A conducting bar moves along a circuit with a
1 answer:
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Work done by force is given by formula
now here work done against friction is given so force of friction here is
It moved by three corridors with each measures
so total distance will be
now from above formula of work done we will say
so above is the work done to move three corridors
Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,
Mass of block 2,
Velocity of block 1,
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
Final kinetic energy after collision
Now, he ratio of the total kinetic energy after the collision to that before the collision
=
=0.00169