The historical method includes what steps?
Answer: 24 hours
Explanation: One day is one rotation of the Earth, but it takes the earth 365 days to do one revolution or one year
Answer: I organized the data collected from the lab by using a table for each test. Table A contains the data collected from the table tennis ball and the golf ball, Table B contains the data collected from the inflated football and the deflated football, and Table C contains data collected from the question baseball and the data from Table C helps to find out whether or not the baseball is legitimate or altered.
Explanation:
i hope this like helps you because i wonldnt like you to get in trouble so see if this help spark sum
The calculated total mechanical energy will reduce if the oscillation is not perpendicular to the photogate.
<h3>Mechanical energy at the lowest position of the pendulum</h3>
The mechanical energy at the lowest position of the pendulum is calculated as follows;
![U = K.E = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=U%20%3D%20K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
<h3>When the direction of the motion changes</h3>
let the velocity of the pendulum = vsin(θ)
when the velocity is perpendicular = vsin(90) = v
At any direction different from perpendicular direction, the mechanical energy reduces by;
![\Delta U = \frac{1}{2} mv^2 - \frac{1}{2} m(vsin\theta)^2\\\\\Delta U = \frac{1}{2} m(v^2 - v^2sin^2\theta)\\\\\Delta U = \frac{1}{2} mv^2(1 - sin^2\theta )](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20m%28vsin%5Ctheta%29%5E2%5C%5C%5C%5C%5CDelta%20U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28v%5E2%20-%20v%5E2sin%5E2%5Ctheta%29%5C%5C%5C%5C%5CDelta%20U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%281%20-%20sin%5E2%5Ctheta%20%29)
Thus, the calculated total mechanical energy will reduce if the oscillation is not perpendicular to the photogate.
Learn more about mechanical energy here: brainly.com/question/24443465
Answer:
1)
![v_{oy}=11.29\ m/s](https://tex.z-dn.net/?f=v_%7Boy%7D%3D11.29%5C%20m%2Fs)
2)
![y=7.39\ m](https://tex.z-dn.net/?f=y%3D7.39%5C%20m)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle
with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
![\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2BV_%7Boy%7Dt-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where
is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
![v_y=v_{oy}-gt](https://tex.z-dn.net/?f=v_y%3Dv_%7Boy%7D-gt)
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of ![v_o](https://tex.z-dn.net/?f=v_o)
The object will reach the maximum height when
. It allows us to compute the time to reach that point
![v_{oy}-gt_m=0](https://tex.z-dn.net/?f=v_%7Boy%7D-gt_m%3D0)
Solving for ![t_m](https://tex.z-dn.net/?f=t_m)
![\displaystyle t_m=\frac{v_{oy}}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t_m%3D%5Cfrac%7Bv_%7Boy%7D%7D%7Bg%7D)
Thus, the maximum heigh is
![\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3Dy_o%2B%5Cfrac%7Bv_%7Boy%7D%5E2%7D%7B2g%7D)
We know this value is 8 meters
![\displaystyle y_o+\frac{v_{oy}^2}{2g}=8](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_o%2B%5Cfrac%7Bv_%7Boy%7D%5E2%7D%7B2g%7D%3D8)
Solving for ![v_{oy}](https://tex.z-dn.net/?f=v_%7Boy%7D)
![\displaystyle v_{oy}=\sqrt{2g(8-y_o)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D%5Csqrt%7B2g%288-y_o%29%7D)
Replacing the known values
![\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D%5Csqrt%7B2%289.8%29%288-1.5%29%7D)
![\displaystyle v_{oy}=11.29\ m/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D11.29%5C%20m%2Fs)
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
![\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2BV_%7Boy%7Dt-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
![\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D1.5%2B%2811.29%29%281.505%29-%5Cfrac%7B9.8%281.505%29%5E2%7D%7B2%7D)
![\displaystyle y=1.5\ m+16,991\ m-11.098\ m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D1.5%5C%20m%2B16%2C991%5C%20m-11.098%5C%20m)
![y=7.39\ m](https://tex.z-dn.net/?f=y%3D7.39%5C%20m)