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:
D
Explanation:
The answer is D. Solutions you cannot see the separate substances. however in a mixture you can see the different substances. So the correct choice is D
so, 444 eggs would have been released in 37yrs
Answer:
Explanation:
Check the attachment for solution
Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire = 
= 
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x = 
= 0.25 cm
The wire is 0.25 cm thick.
