70 mph
= 70 * 1600 m/h
= <span>112000 m/h
-> 112000 m/h
= 112000 /3600 m/s
= 31.111... m/s
= <u>31 m/s (2s.f)</u></span>
The main units for acceleration are <span>the meter per squared second as told by Galileo Galilei. Although there can be more than one example, I consider this one to be correct. Hope it will help you in some measure.</span>
When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.
p1+p2+...=pf
In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.
m1v1+m2v2=(m1+m2)vf
To find the velocity of the combined mass we simply rearrange the terms.
vf=m1v1+m2v2m1+m2
Plug in the values given in the problem.
vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg
vf=.84m/s
I'd say d.) Concentrated in the nucleus.
This is because Nucleus consists of protons and neutrons; contributing to some mass, whereas Electron's mass is almost negligible ( 9.1 x 10^-31 kg)
Hope this answers your question!
