Answer:
The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.
Explanation:
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Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
solution:
When an uncharged conducting object brought near to a charged insulating object there is a force on the conducting object to move the electrons within it to opposite sides of the conductor. That means there is a separation of charges in the conducting object in the presence of the charged insulating object near to it but the charge on the conducting object is neutral.
Thus, the conducting object is uncharged.
There is a force of attraction between the uncharged conducting object and the insulating object when it brought near to the insulating object.
Thus, there is a force on the conducting object.
The conductor remains uncharged and a force is exerted on it.
Answer:
Value that the spring constant k = 12Mg / h
Explanation:
According to 2nd law of Newton:
upward force of the spring= F
The weight of the elevator W = mg
F = Mg = M(5g)
==> F =6Mg.
As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence
F =ks = 6Mg
==> s = 6Mg/k
We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:
Mg(h+s) = 1/2ks2
And plugging in our expression for s:
Mg(h+6Mg/k)= 1/2k(6Mg / k)2
gh + 6M2g2/k = 1/2k(36M2g2 /k2)
Mgh +6M2g2/k = 1/2k(36M2g2 /k2)
gh + 6Mg2/k = 18Mg2 / k
gh = 12Mg2 / k
h = 12Mg / k
k = 12Mg / h
