The answer is methamphetamine
Answer:
only the weight of the ball will act on the ball
Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a
x₂ = 0.5*a*t² = 0.5*v°²/a
The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7
You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀.
#A
Mass=4.4kg
acceleration=-1.74m/s^2
Use newtons second law



#B
initial velocity=u
Final velocity=v=0
Acceleration=a=-1.74m/s^2
Time=t=1.27s





Answer:
-0.4 m/s
Explanation:
According to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the shot is zero, since they are both at rest:

While the total momentum after the shot can be written as:

where
m = 10 g = 0.010 kg is the mass of the bullet
M = 5 kg is the mass of the rifle
v = 200 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
Since the total momentum is conserved, we can write:

So

And solving for V, we find the recoil velocity:

and the negative sign indicates that the velocity is opposite to the bullet.