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)
Answer:
The pressure is ![p_1 = 4051.4 \ Pa](https://tex.z-dn.net/?f=p_1%20%3D%204051.4%20%5C%20Pa)
Explanation:
From the question we are told that
The gauge pressure at the mouth is ![p_1](https://tex.z-dn.net/?f=p_1)
The radius of the column is ![r_2 = 4 \ mm = 0.004 \ m](https://tex.z-dn.net/?f=r_2%20%3D%20%204%20%5C%20mm%20%20%3D%20%200.004%20%5C%20m)
The speed of the liquid outside the body is ![v_2 = 3.1 \ m/s](https://tex.z-dn.net/?f=v_2%20%3D%20%203.1%20%5C%20m%2Fs)
The area of the column is ![A_2](https://tex.z-dn.net/?f=A_2)
The area inside the mouth ![A_1 = 10 A_2](https://tex.z-dn.net/?f=A_1%20%3D%2010%20A_2)
Generally according to continuity equation
![v_1 A_1 = v_2 A_2](https://tex.z-dn.net/?f=v_1%20A_1%20%3D%20%20v_2%20A_2)
=> ![v_ 1 = v_2 * \frac{A_2}{A_1}](https://tex.z-dn.net/?f=v_%201%20%3D%20v_2%20%2A%20%20%5Cfrac%7BA_2%7D%7BA_1%7D)
=> ![v_ 1 = 3.1 * \frac{1}{10}](https://tex.z-dn.net/?f=v_%201%20%3D%203.1%20%2A%20%20%5Cfrac%7B1%7D%7B10%7D)
=> ![v_ 1 = 0.31 \ m/s](https://tex.z-dn.net/?f=v_%201%20%3D%200.31%20%5C%20m%2Fs)
So
![A_1 = 10A_2](https://tex.z-dn.net/?f=A_1%20%3D%2010A_2)
=> ![\pi * r_1^2 = 10(\pi * r_2^2)](https://tex.z-dn.net/?f=%5Cpi%20%2A%20r_1%5E2%20%3D%2010%28%5Cpi%20%2A%20r_2%5E2%29)
=> ![r_1 = 10 * r_2](https://tex.z-dn.net/?f=r_1%20%3D%2010%20%2A%20r_2)
substituting values
![r_1 = 10 * 0.004](https://tex.z-dn.net/?f=r_1%20%3D%2010%20%2A%200.004)
![r_1 =0.04 \ m](https://tex.z-dn.net/?f=r_1%20%3D0.04%20%5C%20m)
Now the height of inside the mouth is ![h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m](https://tex.z-dn.net/?f=h_1%20%3D%20%20d%20%3D%20%202r_1%20%3D%20%202%2A%200.04%20%3D%20%200.08%5C%20m)
Now the height of the column is ![h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m](https://tex.z-dn.net/?f=h_2%20%3D%20%20d%20%3D%20%202r_2%20%3D%20%202%2A%200.004%20%3D%20%200.008%5C%20m)
Generally according to Bernoulli's equation
![p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%5Cfrac%7B1%7D%7B2%7D%20%20%5Crho%20v_2%5E2%20%2B%20h_2%20%5Crho%20g%20%2Bp_2%5D%20-%5B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%2A%20v_1%5E2%20%2B%20h_1%20%5Crho%20g%20%5D)
Now
which is the density of water
is the gauge pressure of the atmosphere which is zero
So
![p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%280.5%20%2A%201000%20%2A%20%283.1%29%5E2%29%20%2B%280.008%20%2A%201000%20%2A%209.8%29%20%2B%200%5D-)
![p_1 = 4051.4 \ Pa](https://tex.z-dn.net/?f=p_1%20%3D%204051.4%20%5C%20Pa)
Try to go to Google and type the Answer and it will show you
Answer:
False
Explanation:
Atomic mass (Also called Atomic Weight, although this denomination is incorrect, since the mass is property of the body and the weight depends on the gravity) Mass of an atom corresponding to a certain chemical element). The uma (u) is usually used as a unit of measure. Where u.m.a are acronyms that mean "unit of atomic mass". This unit is also usually called Dalton (Da) in honor of the English chemist John Dalton.
It is equivalent to one twelfth of the mass of the nucleus of the most abundant isotope of carbon, carbon-12. It corresponds roughly to the mass of a proton (or a hydrogen atom). It is abbreviated as "uma", although it can also be found by its English acronym "amu" (Atomic Mass Unit). However, the recommended symbol is simply "u".
<u>
The atomic masses of the chemical elements are usually calculated with the weighted average of the masses of the different isotopes of each element taking into account the relative abundance of each of them</u>, which explains the non-correspondence between the atomic mass in umas, of an element, and the number of nucleons that harbors the nucleus of its most common isotope.
Answer:
10seconds
Explanation:
use the formula a=v final_v inital/time