Answer:

Explanation:
As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry
La longitud <em>final</em> del puente de acero es 100.018 metros.
Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (
), en metros:
(1)
Donde:
- Longitud inicial del puente, en metros.
- Coeficiente de dilatación, sin unidad.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si tenemos que
,
,
y
, entonces la longitud final del puente de acero es:
![L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]](https://tex.z-dn.net/?f=L%20%3D%20%28100%5C%2Cm%29%5Ccdot%20%5B1%2B%2811.5%5Ctimes%2010%5E%7B-6%7D%29%5Ccdot%20%2824%5C%2C%5E%7B%5Ccirc%7DC%20-%208%5C%2C%5E%7B%5Ccirc%7DC%29%5D)

La longitud <em>final</em> del puente de acero es 100.018 metros.
Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416
Explanation:
Below is an attachment containing the solution.
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so

=> 

integrating both sides

Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have

![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
