The first right-hand rule determines the directions of magnetic force, conventional current and the magnetic field. Given any two of theses, the third can be found.
The second Right-Hand Rule determines the direction of the magnetic field around a current-carrying wire and vice-versa<span> </span>
So, assuming that a magnetic field <span>exists and its direction is known and assuming that a charged particle moves in a specific direction through that field with velocity (v(, to determine the direction of force on the particle we should use the second right-hand rule.</span>
Answer: D. The object would have an acceleration of 6 m/s/s to the right.
Explanation:
I got it right on USA test prep
Answer:
Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
