A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firml
y in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g
1 answer:
Answer:
d) greater than g
Explanation:
You have to apply Newton's Second Law to have an idea about the value of <em>the block's aceleration.</em>
∑F = ma
Where m is the mass, a is the acceleration and F represents all the forces involved.
You have to draw a free body diagram for the block at the top of the loop. (See the attachment)
Applying Newton's Second Law:
y: N+mg=ma
Where N is the <em>normal force</em> (the contact force that the loop's surface applies to the block) and mg is the <em>weight.</em>
Therefore a= + g
So, you can notice that <em>the magnitude of the block's acceleration</em> is greater than the acceleration of gravity because of the effect of the magnitude N/m.
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Answer:
(A) Capacitance per unit length =
(B) The magnitude of charge on both conductor is C and the sign of charge on inner conductor is
and the sign on outer conductor is
Explanation:
Given :
Radius of inner part of conductor =
m
Radius of outer part of conductor =
m
The length of the capacitor =
m
(A)
Capacitance is purely geometrical property. It depends only on length, radius of conductor.
From the formula of cylindrical capacitor,
Where,
But we need capacitance per unit length so,
capacitance per unit length =
(B)
The charge on both conductors is given by,
Where, C = capacitance of cylindrical capacitor and value of F,
V
∴ C
The magnitude of charge on both conductor is same as above but the sign of charge is different.
Charge on inner conductor is and Charge on outer conductor is
.