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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Best Answer:<span> </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km
The work done by the aircraft overcoming the wind is
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
If the solution is treated as an ideal solution, the extent of freezing
point depression depends only on the solute concentration that can be
estimated by a simple linear relationship with the cryoscopic constant:
ΔTF = KF · m · i
ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF
(solution).
KF, the cryoscopic constant, which is dependent on the properties of the
solvent, not the solute. Note: When conducting experiments, a higher KF
value makes it easier to observe larger drops in the freezing point.
For water, KF = 1.853 K·kg/mol.[1]
m is the molality (mol solute per kg of solvent)
i is the van 't Hoff factor (number of solute particles per mol, e.g. i =
2 for NaCl).
Answer:
80 ft/s
Explanation:
Use III equation of motion
V^2 = U^2 + 2g h
Here, U = 0, g = 32 ft/s^2, h = 100 ft
V^2 = 0 + 2 × 32 ×100
V^2 = 6400
V = 80 ft/s
I believe it she should use the first aid kit next
