Pure water.
A salt solution contains impurities whereas pure water will not contain any impurities.
Impurities increase the boiling point (freezing point) of a substance.
Thus, I would expect the pure water solution to freeze faster than the salt solution.
Sound waves travel faster through <em>solids</em> than they do through gases or liquids. <em>(C) </em>They don't travel through vacuum at all.
Example:
Speed of sound in normal air . . . around 340 m/s
Speed of sound in water . . . around 1,480 m/s
Speed of sound in iron . . . around 5,120 m/s
Answer:
6m/s
Explanation:
Given parameters:
Initial velocity = 0m/s
Acceleration = 2m/s²
Distance = 9m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use the expression below:
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
v² = 0² + (2 x 2 x 9) = 36
v = 6m/s
The question is concerned with the regions found within California, which are the Coastal Region, Mountain Region, Central Valley Region, and the Desert Region.
The Coastal Region is located furthest to the west out of all of these regions. The Coastal Region is where the California meets the Pacific Ocean, and it has a somewhat moderate and constant climate throughout the year due to its location near the ocean.
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)
=> 
