Answer:
Scientific evidence is evidence that serves to either support or counter a scientific theory or hypothesis. Such evidence is expected to be empirical evidence and interpretable in accordance with scientific method.
Answer:
Kinetic energy of diver at 90% of the distance to the water is 9000 J
Explanation:
Let d is the distance between the position of the diver and surface of the pool.
Initially, the diver is at rest and only have potential energy which is equal to 10000 J.
As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.
Energy before diving = Energy during diving
(Potential Energy + Kinetic Energy) = (Kinetic Energy + Potential Energy)
When the diver reaches 90% of the distance to the water, its kinetic energy
is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,
K.E. = ![\frac{90}{100}\times10000](https://tex.z-dn.net/?f=%5Cfrac%7B90%7D%7B100%7D%5Ctimes10000)
K.E. = 9000 J
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
Read more on waves here
brainly.com/question/25699025
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer:
net force is positive downward..B
Explanation:
It is given that,
A particle starts from rest and has an acceleration function as :
![a(t)=(5-10t)\ m/s^2](https://tex.z-dn.net/?f=a%28t%29%3D%285-10t%29%5C%20m%2Fs%5E2)
(a) Since, ![a=\dfrac{dv}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D)
v = velocity
![dv=a.dt](https://tex.z-dn.net/?f=dv%3Da.dt)
![v=\int(a.dt)](https://tex.z-dn.net/?f=v%3D%5Cint%28a.dt%29)
![v=\int(5-10t)(dt)](https://tex.z-dn.net/?f=v%3D%5Cint%285-10t%29%28dt%29)
![v=5t-\dfrac{10t^2}{2}=5t-5t^2](https://tex.z-dn.net/?f=v%3D5t-%5Cdfrac%7B10t%5E2%7D%7B2%7D%3D5t-5t%5E2)
(b) ![v=\dfrac{dx}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bdx%7D%7Bdt%7D)
x = position
![x=\int v.dt](https://tex.z-dn.net/?f=x%3D%5Cint%20v.dt)
![x=\int (5t-5t^2)dt](https://tex.z-dn.net/?f=x%3D%5Cint%20%285t-5t%5E2%29dt)
![x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B5%7D%7B2%7Dt%5E2-%5Cdfrac%7B5%7D%7B3%7Dt%5E3)
(c) Velocity function is given by :
![v=5t-5t^2](https://tex.z-dn.net/?f=v%3D5t-5t%5E2)
![5t-5t^2=0](https://tex.z-dn.net/?f=5t-5t%5E2%3D0)
t = 1 seconds
So, at t = 1 second the velocity of the particle is zero.