Answer: Acceleration will have 2 components, vertical and horizontal.
Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.
Net-horizontal acceleration will be equal to the horizontal component of the applied acceleration.
Explanation:
Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.
<u>For vertical acceleration:</u>
If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.
In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.
<u>For horizontal acceleration:</u>
This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.
But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.
I’m not sure but I think it’s
△ m=5 and △= -3 and so
Answer: 5/△-3 m/s
So sorry if it’s wrong
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)
=> 

The number of cycles of a periodic wave per unit time is called the wave's "frequency". The height at which the peaks of the waves reside is the "amplitude".