Calculate eenrgy of the spring ke 1
Ke = ½ k₁ x₁²
Ke = ½ 2000 0.60²
Ke = 360 J