Answer:
B. how hard the string was plucked
Explanation:
Answer: Your question is missing below is the question
Question : What is the no-friction needed speed (in m/s ) for these turns?
answer:
20.1 m/s
Explanation:
2.5 mile track
number of turns = 4
length of each turn = 0.25 mile
banked at 9 12'
<u>Determine the no-friction needed speed </u>
First step : calculate the value of R
2πR / 4 = πR / 2
note : πR / 2 = 0.25 mile
∴ R = ( 0.25 * 2 ) / π
= 0.159 mile ≈ 256 m
Finally no-friction needed speed
tan θ = v^2 / gR
∴ v^2 = gR * tan θ
v = √9.81 * 256 * tan(9.2°) = 20.1 m/s
The answer is C.endothermic change.
♡♡Hope I helped!!! :)♡♡
A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.
Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.
Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.
For an elastic collision, we use the formula,
m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f
For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.
Final velocity = 0.5/2
=0.25 m/s.
To know more about elastic collision, refer
brainly.com/question/7694106
#SPJ4
Answer: Heat waves? I’m not %100 sure
