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)
=> 

Two types of mechanical waves: longitudinal<span> waves and </span>transverse<span> waves; the medium movement differs between the two.
</span>In a longitudinal wave the medium particle movement is parallel to the direction of wave propagation; example is sound wave in air.
I<span>n a transverse wave the medium particle movement is perpendicular to the direction of wave propagation; example is mechanical wave on a string.
</span><span>
</span>
Answer:
B. When the ball is released, the thrower's arm transfers its energy to the ball.
Jack------------ force of 92.5 n eastward-------Fjack(X)=92.5 n Fjack(Y)=0
<span>jill ------------------------------- force of 89.9 n northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>
</span>jane -----------------------------force of 163 n southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n
Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n
Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast
the magnitude of the net force the people exert on the donkey. is 294.80 n southeast