B. only its velocity should change
Answer:
Acceleration of the second particle at that moment is given as
Explanation:
As we know that both cars are connected by same spring
So on this system of two cars there is no external force
So we will have
now we have
now we have
so we have
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
<h2>Answer:</h2>
0
<h2>Explanation:</h2>
Since the current carrying wire is placed along the axis of the cylinder, according to the right hand rule, the magnetic field will be tangent to the surface of the cylinder. Therefore, there is no magnetic field through the cylinder.
Remember that the magnetic flux through a given area is the total magnetic field passing through that area. Since there is not magnetic field through the cylinder, the total magnetic flux is therefore zero (0).
