Required value of initial speed of the bullet be ( 4M/m)√(gL).
Given parameters:
Mass of the bullet =m.
Mass of the bob of the pendulum = M.
speed of the bullet before collision = v
Speed of the bullet after collision = v/2.
Length of the pendulum stiff rod = L.
Let speed transmitted to the pendulum be u.
Using principle of conservation of momentum:
mv = Mu + mv/2
⇒ Mu = mv/2
⇒ u = (m/M)v/2
We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]
To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:
u = √(4gL)
⇒ (m/M)v/2 = √(4gL)
⇒ v =( 4M/m)√(gL).
Hence, minimum required speed of the bullet be ( 4M/m)√(gL).
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Answer:
It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.
Explanation:
Answer:
Here is the complete question:
https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632
a) Current for long straight wire 
b) Current at the center of the circular coil 
c) Current near the center of a solenoid 
Explanation:
⇒ Magnetic Field due to long straight wire is given by (B),where

Plugging the values,
Conversion
,and 

⇒Magnetic Field at the center due to circular coil (at center) is given by,
So 
⇒Magnetic field due to the long solenoid,
Then
So the value of current are
,
and
respectively.
Answer:
incomplete question, resistor must be there
Explanation: