Answer:
if i do give me brainliest ok ok
Explanation:
Potential energy = mass x gravity x height
P.E = 4 x 9.8 x 3
P.E = 117.6 J
Answer:
The answer is "
".
Explanation:
Cavity and benzene should be extended in equal quantities.



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)
=> 

Answer:
70.5 mph
Explanation:
A passenger jet travels from Los Angeles to Bombay, India, in 22h.
The return flight takes 17 h.
The difference in flight times is caused by winds over the Pacific Ocean that
blow primarily from west to east.
If the jet's average speed in still air is 550 mi/h what is the average speed
of the wind during the round trip flight? Round to the nearest mile per hour.
Is your answer reasonable?
:
Let w = speed of the wind
:
Write a distance equation (dist is the same both ways
17(550+w) = 22(550-w)
9350 + 17w = 12100 - 22w
17w + 22w = 12100 - 9350
39w = 2750
W = 2750/39
w = 70.5 mph seems very reasonable
:
Confirming if the solution by finding the distances using these value
17(550+70.5) = 10549 mi
22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph