Answer:
-2.8 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²
Using the equation of motion,
v² = u² + 2as................... Equation 1
Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,
Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.
Substituting into equation 1
6² = 8²+2(a)5
36 = 64 + 10a
10a = 36-64
10a = -28
10a/10 = -28/10
a = -2.8 m/s²
Note: a is negative because because the skater decelerate on the rough ice
Hence the magnitude of her acceleration is = -2.8 m/s²
F=K*X,
F=M*a
M*a=K*X
2.5*9.81=K*0.0276
24.525=K*0.0276
24.525/0.0276=K
K= 888.6 N/m ---- force constant
assuming 2.5 refers to the new extension, just divide F/ 0.025
to get
981N/m
I think the correct answer is the second option. A circuit describes a closed conducting loop through which an electrical current can flow. It is a path that an electrical current could flow. A circuit could be a closed one or an open circuit. A closed circuit would be a circuit where the current could flow continuously. An open circuit would be a type of circuit where the flow current would only go once and stopped at a particular point since the current has nowhere to go. For a circuit to work, an electric supply should be available to supply the electric current.
Water is treated by purifying it by adding slaked lime or potash alum
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)
=> 
