Answer:
Given: Vi = 382 km/h, Vf = 0 km/h, Mc = 705 kg, Md = 65 kg, Δt = 12
Required: Δx
F = Δp / Δt
= ![\frac{(Mc+Md)Vf-(Mc+Md)Vi}{t} \\\\= 6.81 * 10x^{3} N [left]\\\\x=\frac{1}{2} (Vi+Vf)\\ \\ = 637m[right]](https://tex.z-dn.net/?f=%5Cfrac%7B%28Mc%2BMd%29Vf-%28Mc%2BMd%29Vi%7D%7Bt%7D%20%5C%5C%5C%5C%3D%206.81%20%2A%2010x%5E%7B3%7D%20N%20%5Bleft%5D%5C%5C%5C%5Cx%3D%5Cfrac%7B1%7D%7B2%7D%20%28Vi%2BVf%29%5C%5C%20%5C%5C%20%3D%20637m%5Bright%5D)
The formula for acceleration is force divided by mass.
Force = 50 N
Mass = 200 kg
Acceleration = ???


~~
Acceleration = 0.25 m/s²
~~
Next time use, f = ma in a triangle,
F
M A
~~
I hope that helps you out!!
Any more questions, please feel free to ask me and I will gladly help you out!!
~Zoey
Answer:
6 m/s²
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
Acceleration is defined mathematically as:
Acceleration (a) = Velocity (v) /time (t)
a = v /t
With the above formula, we can obtain the acceleration of the object as follow:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
a= v/t
a= 30/5
a = 6 m/s²
Therefore, the acceleration of the object is 6 m/s² due East.
Answer:
Final momentum after a head on collision is -2kgm/
Explanation:
One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction
Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s
Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s