Answer:
<em>h = 20 m</em>
Explanation:
<u>Gravitational Potential Energy</u>
Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.
It can be calculated with the equation:
U=m.g.h
Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or
.
The weight of an object of mass m is:
W = m.g
Thus, the GPE is:
U=W.h
Solving for h:

The weight of the owl is W=40 N and its GPE is U=800 J.

h = 20 m
1. Encoding Information
2. Storing Information
3. Retrieval Information
As we know by the first law of thermodynamics

here we know that
Q = heat given to the system

W = work done by the system
now here we can say


now we can say that heat will be given as

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy
So best describe Jin's Error is
<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>
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:
No. Neither the USGS nor any other scientists have ever predicted a major earthquake. We do not know how, and we do not expect to know how any time in the foreseeable future.