Potential energy is an object’s stored energy when it is not moving. Kinetic energy is an object’s energy in motion. You can’t be both moving and not moving at the same time, so PE and KE cannot be equal.
The distance covered is 115 m
Explanation:
The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

where
s is the distance covered
u is the initiaal velocity
v is the final velocity
t is the time elapsed
In this problem, we have:
u = 4.20 m/s
v = 5.00 m/s
t = 25.0 s
Therefore, we can re-arrange the equation to find the distance covered:

Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
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#LearnwithBrainly
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀
What is an example of how you can use scientific inquiry to solve a real life problem.
The sound wave would behave differently in a swimming pool than in his bedroom because sound waves travel faster in more dense mediums; such as water. The wave will travel faster in water, and slower in air.