Answer:
<h3>The answer is 2.5 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question we have

We have the final answer as
<h3>2.5 g/cm³</h3>
Hope this helps you
The intense heat in the earth's core that causes molten rock in the mantle layer to move.
Answer:
h=2.86m
Explanation:
In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

There is no kinetic energy in the initial state, nor potential energy in the end,

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

The inertia of the bodies is given by the equation,



On the other hand the angular velocity is given by

Replacing these values in the equation,

Solving for h,

If something is traveling at 20 m/s constant speed AND its direction isn't changing, then its velocity is constant. Another way to say that is: Its acceleration is zero. Zero acceleration means zero NET force acting on the object, or a group of BALANCED forces acting on it, also called EQUILIBRIUM. The required answer is: YES.
If a real projectile is launched, the force of gravity acts on it vertically downward. There's no upward force acting on it to balance gravity. Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium. Too bad.
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)
=> 
