<span>The medium in which it travels through</span>
Let you be at point A while your friend be at point B,
while you are still having lunch,
you friend travels= 27/60*75miles/h=33.75miles
Remaining Distance= 135miles-33.75miles=101.25miles
Take note here is the tricky part, for the remaining distance both you and your friend are driving towards each other,
So first,
Total speed =Adding both ur friend and ur speed= 60+75=135miles/h
Time taken to meet= 101.25miles/135miles/h=0.75h
Total hours ur friend drives= 0.75h+27/60h=1.2h
total distance ur friend drove= 1.2h * 75miles/h= 90km
for the part how far u travelled,
simply take 135miles -90miles=45miles
Answer:
<em>Since I can see no choices, I answered it in my own understanding.</em>
Brian - amplitude and frequency
Marcia - amplitude and longitudinal wave
Explanation:
"Sound" and "sound waves" are essential part of a person's life. They can be used for<u> communicating</u> and <u>detecting some object</u>s.
Brian loves singing in the shower which means that he is using a greater amplitude. Amplitude refers to the<em> intensity of the sound </em>or the amount of energy that a sound carries. When one sings in the shower, the sound cannot travel very far. It bounces immediately back to the person singing thus, making the sound bigger. Brian is also using a <em>different range of </em><em>frequency</em><em> compared to his normal way of talking.</em> The frequency of a normal male voice is normally 85 to 180 Hz. A person singing may have a frequency as high as 1,500 Hz.
Marcia talks loudly on the phone. This means that she is also using a greater amplitude because the intensity of her voice is big. Since she is using the telephone, this means that her voice travels in a longitudinal wave through the telephone. This allows her voice to reach to the person on the other end of the line.
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)
element z are isotopes because number of netron are different and number of protons are same