If the girl is also near the source of the sound, two alike sets of sounds will be heard.
The correct answer is the Sun.i hope that helped! if you have any questions or concerns about the answer i gave you please let me know!!
"physical science" is a branch of science that is based on practical tests and explanations of the different phenomena. It is based on scientific evidence and tests/experiments.
Some of the branches that are based on physical science are:
1- Astronomy
2- Electronics
3- Engineering
4- Radiology
An upwards acceleration of the elevator would cause this greater reading since the scale displays the Normal Force of the object on it. The inertia of the person would prefer to stay stationary, so the scale must push up on the person to accelerate him upward along with the elevator. This results in the following equation:
W = mg + ma
Where mg represents the weight reading without external acceleration and ma as the weight reading due to the external acceleration.
Answer:
<em>t=14.96 sec</em>
Explanation:
<u>Diagonal Launch
</u>
It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.
The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is
:
![\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2Bv_osin%5Ctheta%20%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where vo is the initial speed,
is the angle, t is the time and g is the acceleration of gravity
.
In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus
, and:
![\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dv_osin%5Ctheta%20%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
The value of y is zero twice: when t=0 (at launching time) and in t=
when it goes back to the ground. We need to find that time
by making
![\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3Dv_osin%5Ctheta%5C%20t_f-%5Cfrac%7Bgt_f%5E2%7D%7B2%7D)
Dividing by ![t_f](https://tex.z-dn.net/?f=t_f)
![\displaystyle v_osin\theta=\frac{gt_f}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_osin%5Ctheta%3D%5Cfrac%7Bgt_f%7D%7B2%7D)
Then we find the total flight time as
![\displaystyle t_f=\frac{2v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t_f%3D%5Cfrac%7B2v_osin%5Ctheta%7D%7Bg%7D)
![\displaystyle t_f=14.96\ sec](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t_f%3D14.96%5C%20sec)