So you subtract the numbers that are on the same axis. So if your gravitational force is 10 and your normal force is 5 you do 5-10 to get -5 since gravity acts downward
        
             
        
        
        
Answer:
v = 12.12 m/s
Explanation:
Given that,
The mass of the cart, m = 75 kg
The roller coaster begins 15 m above the ground.
We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

So, the velocity of the cart is 12.12 m/s.
 
        
             
        
        
        
Answer:
Approximately  (assuming that the projectile was launched at angle of
 (assuming that the projectile was launched at angle of  above the horizon.)
 above the horizon.)
Explanation:
Initial vertical component of velocity:
 .
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing  is the same as the altitude
 is the same as the altitude  at which this projectile was launched:
 at which this projectile was launched:  .
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is  (upwards,) the vertical velocity right before landing would be
 (upwards,) the vertical velocity right before landing would be  (downwards.) The change in vertical velocity is:
 (downwards.) The change in vertical velocity is:
 .
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be  . In other words,
. In other words,  .
.
Hence, the time it takes to achieve a (vertical) velocity change of  would be:
 would be:
 .
.
Hence, this projectile would be in the air for approximately  .
.
 
        
                    
             
        
        
        
They interact with one another with their facial expressions and body movements all the time
        
             
        
        
        
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration) 
Velocity of second car; v2 = 0 m/s (since the second car starts from rest) 
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²