<h2>Given that,</h2>
Mass of two bumper cars, m₁ = m₂ = 125 kg
Initial speed of car X is, u₁ = 10 m/s
Initial speed of car Z is, u₂ = -12 m/s
Final speed of car Z, v₂ = 10 m/s
We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :
v₁ is the final speed of car X.
So, car X will move with a velocity of -12 m/s.
Explanation:
The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.
It would be either A or C if its still moving and not stopping
Answer:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:
![E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DQ%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D%5C%5CQ%3Dthermal%20energy%20%5BJ%5D%5C%5CRe-employment%20values%20and%20equalizing%20equations%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%3Dm%2AC_%7Bp%7D%2AdT%20%20%5C%5CThe%20masses%20are%20canceled%20%5C%5C%5C%5CdT%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7BC_%7Bp%7D%20%2A2%7D%20%5C%5CdT%3D2.1%20%5BC%5D)
<h2>♨ANSWER♥</h2>
14mg = 14 × 10^-3 g
= 14 × 10^-3 / 10^3 kg
= 14 × 10^-6 kg
= 0.014 kg
<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>
_♡_<em>mashi</em>_♡_
